Question Bank - Set 5 CHEM 107

CHEM 107 - ESSENTIALS OF GENERAL AND ORGANIC CHEMISTRY - Calorimetry Question Bank - Set 5 Liberty University Question 1 Question A 50 g aluminum block at 100°C is placed in 200 g of water at 20°C in a calorimeter. The final temperature of the system is 25°C. Calculate the specific heat capacity of the aluminum block. The specific heat capacity of water is cw = 4.18 J/g°C. Solution Step 1: Calculate the heat gained by the water. The heat gained by the water can be calculated using the formula: Qwater = mwater · cw · ∆T where - mwater = 200 g is the mass of water, - cw = 4.18 J/g°C is the specific heat capacity of water, - ∆T = Tfinal − Tinitial = 25C − 20C = 5C is the change in temperature. Plugging in the values: Qwater = 200 g × 4.18 J/g°C × 5 C = 4180 J Step 2: Calculate the heat lost by the aluminum block. The heat lost by the aluminum block is equal in magnitude to the heat gained by the water: Qwater = Qaluminum Using the formula for heat: Qaluminum = maluminum · caluminum · ∆T We know: - maluminum = 50 g is the mass of the aluminum block, - ∆T = 75C (since the final temperature of the system is 25°C), - caluminum is the specific heat capacity of aluminum (to be determined). Plugging in the values: 4180 J = 50 g × caluminum × 75C Step 3: Solve for the specific heat capacity of aluminum. caluminum = 4180 J = 1.114 J/g°C 3750 g°C Therefore, the specific heat capacity of the aluminum block is 1.114 J/g°C. Question 2 Question A 50.0 g piece of aluminum at 120◦ C is placed in 100.0 g of water at 25.0◦ C. Assuming no heat is lost to the surroundings, calculate the final temperature of the system after thermal equilibrium is reached. The specific heat capacity of aluminum is 0.900 J/g◦ C and the specific heat capacity of water is 4.184 J/g◦ C. Solution Step 1: Calculate the heat lost by aluminum and gained by water using the formula: q = m · c · ∆T where q is the heat energy, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. For aluminum: qAl = 50.0 g × 0.900 J/g◦ C × (Tfinal − 120◦ C) For water: qH2 O = 100.0 g × 4.184 J/g◦ C × (Tfinal − 25◦ C) Step 2: Since energy is conserved, the heat lost by aluminum is equal to the heat gained by water: qAl = −qH2 O Step 3: Substitute the expressions for qAl and qH2 O into the energy conservation equation and solve for Tfinal : 2 50.0 g×0.900 J/g◦ C×(Tfinal −120◦ C) = −100.0 g×4.184 J/g◦ C×(Tfinal −25◦ C) Step 4: Solve for Tfinal : 50.0 × 0.900 × (Tfinal − 120) = −100.0 × 4.184 × (Tfinal − 25) 45(Tfinal − 120) = −418.4(Tfinal − 25) 45Tfinal − 5400 = −418.4Tfinal + 10460 463.4Tfinal = 15860 Tfinal = 34.26◦ C Therefore, the final temperature of the system after thermal equilibrium is reached is 34.26◦ C. Question 3 Question A student conducts a calorimetry experiment by mixing 200 mL of water at 25°C with 100 mL of methanol at 60°C in a calorimeter. Assuming no heat is lost to surroundings, calculate the final temperature of the system. The specific heat capacity of water is 4.18 J/g°C and the specific heat capacity of methanol is 2.51 J/g°C. The density of water is 1 g/mL and the density of methanol is 0.791 g/mL. Solution Step 1: Calculate the initial heat content of water and methanol. The heat content (q) for a substance is given by the formula: q = m × c × ∆T where: - m is the mass of the substance, - c is the specific heat capacity of the substance, and - ∆T is the change in temperature. For water: - mwater = 200 mL × 1 g/mL = 200 g - cwater = 4.18 J/g°C ∆Twater = Tfinal − 25C For methanol: - mmethanol = 100 mL × 0.791 g/mL = 79.1 g - cmethanol = 2.51 J/g°C - ∆Tmethanol = Tfinal − 60C Step 2: Since heat is conserved in the system (no heat lost to the surroundings), the total initial heat content must equal the total final heat content. 3 qwater + qmethanol = 0 mwater × cwater × ∆Twater + mmethanol × cmethanol × ∆Tmethanol = 0 Substitute in the known values and solve for Tfinal . Question 4 Question A 50 g piece of aluminum at 80°C is placed in 200 g of water at 20°C. The final temperature of the system is 22.5°C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity of aluminum. Solution Step 1: Calculate the heat lost by the aluminum and the heat gained by the water using the formula: Q = mc∆T where: - Q is the heat energy, - m is the mass of the substance, - c is the specific heat capacity of the substance, and - ∆T is the change in temperature. For the aluminum: Qaluminum = maluminum caluminum ∆Taluminum Qaluminum = 50 × caluminum × (22.5 − 80) Qaluminum = −3575caluminum For the water: Qwater = mwater cwater ∆Twater Qwater = 200 × 4.18 × (22.5 − 20) Qwater = 4388 J Step 2: Since heat is conserved in this closed system, the heat lost by the aluminum is equal to the heat gained by the water: Qaluminum = Qwater −3575caluminum = 4388 4388 caluminum = 3575 caluminum ≈ 1.226 J/g°C Therefore, the specific heat capacity of aluminum is approximately 1.226 J/g°C. 4 Question 5 Question A 50 g aluminum block at 80°C is dropped into 200 g of water at 20°C in a calorimeter. The final temperature of the system is 25°C. Assuming no heat is lost to the surroundings, find the specific heat capacity of aluminum. The specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat lost by the aluminum block as it cools down from 80°C to 25°C. The specific heat capacity of aluminum is 0.897 J/g°C. The formula to calculate heat energy is Q = mc∆T , where: - Q is the heat energy, - m is the mass, - c is the specific heat capacity, - ∆T is the change in temperature. Given: m = 50 g, c = 0.897 J/g°C, and ∆T = 80C − 25C = 55C. Using the formula, we have: Qaluminum = (50 g)(0.897 J/g°C)(55C). Calculating Qaluminum , we get: Qaluminum = 2481.75 J. Step 2: Calculate the heat gained by the water and the calorimeter. Using the same formula as before with the given values: Qwater = (200 g)(4.18 J/g°C)(25C− 20C). Calculating Qwater , we get: Qwater = 418 J. Step 3: Since the total heat lost by the aluminum block is equal to the heat gained by the water and the calorimeter (assuming no heat loss to surroundings), we have: Qaluminum = Qwater + Qcalorimeter . Substitute the values we found into the equation: 2481.75 J = 418 J + Qcalorimeter . Solving for Qcalorimeter , we get: Qcalorimeter = 2063.75 J. Step 4: The heat gained by the calorimeter is due to the temperature change of the water and calorimeter together. Let the specific heat capacity of the calorimeter be C. Since the water and calorimeter have the same temperature change, we have: Qcalorimeter = (200 g + mcalorimeter )C(25C − 20C). Given that the mass of water is 200 g, the specific heat capacity of water is 4.18 J/g°C, and the final temperature is 25C. Substitute the known values into the equation: 2063.75 J = (200 g+mcalorimeter )4.18 J/g°C× 5C. Solving for mcalorimeter , we find: mcalorimeter = 105 g. Step 5: Calculate the specific heat capacity of aluminum. Given that the mass of the aluminum block is 50 g and the specific heat capacity of the aluminum is 0.897 J/g°C, we have: Qaluminum = (50 g)(0.897 J/g°C)(55C). Substitute the values we know: 2481.75 J = (50 g)(0.897 J/g°C)(55C). Solving for the specific heat capacity of aluminum, we get: caluminum = 0.897 J/g°C. Therefore, the specific heat capacity of aluminum is 0.897 5 Question 6 Question A 50 g piece of aluminum at 100◦ C is dropped into 200 g of water at 20◦ C in a calorimeter. The final temperature of the system is 25◦ C. Assuming all the heat lost by the aluminum is gained by the water and the calorimeter, what is the heat capacity of the calorimeter? The specific heat capacity of aluminum is 0.90 J/g◦ C and that of water is 4.18 J/g◦ C. Solution Step 1: Calculate the heat lost by the aluminum. The heat lost by the aluminum can be calculated using the formula: Qaluminum = maluminum × caluminum × ∆T where: - maluminum is the mass of aluminum (50 g) - caluminum is the specific heat capacity of aluminum (0.90 J/g◦ C) - ∆T is the change in temperature (final temperature - initial temperature) ◦ Qaluminum = 50 g × 0.90 J/g × (25 − 100)◦ CQaluminum = 50 × 0.90 × (−75) Qaluminum = −3375 J Step 2: Calculate the heat gained by the water and the calorimeter. Since the heat lost by the aluminum is equal to the heat gained by the water and the calorimeter, we have: Qaluminum = Qwater + Qcalorimeter Substitute the values we know: ◦ −3375 J = (200 g + Ccalorimeter ) × 4.18 J/g × (25 − 20)◦ C −3375 J = (200 + Ccalorimeter ) × 4.18 × 5 −3375 = (200 + Ccalorimeter ) × 20.9 −3375 = 4180 + 20.9Ccalorimeter 20.9Ccalorimeter = −7555 Ccalorimeter = −361.2 J/◦ CJ/◦ C J/◦ C J/◦ C Therefore, the heat capacity of the calorimeter is -361.2 J/◦ C. Question 7 Question A 50 gram piece of iron at 80°C is placed in 200 grams of water at 20°C. If the final temperature of the system is 25°C, what is the specific heat capacity of iron? Assume no heat is lost to the surroundings. 6 Solution Step 1: Calculate the heat lost by the iron and the heat gained by the water using the formula: qiron = −qwater where q = mc∆T (m = mass, c = specific heat capacity, ∆T = change in temperature). Step 2: Calculate the heat lost by the iron: qiron = miron ciron ∆Tiron qiron = (0.05 kg)(ciron )(80 − 25) Step 3: Calculate the heat gained by the water: qwater = mwater cwater ∆Twater ◦ qwater = (0.2 kg)(4190 J/kg C)(25 − 20) Step 4: Set the two equations equal and solve for the specific heat capacity of iron: ◦ (0.05 kg)(ciron )(55) = (0.2 kg)(4190 J/kg C)(5) ◦ ciron = (0.2 kg)(4190 J/kg C)(5) (0.05 kg)(55) ◦ ciron = 3821 J/kg C Question 8 Question A 50 g aluminum block at 80°C is dropped into 200 g of water at 20°C in a calorimeter. If the final temperature of the system is 25°C, calculate the specific heat capacity of aluminum. Assume that the specific heat capacity of water is 4.18 J/g°C and neglect any heat loss to the surroundings. 7 Solution Step 1: Calculate the heat lost by the aluminum block as it cools down from 80°C to 25°C. The formula for heat transfer is given by: q = mc∆T Where: m = mass of the substance (Aluminum block) in grams = 50 g c = specific heat capacity of the substance (Aluminum) in J/g°C ∆T = change in temperature = 25C − 80C = −55C (negative since it is cooling down) Substitute the values into the formula: qAluminum = 50 × c × (−55) Step 2: Calculate the heat gained by the water as it warms up from 20°C to 25°C. Since the water gains the same amount of heat that the aluminum loses, we can write: qAluminum = qWater The formula for heat transfer in water is similar to that of the aluminum block: qWater = mc∆T Where: m = mass of the substance (Water) in grams = 200 g c = specific heat capacity of the substance (Water) = 4.18 J/g°C ∆T = change in temperature = 25C − 20C = 5C Substitute the values into the formula: 50 × c × (−55) = 200 × 4.18 × 5 Step 3: Solve the equation obtained in Step 2 to find the specific heat capacity of aluminum (c). 50c × (−55) = 200 × 4.18 × 5 200 × 4.18 × 5 50 × 55 4180 c= 55 c ≈ 76.0 J/g°C c= Therefore, the specific heat capacity of aluminum is approximately 76.0 J/g°C. Question 9 Question A 50 g lead bullet at 100°C is fired into a large block of ice at 0°C. If the bullet comes to rest after melting some ice, what is the mass of ice melted? (Specific heat of lead = 0.128 cal/g°C, heat of fusion of ice = 79.7 cal/g, specific heat of ice = 0.5 cal/g°C) 8 Solution Step 1: Calculate the heat lost by the lead bullet as it cools down to the final temperature. The specific heat formula Q = mc∆T can be used where Q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. Given: mlead = 50 g, clead = 0.128 cal/g°C, Ti-lead = 100C, Tf-lead = x Heat lost by lead = −(50 g)(0.128 cal/g°C)(x − 100) cal Step 2: Calculate the heat required to melt the ice. The heat required for a phase change (melting/freezing) is given by Q = mL where Q is the heat absorbed or released, m is the mass, and L is the latent heat of fusion. Given: Lice = 79.7 cal/g, mice = y Heat needed to melt ice = (y)(79.7) cal Step 3: Calculate the heat gained by the melted ice as it warms up to its final temperature. The specific heat formula can be used again. Given: mice = y, cice = 0.5 cal/g°C, Ti-ice = 0C, Tf-ice = x Heat gained by ice = (y)(0.5)(x − 0) cal Step 4: Setting up the conservation of energy equation. The heat lost by the lead bullet equals the heat gained by the melted ice as well as the heat required to melt the ice. −(50)(0.128)(x − 100) = (y)(79.7) + (y)(0.5)(x) Step 5: Solve for y. Solving the equation found in Step 4 will give the mass of ice melted. Let’s solve this equation. −6.4x + 640 = 79.7y + 0.5yx Since the initial temperature of the ice is 0°C, we can simplify this further to only consider the ice reaching the final temperature: −6.4x + 640 = 79.7y + 0.5yx −6.4x + 640 = y(79.7 + 0.5x) −6.4x + 640 = y(79.7 + 0.5x) Hence, the mass of ice melted is given by y = 9 −6.4x+640 79.7+0.5x . Question 10 Question A student performs a calorimetry experiment by mixing 200 g of water at 20◦ C with 100 g of an unknown metal at 100◦ C. The final temperature of the mixture is 25◦ C. If the specific heat capacity of water is 4.18 J/(g·◦ C), determine the specific heat capacity of the metal. Solution Step 1: Calculate the heat lost by the hot metal and gained by the cold water using the formula: Qmetal = −Qwater where Q = mc∆T Step 2: Calculate the heat lost by the metal: Qmetal = mmetal cmetal ∆Tmetal where - mmetal = 100 g (mass of metal) - cmetal (specific heat capacity of the metal, to be determined) - ∆Tmetal = Tf − Tmetal = 25 − 100 = −75 ◦ C Step 3: Calculate the heat gained by the water: Qwater = mwater cwater ∆Twater where - mwater = 200 g (mass of water) - cwater = 4.18 J/(g·◦ C) (specific heat capacity of water) - ∆Twater = Tf − Twater = 25 − 20 = 5 ◦ C Step 4: Set up the equation: 100 × cmetal × −75 = −200 × 4.18 × 5 Step 5: Solve for cmetal : −200 × 4.18 × 5 100 × −75 Step 6: Calculate the specific heat capacity of the metal: cmetal = cmetal = 1.39 J/(g·◦ C) Therefore, the specific heat capacity of the metal is 1.39 J/(g·◦ C). Question 11 Question A 50.0 g piece of aluminum at 80.0°C is dropped into 100.0 g of water at 20.0°C. Assuming no heat loss to the surroundings, what will be the final temperature of the system? (Specific heat capacity of aluminum = 0.900 J/g°C, specific heat capacity of water = 4.18 J/g°C, and the heat of fusion of water = 334 J/g) 10 Solution Step 1: Calculate the heat lost by the aluminum as it cools down to the final temperature. The heat lost is given by the formula: Qlost, Al = m · c · ∆T where: - m = 50.0 g is the mass of the aluminum, - c = 0.900 J/g°C is the specific heat capacity of aluminum, - ∆T = Tinitial − Tfinal , - Tinitial = 80.0C is the initial temperature of the aluminum, and - Tfinal is the final temperature of the system. Substitute the values: Qlost, Al = 50.0 g × 0.900 J/g°C × (80.0 − Tfinal )C Step 2: Calculate the heat gained by the water as it warms up to the final temperature. The heat gained by the water is given by the formula: Qgain, water = m · c · ∆T where: - m = 100.0 g is the mass of the water, - c = 4.18 J/g°C is the specific heat capacity of water, - ∆T = Tfinal − Tinitial , - Tinitial = 20.0C is the initial temperature of the water. Substitute the values: Qgain, water = 100.0 g × 4.18 J/g°C × (Tfinal − 20.0)C Step 3: Set up the equation based on the law of conservation of energy. The heat lost by the aluminum must equal the heat gained by the water since no energy is lost to the surroundings: Qlost, Al = Qgain, water Step 4: Solve the equation to find the final temperature of the system. 50.0 g × 0.900 J/g°C × (80.0 − Tfinal )C = 100.0 g × 4.18 J/g°C × (Tfinal − 20.0)C Solve for Tfinal to find the final temperature of the system. Question 12 Question A student wants to determine the specific heat capacity of a metal block. The student immerses the metal block, initially at a temperature of 200°C, into a 1.5 kg of water at 20°C contained in a calorimeter. The final equilibrium temperature of the system is found to be 40°C. If the specific heat capacity of ◦ water is 4186 J/kg C, determine the specific heat capacity of the metal block. 11 Solution Let’s denote the specific heat capacity of the metal block as cm and the mass of the metal block as mm . We can use the principle of conservation of energy to solve for cm . Step 1: Calculate the heat lost by the metal block and the heat gained by the water. The heat lost by the metal block is equal to the heat gained by the water. Therefore, we can write: mm cm ∆T = mw cw ∆T ◦ where: mm = 1.5 kg (mass of water), cw = 4186 J/kg C (specific heat capacity of water), ∆T = 20◦ C (change in temperature). Step 2: Calculate the change in temperature for the metal block. The temperature of the metal block changes from 200°C to 40°C, thus ∆Tm = 40◦ C − 200◦ C = −160◦ C. Step 3: Substitute known values into the heat transfer equation. ◦ Substitute mm = 1.5 kg, cw = 4186 J/kg C, ∆T = 20◦ C, and ∆Tm = −160◦ C into the heat transfer equation: 1.5cm (−160) = 1.5 · 4186 · 20 −240cm = 125580 125580 cm = −240 ◦ cm = −523.25 J/kg C Step 4: Conclusion The specific heat capacity of the metal block is ◦ −523.25 J/kg C. It is negative because heat transfer from the metal block causes a temperature decrease of the metal block. Question 13 Question A 50.0 g sample of aluminum at 90.0°C is added to a calorimeter containing 100.0 g of water at 25.0°C. If the final temperature of both the aluminum and water is 28.0°C, what is the specific heat capacity of aluminum? Assume the specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat absorbed by the water using the formula qwater = mwater × cwater × ∆T 12 where - mwater = 100.0 g is the mass of water, - cwater = 4.18 J/g°C is the specific heat capacity of water, and - ∆T = Tfinal − Tinitial = 28.0C − 25.0C = 3.0C is the change in temperature. Substitute the values into the formula to find qwater . Step 2: Calculate the heat lost by the aluminum using the formula qaluminum = −qwater because heat lost by the aluminum is equal in magnitude but opposite in sign to the heat gained by the water. Step 3: Calculate the heat lost by the aluminum using its specific heat capacity with the formula qaluminum = maluminum × caluminum × ∆T where - maluminum = 50.0 g is the mass of aluminum, - caluminum is the specific heat capacity of aluminum, and - ∆T = Tfinal − Tinitial = 28.0C − 90.0C = −62.0C is the change in temperature. Step 4: Set up and solve an equation to find caluminum using the relationships between the heat gained by the water and the heat lost by the aluminum. This equation can be written as mwater × cwater × ∆T = maluminum × caluminum × ∆T Substitute the known values and solve for caluminum . Question 14 Question A 50 g block of copper initially at 150◦ C is dropped into 200 g of water at 20◦ C. The final temperature of the system is 22◦ C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity of copper. The specific heat capacity of water is 4.18 J/g◦ C. Solution Step 1: Determine the heat transfer from the copper block to the water. The heat lost by the copper block equals the heat gained by the water. Let qCu be the heat lost by the copper block and qH2O be the heat gained by the water. qCu = −qH2O Step 2: Calculate the heat lost by the copper block. The heat lost by the copper block is given by the formula: qCu = −mc∆T where m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. 13 Substitute the given values: m = 50 g, c = specific heat capacity of copper, ∆T = (22 − 150) ◦ C = −128 ◦ C. qCu = 50 · c · (−128) = −6400c J Step 3: Calculate the heat gained by the water. The heat gained by the water is also given by the formula: qH2O = mc∆T Substitute the given values: m = 200 g, c = 4.18 J/g◦ C, ∆T = (22 − 20) ◦ C = 2 ◦ C. qH2O = 200 · 4.18 · 2 = 1672 J Step 4: Set up the equation and solve for the specific heat capacity of copper. Since qCu = −qH2O , we have −6400c = 1672 Solving for c: 1672 = 0.2615 J/g◦ C 6400 Therefore, the specific heat capacity of copper is 0.2615 J/g◦ C. c= Question 15 Question A 200 g aluminum calorimeter initially at 20°C contains 500 g of water at 40°C. A 50 g iron ball at 100°C is placed into the calorimeter, causing the temperature to stabilize at 30°C. Assuming no heat is lost to the surroundings, what is the specific heat capacity of the iron ball? (Specific heat capacities: aluminum = 0.902 J/g°C, water = 4.18 J/g°C) Solution Step 1: Calculate the heat lost by the iron ball as it cools to 30°C. The heat lost by the iron ball is equal to the heat gained by the aluminum calorimeter, water, and the final mixture. Let Ciron be the specific heat capacity of iron. The heat lost by the iron ball is given by: Qiron = miron · Ciron · (Tf − Ti ) Qiron = 50 g · Ciron · (30 − 100) Qiron = −3500 · Ciron J 14 Step 2: Calculate the heat gained by the water and aluminum calorimeter. The heat gained by the water and aluminum calorimeter is given by: Qwater = mwater · Cwater · (Tf − Ti ) Qaluminum = maluminum · Caluminum · (Tf − Ti ) We know that the total heat gained is equal to the heat lost by the iron ball: Qwater + Qaluminum = −3500 · Ciron J Substitute the given values and solve for Ciron : 500 g · 4.18 J/g°C + 200 g · 0.902 J/g°C = −3500 · Ciron 2090 + 180 ≈ −3500 · Ciron 2270 ≈ −3500 · Ciron Ciron ≈ −0.648 J/g°C Therefore, the specific heat capacity of the iron ball is approximately 0.648 J/g°C. Question 16 Question A piece of iron with a mass of 150 grams at a temperature of 200◦ C is placed into 500 grams of water at 20◦ C in a calorimeter. The final temperature of the system is 25◦ C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity of iron. (Specific heat capacity of water = 4.18 J/g◦ C) Solution Step 1: Calculate the heat lost by the iron and the heat gained by the water using the formula for heat transfer: q = mc∆T where q is the heat transferred, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. For the iron: qiron = miron ciron ∆Tiron qiron = (0.150 kg)(ciron )(200 − 25) For the water: qwater = mwater cwater ∆Twater ◦ qwater = (0.500 kg)(4.18 J/g C)(25 − 20) 15 Step 2: Since the heat lost by the iron is equal to the heat gained by the water (assuming no heat loss to the surroundings): qiron = qwater (0.150)(ciron )(175) = (0.500)(4.18)(5) Step 3: Solve for the specific heat capacity of iron, ciron : ciron = (0.500)(4.18)(5) (0.150)(175) 10.45 26.25 ◦ = 0.398 J/g C ciron = ciron Question 17 Question A 50.0 g piece of aluminum at 90.0°C is placed in 100.0 g of water at 20.0°C. The final temperature of the mixture is 22.5°C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity of aluminum. Given: specific ◦ heat capacity of water = 4.18 J/(g C). Solution Step 1: Calculate the heat lost by aluminum and the heat gained by water. The heat lost by aluminum is equal to the heat gained by water. Heat lost by aluminum = Heat gained by water m1 · c1 · ∆T1 = m2 · c2 · ∆T2 where: m1 = 50.0 g (mass of aluminum), c1 =? (specific heat capacity of aluminum), ∆T1 = −67.5◦ C (temperature change for aluminum), m2 = 100.0 g ◦ (mass of water), c2 = 4.18 J/(g C) (specific heat capacity of water), ∆T2 = 2.5◦ C (temperature change for water). Step 2: Substitute the values and solve for the specific heat capacity of aluminum. ◦ 50.0 g · c1 · (−67.5◦ C) = 100.0 g · 4.18 J/(g C) · 2.5◦ C ◦ c1 = 100.0 g · 4.18 J/(g C) · 2.5◦ C 50.0 g · (−67.5◦ C) 1045.0 J c1 = −3375 g ·◦ C ◦ c1 = −0.309 J/(g C) ◦ Therefore, the specific heat capacity of aluminum is −0.309 J/(g C). 16 Question 18 Question A 50.0 g piece of copper is heated to 95.0°C and then placed into 100.0 g of water at 25.0°C. The final temperature of the system is 26.0°C. Assuming all heat is transferred to the water, determine the specific heat of copper. Solution Step 1: Calculate the heat absorbed by the water The heat absorbed by the water can be calculated using the formula: qwater = mwater · cwater · ∆T where: - mwater is the mass of water (100.0 g), - cwater is the specific heat capacity of water (4.18 J/g°C), and - ∆T is the temperature change of the water (final temperature - initial temperature). Substitute the values into the formula: qwater = (100.0 g) · (4.18 J/g°C) · (26.0 − 25.0)C qwater = 100 · 4.18 · 1 = 418 J Step 2: Calculate the heat lost by the copper The heat lost by the copper is equal to the heat gained by the water, so: qcopper = −qwater = −418 J Step 3: Calculate the specific heat of copper The heat lost by the copper can be calculated using the formula: qcopper = mcopper · ccopper · ∆T where: - mcopper is the mass of copper (50.0 g), - ccopper is the specific heat capacity of copper (unknown), and - ∆T is the temperature change of the copper (final temperature - initial temperature). Substitute the values into the formula and solve for ccopper : −418 = (50.0 g) · ccopper · (26.0 − 95.0)C ccopper = −418 50 · (26.0 − 95.0) ccopper ≈ 0.39 J/g°C Therefore, the specific heat of copper is approximately 0.39 J/g°C. 17 Question 19 Question A 50 g piece of copper at 95°C is placed in 200 g of water at 25°C in an insulated container. Assuming no heat is lost to the surroundings, what will be the final temperature of the system? (Specific heat capacity of copper is 0.385 J/g°C and specific heat capacity of water is 4.18 J/g°C) Solution Step 1: Calculate the heat gained by the water. The heat gained by the water can be calculated using the formula: Qwater = mwater × cwater × ∆T where: - mwater = 200 g (mass of water), - cwater = 4.18 J/g°C (specific heat capacity of water), - ∆T = Tfinal − Tinitial . The initial temperature of water, Tinitial = 25C. Step 2: Calculate the heat lost by the copper. The heat lost by the copper can be calculated using the formula: Qcopper = mcopper × ccopper × ∆T where: - mcopper = 50 g (mass of copper), - ccopper = 0.385 J/g°C (specific heat capacity of copper), - ∆T = Tfinal − Tinitial . The initial temperature of copper, Tinitial = 95C. Step 3: Set up the energy conservation equation. Since no heat is lost to the surroundings, the heat gained by the water must be equal to the heat lost by the copper. Qwater = Qcopper Step 4: Solve for the final temperature. Substitute the expressions for Qwater and Qcopper into the energy conservation equation and solve for the final temperature, Tfinal . mwater × cwater × ∆T = mcopper × ccopper × ∆T 200 × 4.18 × (Tfinal − 25) = 50 × 0.385 × (95 − Tfinal ) Solve this equation to find the final temperature, Tfinal . 18 Question 20 Question A 50 g piece of copper is heated to 100°C and then placed in 200 g of water at 20°C. The final temperature of the system is 25°C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity of copper. (Specific heat capacity of water = 4.18 J/g°C) Solution Step 1: Calculate the heat lost by the copper piece as it cools down from 100°C to 25°C. The heat lost is given by the formula: Q = mc∆T , where: m is the mass of the copper piece (in grams), - c is the specific heat capacity of copper (in J/g°C), - ∆T is the change in temperature (final temperature initial temperature). Given: - mcopper = 50 g, - Tinitial = 100 °C, - Tfinal = 25 °C. Substitute these values into the formula: Qcopper = (50 g)(c)(25 °C) Step 2: Calculate the heat gained by the water as it heats up from 20°C to 25°C. The heat gained is given by the same formula: Q = mc∆T , where: - m is the mass of the water (in grams), - c is the specific heat capacity of water (in J/g°C), - ∆T is the change in temperature (final temperature - initial temperature). Given: - mwater = 200 g, - cwater = 4.18 J/g°C, - Tinitial = 20 °C, - Tfinal = 25 °C. Substitute these values into the formula: Qwater = (200 g)(4.18 J/g°C)(5 °C) Step 3: Since heat lost by the copper equals heat gained by the water (assuming no heat loss to the surroundings), we can set the two equations equal to each other: (50 g)(c)(25 °C) = (200 g)(4.18 J/g°C)(5 °C) J/g°C)(5 °C) Step 4: Solve for the specific heat capacity of copper, c: c = (200 g)(4.18 (50 g)(25 °C) 4180 J c = 1250 g°C c = 3.344 J/g°C Therefore, the specific heat capacity of copper is 3.344 J/g°C. Question 21 Question A 50 g aluminum block at 80°C is dropped into 200 g of water at 20°C in a calorimeter. The specific heat capacity of aluminum is 0.90 J/g°C and the specific heat capacity of water is 4.18 J/g°C. If the final temperature of the system is 24.5°C, determine the heat capacity of the calorimeter. 19 Solution Step 1: Calculate the heat lost by the aluminum block as it cools down to the final temperature of the system. The formula for heat transfer is given by Q = mc∆T, where Q is the heat transfer, m is the mass of the substance, c is the specific heat capacity, and ∆T is the change in temperature. The heat lost by the aluminum block is calculated as follows: QAluminum = (50 g)(0.90 J/g°C)(80 − 24.5) = 2075 J. Step 2: Calculate the heat gained by the water and the calorimeter. The heat gained by the water and calorimeter is given by Qwater+calorimeter = (50 g + 200 g)(4.18 J/g°C)(24.5 − 20) = 3503 J. Step 3: Set up an equation to determine the heat capacity of the calorimeter. Since the heat lost by the aluminum block is equal to the heat gained by the water and calorimeter (law of conservation of energy), we have QAluminum = Qwater+calorimeter . Step 4: Solve for the heat capacity of the calorimeter. 2075 J = 3503 J + Ccalorimeter , where Ccalorimeter is the heat capacity of the calorimeter. Therefore, the heat capacity of the calorimeter is Ccalorimeter = 2075 J − 3503 J = −1428 J. The negative sign indicates that the calorimeter releases 1428 J of heat, which makes sense since the calorimeter would lose heat to the surroundings during the experiment. Question 22 Question A 50 g piece of aluminum at 80°C is dropped into 200 g of water at 20°C in a calorimeter. The final temperature of the system is 25°C. Calculate the specific heat capacity of aluminum. Assume the specific heat capacity of water is 4.18 J/g°C. 20 Solution Step 1: Calculate the heat lost by the aluminum piece. The formula for calculating heat transfer is: Q = mc∆T where: Q = heat transfer (in J) m = mass of the substance (in g) c = specific heat capacity of the substance (in J/g°C) ∆T = change in temperature (in °C) Given: maluminum = 50 g cwater = 0.903 J/g°C ∆Taluminum = (80 − 25) °C = 55 °C Substitute the values into the formula: Qaluminum = 50 g × caluminum × 55 °C Step 2: Calculate the heat gained by the water. Since the system is isolated, the heat lost by the aluminum piece is equal to the heat gained by the water. Therefore: Qaluminum = Qwater Using the same formula and the given values: Qwater = 200 g × 4.18 J/g°C × (25 − 20) °C Step 3: Set the two equations equal to each other: 50 g × caluminum × 55 °C = 200 g × 4.18 J/g°C × 5 °C Step 4: Solve for caluminum . caluminum = 200 g × 4.18 J/g°C × 5 °C 50 g × 55 °C Question 23 Question A 50 g piece of aluminum at 80◦ C is dropped into 200 g of water at 20◦ C in a calorimeter. The final temperature of the system is 25◦ C. Assuming the specific heat capacity of water is 4.18 J/g◦ C and that of aluminum is 0.897 J/g◦ C, calculate the specific heat capacity of the calorimeter. Solution Step 1: Calculate the heat gained by water: The heat gained by the water is given by the equation Q = mc∆T where: - m is the mass of water (200 g), - c is the specific heat capacity of water (4.18 J/g◦ C), - ∆T is the change in temperature of the water. 21 The change in temperature of the water is ∆T = Tf − Ti = 25◦ C − 20◦ C = 5◦ C. Substitute the values in: ◦ Q = (200 g)(4.18 J/g C)(5◦ C) Q = 4180 J Step 2: Calculate the heat lost by the aluminum: The heat lost by the aluminum is given by the equation Q = mc∆T where: - m is the mass of aluminum (50 g), - c is the specific heat capacity of aluminum (0.897 J/g◦ C), - ∆T is the change in temperature of the aluminum. The change in temperature of the aluminum is ∆T = Tf − Ti = 25◦ C − 80◦ C = −55◦ C. Notice that the negative sign indicates that the aluminum is losing heat energy. Substitute the values in: ◦ Q = (50 g)(0.897 J/g C)(−55◦ C) Q = −2456.25 J Step 3: Calculate the heat absorbed by the calorimeter (water + aluminum): Since we assume no heat is lost to the surroundings, the heat lost by the aluminum is equal to the heat gained by the water and the calorimeter: Qcalorimeter = −Qaluminum = Qwater Qcalorimeter = 4180 J Step 4: Calculate the heat capacity of the calorimeter: We know the heat capacity formula is Q = mc∆T where: - m is the mass of the calorimeter, - c is the specific heat capacity of the calorimeter, - ∆T is the change in temperature of the calorimeter. Substitute the values in: 4180 J = (m)(c)(5◦ C) We are looking for c, the specific heat capacity of the calorimeter. Solving for c, we get 4180 J c= (m)(5◦ C) Since we don’t have the mass of the calorimeter, further calculation is not possible based on the information provided. 22 Question 24 Question A piece of metal of mass 120.0 g at a temperature of 200.0°C is placed in 250.0 g of water at 20.0°C. The final equilibrium temperature of the system is 26.5°C. Assume the specific heat capacities of the metal and water are constant at 0.385 J/g°C and 4.18 J/g°C, respectively. Determine the specific heat capacity of the metal. Solution Step 1: Calculate the heat gained by the metal as it cools down to the final temperature. The formula for heat gained or lost is given by: q = mc∆T Where: - q is the heat gained or lost - m is the mass of the object - c is the specific heat capacity of the substance - ∆T is the change in temperature Given that the metal is cooling down, the heat gained by the metal is equal to the heat lost by the water. Thus, we can set up the equation: mmetal cmetal ∆Tmetal = −mwater cwater ∆Twater Substitute the values to find cmetal : 120.0 g · cmetal · (26.5C − 26.5C) = −250.0 g · 4.18 J/g°C · (26.5C − 20.0C) 120.0 g · cmetal · 0 = −250.0 g · 4.18 J/g°C · 6.5C 0 = −250.0 · 4.18 · 6.5 0 = −6505.0 Since our calculated value is negative, there seems to be an error in our calculation. Let’s go back and verify our computations. Question 25 Question A 50 g piece of aluminum at 100°C is placed in 150 g of water at 20°C. If the final temperature of the system is 25°C, calculate the specific heat capacity of aluminum. (Specific heat capacity of water = 4.18 J/g°C) 23 Solution Step 1: First, we find the heat absorbed by the aluminum and the heat released by the water. The heat lost by the aluminum equals the heat gained by the water since no heat is lost to the surroundings. Let the specific heat capacity of aluminum be cAl . Step 2: The heat absorbed by the aluminum can be calculated using the formula: qAl = mAl · cAl · ∆TAl where mAl = 50 g ∆TAl = (25 − 100) = −75 °C Therefore, qAl = 50 g · cAl · (−75) Step 3: The heat released by the water can be calculated using the formula: qH2O = mH2O · cH2O · ∆TH2O where mH2O = 150 g cH2O = 4.18 J/g°C ∆TH2O = (25 − 20) = 5 °C Therefore, qH2O = 150 g · 4.18 J/g°C · 5 °C Step 4: Since the heat lost by the aluminum equals the heat gained by the water: qAl = qH2O 50 g · cAl · (−75) = 150 g · 4.18 J/g°C · 5 Step 5: Solve for cAl to find the specific heat capacity of aluminum. 50 g · cAl · (−75) = 150 g · 4.18 J/g°C · 5 cAl = 150 g · 4.18 J/g°C · 5 50 g · (−75) cAl = 3135 J −3750 g Step 6: Calculate the specific heat capacity of aluminum. cAl = −0.836 J/g°C Therefore, the specific heat capacity of aluminum is −0.836 J/g°C. 24 Question 26 Question A student conducts an experiment to determine the specific heat capacity of a metal by immersing it in a container of water. The metal is heated to a high temperature and then quickly transferred to the water, causing the water’s temperature to rise. The student records the initial temperature of the water as 20°C and the final temperature as 32°C. The mass of the metal is 0.2 kg and its initial temperature is 100°C. The mass of the water is 0.5 kg and its specific heat capacity is 4186 J/kg · K. If the final temperature of the metal and water is 30°C, what is the specific heat capacity of the metal? Solution Step 1: Calculate the heat absorbed by the water. The heat absorbed by the water can be calculated using the formula: Qwater = mc∆T where: m = mass of water = 0.5 kg, c = specific heat capacity of water = 4186 J/kg · K, ∆T = change in temperature = final temperature - initial temperature = 32C − 20C = 12C. Substitute the values into the formula: Qwater = 0.5 kg × 4186 J/kg · K × 12C = 25116 J Step 2: Calculate the heat lost by the metal. The heat lost by the metal can be calculated using the formula: Qmetal = mc∆T where: m = mass of metal = 0.2 kg, c = specific heat capacity of the metal (to be determined), ∆T = change in temperature = initial temperature - final temperature = 100C − 30C = 70C. Substitute the values into the formula: Qmetal = 0.2 kg × c × 70C Step 3: Set up the heat balance equation. Since the system is isolated and there is no heat lost to the surroundings, the heat lost by the metal is equal to the heat gained by the water. Qwater = Qmetal 0.2 kg × c × 70C = 25116 J Step 4: Solve for the specific heat capacity of the metal. 0.2 kg × c × 70C = 25116 J 25 c= 25116 J 0.2 kg × 70C c = 179.4 J/kg · K Therefore, the specific heat capacity of the metal is 179.4 J/kg · K. Question 27 Question A 50 g piece of iron at 80°C is placed into 200 g of water at 20°C in a calorimeter. If the final temperature of the system is 25°C, what is the specific heat capacity of the iron? Assume the specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat gained or lost by the water. Given: Mass of water, mw = 200 g Initial temperature of water, Tin = 20 °C Final temperature of the system, Tfinal = 25 °C Specific heat capacity of water, Cw = 4.18 J/g°C The heat gained by the water is given by the formula: Qwater = mw · Cw · ∆T where ∆T = Tfinal − Tin . Substitute the given values: ∆T = 25 °C − 20 °C = 5 °C Qwater = 200 g · 4.18 J/g°C · 5 °C Step 2: Calculate the heat gained or lost by the iron. Given: Mass of iron, miron = 50 g Initial temperature of iron, Tin = 80 °C Final temperature of the system, Tfinal = 25 °C The heat lost by the iron is equal to the heat gained by the water (since this is a closed system). Therefore: Qwater = Qiron mw · Cw · ∆T = miron · Ciron · ∆Tiron Substitute the known values: 200 g · 4.18 J/g°C · 5 °C = 50 g · Ciron · (80 °C − 25 °C) Step 3: Solve for the specific heat capacity of iron. 200 · 4.18 · 5 = 50 · Ciron · 55 Ciron = 200 · 4.18 · 5 50 · 55 26 Calculating the specific heat capacity of iron: Ciron = 4180 = 38 J/g°C 110 Therefore, the specific heat capacity of the iron is 38 J/g°C. Question 28 Question A piece of copper of mass 200 g at a temperature of 100°C is placed in 400 g of water at 20°C. Assuming no heat is lost to the surroundings, calculate the final temperature of the mixture. The specific heat capacity of copper is 0.385 J/g°C and the specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat lost by the copper when it cools down to the final temperature. The heat lost by the copper can be calculated using the formula: Qcopper = mc∆T where: - m is the mass of the copper (200 g), - c is the specific heat capacity of copper (0.385 J/g°C), - ∆T is the temperature change of the copper. The temperature change of the copper can be calculated as follows: ∆T = Tfinal − Tinitial ∆T = Tfinal − 100C Step 2: Calculate the heat gained by the water when it warms up to the final temperature. The heat gained by the water can be calculated using the formula: Qwater = mc∆T where: - m is the mass of the water (400 g), - c is the specific heat capacity of water (4.18 J/g°C), - ∆T is the temperature change of the water. The temperature change of the water can be calculated as follows: ∆T = Tfinal − Tinitial ∆T = Tfinal − 20C Step 3: Set up the equation based on the conservation of energy. Since there is no heat loss to the surroundings, the heat lost by the copper must equal the heat gained by the water. Thus, we have: 27 Qcopper = Qwater Step 4: Solve for the final temperature, Tfinal . Substitute the expressions for Qcopper and Qwater into the equation from Step 3 and solve for Tfinal . mc∆Tcopper = mc∆Twater Finally, solve for Tfinal . Question 29 Question A 150 g aluminum block at 175◦ C is placed into a container of water at 20◦ C. The water has a mass of 200 g. The final temperature of the system is 25◦ C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity ◦ of aluminum. (Specific heat of water = 4.18 J/g C) Solution Step 1: Calculate the heat absorbed by the aluminum block as it cools down: The heat absorbed by the aluminum block can be calculated using the formula: qAl = mAl · cAl · ∆T Where: - mAl = mass of aluminum block = 150 g - cAl = specific heat capacity of aluminum (to be determined) - ∆T = change in temperature of aluminum block = 25◦ C − 175◦ C = −150◦ C (negative sign indicates a decrease in temperature) Substitute the given values into the formula: qAl = 150 g × cAl × (−150◦ C) Step 2: Calculate the heat released by the cooling water as it heats up: The heat released by the water can be calculated using the same formula but with the specific heat capacity of water: qwater = mwater × cwater × ∆T Where: - mwater = mass of water = 200 g - cwater = specific heat capacity of ◦ water = 4.18 J/g C - ∆T = change in temperature of water = 25◦ C − 20◦ C = 5◦ C Substitute the given values into the formula: ◦ qwater = 200 g × 4.18 J/g C × 5◦ C 28 Step 3: Since the system is isolated and assuming no heat is lost to the surroundings, the heat absorbed by the aluminum block is equal to the heat released by the water: qAl = qwater Step 4: Solve for the specific heat capacity of aluminum: Set the two heat values equal to each other and solve for cAl : ◦ 150 g × cAl × (−150◦ C) = 200 g × 4.18 J/g C × 5◦ C ◦ cAl = 200 g × 4.18 J/g C × 5◦ C 150 g × (−150◦ C) cAl = 2090 J −22500 g◦ C ◦ cAl ≈ −0.093 J/g C ◦ The specific heat capacity of aluminum is approximately −0.093 J/g C. Question 30 Question A 50 g block of copper at 100◦ C is dropped into 200 g of water at 20◦ C contained in a 100 g aluminum calorimeter cup. The final temperature of the system is 30◦ C. Given the specific heat capacities of copper, water, and aluminum as 0.385 J/g°C, 4.18 J/g°C, and 0.897 J/g°C, respectively, calculate the initial temperature of the water before the copper block was dropped into it. Solution Step 1: Calculate the heat absorbed by the copper block. The heat absorbed by the copper block can be calculated using the formula: Q = mc∆T where: - m is the mass of the copper block (50 g), - c is the specific heat capacity of copper (0.385 J/g°C), and - ∆T is the change in temperature of the copper block (30 − 100 = −70°C). Substitute the given values into the formula to calculate Q: Q = 50 × 0.385 × (−70) Step 2: Calculate the heat lost by the water. The heat lost by the water can be calculated using the formula: Q = mc∆T 29 where: - m is the mass of the water (200 g), - c is the specific heat capacity of water (4.18 J/g°C), and - ∆T is the change in temperature of the water (30 − x°C, where x is the initial temperature of the water). Substitute the given values into the formula to calculate Q. Step 3: Calculate the heat absorbed by the aluminum calorimeter cup. The heat absorbed by the aluminum calorimeter cup can be calculated using the formula: Q = mc∆T where: - m is the mass of the aluminum calorimeter cup (100 g), - c is the specific heat capacity of aluminum (0.897 J/g°C), and - ∆T is the change in temperature of the aluminum calorimeter cup (30 − 20 = 10°C). Substitute the given values into the formula to calculate Q. Step 4: Apply the principle of conservation of energy. According to the principle of conservation of energy, the heat lost by the copper block is equal to the sum of the heat gained by the water and the aluminum calorimeter cup: Qcopper = Qwater + Qcup Substitute the calculated values of Qcopper , Qwater , and Qcup into the equation and solve for x, the initial temperature of the water. Question 31 Question A 50 g aluminum block at 80°C is placed in 200 g of water at 20°C in a calorimeter. The final temperature of the system is 30°C. Assuming no heat is lost to the surroundings and that the specific heat capacity of water is 4186 J/kg°C, calculate the specific heat capacity of aluminum. Solution Step 1: Calculate the heat lost by the aluminum block Heat lost by aluminum = m × c × ∆T where m = 50 g = 0.05 kg (mass of aluminum block), c (specific heat capacity of aluminum), ∆T = Tf − Ti = 30 − 80 = −50 °C . Therefore, Heat lost by aluminum = 0.05 × c × (−50) Step 2: Calculate the heat gained by water Heat gained by water = m × cw × ∆T 30 where m = 200 g = 0.2 kg (mass of water), cw = 4186 J/kg°C (specific heat capacity of water), ∆T = Tf − Ti = 30 − 20 = 10 °C . Therefore, Heat gained by water = 0.2 × 4186 × 10 Step 3: Since there is no heat loss to the surroundings, the heat lost by the aluminum block is equal to the heat gained by the water. 0.05 × c × (−50) = 0.2 × 4186 × 10 Step 4: Solve for the specific heat capacity of aluminum, c 0.05 × c × (−50) = 0.2 × 4186 × 10 c= 0.2 × 4186 × 10 0.05 × (−50) Question 32 Question A piece of aluminum of mass 250 g at a temperature of 90°C is placed in 500 g of water at 20°C in a calorimeter of negligible heat capacity. The final temperature of the system is 23°C. Assuming no heat loss to the surroundings, calculate the specific heat capacity of aluminum. (Specific heat capacity of water = 4.18 J/g°C) Solution Step 1: Calculate the heat gained by the water. Given: mwater = 500 g ∆Twater = 23−20 = 3 °C cwater = 4.18 J/g°C Using the formula Q = mc∆T , the heat gained by the water can be calculated as: Qwater = (500 g) × (4.18 J/g°C) × 3 °C Step 2: Calculate the heat lost by the aluminum. Given: maluminum = 250 g ∆Taluminum = 23 − 90 = −67 °C (negative as aluminum is losing heat) Let caluminum be the specific heat capacity of aluminum. Using the formula Q = mc∆T , the heat lost by the aluminum can be calculated as: Qaluminum = (250 g) × caluminum × (−67) °C Step 3: Since no heat is lost to the surroundings, the heat gained by the water is equal to the heat lost by the aluminum. Therefore, we have: Qwater = Qaluminum Step 4: Equate Qwater and Qaluminum and solve for caluminum to find the specific heat capacity of aluminum. This is left as an exercise for the reader. 31 Question 33 Question A 50 g piece of copper at 150°C is dropped into 200 g of water at 20°C in a calorimeter. The final temperature of the mixture is 25°C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity of copper. The specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat lost by the copper and the heat gained by the water. The formula for heat transfer in calorimetry is given by: Q = mc∆T where: Q = heat energy transferred, m = mass of the material, c = specific heat capacity of the material, ∆T = change in temperature. Let’s first calculate the heat lost by the copper piece: Given: mcopper = 50 g, cwater = 0.387 J/g°C (specific heat capacity of copper), Tinitial = 150C, Tfinal = 25C. The change in temperature for copper (∆Tcopper ) is: ∆Tcopper = Tfinal − Tinitial = 25C − 150C = −125C (The negative sign indicates the decrease in temperature). Now, calculate the heat lost by the copper: Qcopper = mcopper · ccopper · ∆Tcopper Step 2: Calculate the heat gained by the water. Given: mwater = 200 g, cwater = 4.18 J/g°C (specific heat capacity of water), Tinitial = 20C, Tfinal = 25C. The change in temperature for water (∆Twater ) is: ∆Twater = Tfinal − Tinitial = 25C − 20C = 5C Now, calculate the heat gained by the water: Qwater = mwater · cwater · ∆Twater Step 3: Set up the equation for heat conservation. Since no heat is lost to the surroundings, the heat lost by the copper should be equal to the heat gained by the water: Qcopper = Qwater Step 4: Solve for the specific heat capacity of copper. Equating the two equations: mcopper · ccopper · ∆Tcopper = mwater · cwater · ∆Twater Substitute the known values and solve for ccopper to find the specific heat capacity of copper. 32 Question 34 Question A 50 g aluminum block at 80°C is dropped into 200 g of water at 20°C in a calorimeter. If the final temperature of the system is 25°C, calculate the specific heat capacity of aluminum. Assume no heat is lost to the surroundings. Solution Step 1: Calculate the heat absorbed by the aluminum block. The heat absorbed by the aluminum block can be calculated using the formula: Qaluminum = mc∆T Where: - m is the mass of the aluminum block (50 g) - c is the specific heat capacity of aluminum - ∆T is the temperature change of the aluminum block (final temperature - initial temperature) Given that the initial temperature of the aluminum block is 80°C and the final temperature is 25°C, we have: ∆T = 25C − 80C = −55C Substitute the known values into the formula: Qaluminum = (50 g)(c)(−55C) Step 2: Calculate the heat released by the aluminum block to the water. The heat released by the aluminum block is equal to the heat absorbed by the water, which can be calculated using the formula: Qwater = mc∆T Where: - m is the mass of the water (200 g) - c is the specific heat capacity of water - ∆T is the temperature change of the water (final temperature - initial temperature) Given that the initial temperature of the water is 20°C and the final temperature is 25°C, we have: ∆T = 25C − 20C = 5C Substitute the known values into the formula: Qwater = (200 g)(4.18 J/g°C)(5C) Step 3: Set up the equation for heat transfer. Since the total heat gained by the water is equal to the total heat lost by the aluminum block, we can set up the equation: Qaluminum = Qwater 33 Step 4: Solve for the specific heat capacity of aluminum. Equating Qaluminum and Qwater and solving for c: (50 g)(c)(−55C) = (200 g)(4.18 J/g°C)(5C) c= (200 g)(4.18 J/g°C)(5C) (50 g)(−55C) Question 35 Question A 50.0 g sample of aluminum at 95.0°C is added to 100.0 g of water at 25.0°C. The final temperature of the system is 30.0°C. Assuming no heat is lost to the surroundings and the specific heat capacity of water is 4.18 J/g°C, determine the specific heat capacity of aluminum. Solution Step 1: Calculate the heat lost by the aluminum and gained by the water. The heat lost by the aluminum is equal to the heat gained by the water: (m · c · ∆T )aluminum = (m · c · ∆T )water where maluminum = 50.0 g (mass of aluminum), cwater = 4.18 J/g°C (specific heat capacity of water), Tinitial, aluminum = 95.0 °C (initial temperature of aluminum), Tfinal = 30.0 °C (final temperature of the system), mwater = 100.0 g (mass of water), Tinitial, water = 25.0 °C (initial temperature of water). Therefore, the equation becomes: (50.0 g · caluminum · (30.0 °C − 95.0 °C)) = (100.0 g · 4.18 J/g°C · (30.0 °C − 25.0 °C)) Step 2: Solve for caluminum . Begin by simplifying both sides of the equation: 6750 · caluminum = 20.9 · 100 20.9 · 100 = 0.310 J/g°C 6750 Therefore, the specific heat capacity of aluminum is 0.310 J/g°C. caluminum = 34 Question 5 Question A 50 g aluminum block at 80°C is dropped into 200 g of water at 20°C in a calorimeter. The final temperature of the system is 25°C. Assuming no heat is lost to the surroundings, find the specific heat capacity of aluminum. The specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat lost by the aluminum block as it cools down from 80°C to 25°C. The specific heat capacity of aluminum is 0.897 J/g°C. The formula to calculate heat energy is Q = mc∆T , where: - Q is the heat energy, - m is the mass, - c is the specific heat capacity, - ∆T is the change in temperature. Given: m = 50 g, c = 0.897 J/g°C, and ∆T = 80C − 25C = 55C. Using the formula, we have: Qaluminum = (50 g)(0.897 J/g°C)(55C). Calculating Qaluminum , we get: Qaluminum = 2481.75 J. Step 2: Calculate the heat gained by the water and the calorimeter. Using the same formula as before with the given values: Qwater = (200 g)(4.18 J/g°C)(25C− 20C). Calculating Qwater , we get: Qwater = 418 J. Step 3: Since the total heat lost by the aluminum block is equal to the heat gained by the water and the calorimeter (assuming no heat loss to surroundings), we have: Qaluminum = Qwater + Qcalorimeter . Substitute the values we found into the equation: 2481.75 J = 418 J + Qcalorimeter . Solving for Qcalorimeter , we get: Qcalorimeter = 2063.75 J. Step 4: The heat gained by the calorimeter is due to the temperature change of the water and calorimeter together. Let the specific heat capacity of the calorimeter be C. Since the water and calorimeter have the same temperature change, we have: Qcalorimeter = (200 g + mcalorimeter )C(25C − 20C). Given that the mass of water is 200 g, the specific heat capacity of water is 4.18 J/g°C, and the final temperature is 25C. Substitute the known values into the equation: 2063.75 J = (200 g+mcalorimeter )4.18 J/g°C× 5C. Solving for mcalorimeter , we find: mcalorimeter = 105 g. Step 5: Calculate the specific heat capacity of aluminum. Given that the mass of the aluminum block is 50 g and the specific heat capacity of the aluminum is 0.897 J/g°C, we have: Qaluminum = (50 g)(0.897 J/g°C)(55C). Substitute the values we know: 2481.75 J = (50 g)(0.897 J/g°C)(55C). Solving for the specific heat capacity of aluminum, we get: caluminum = 0.897 J/g°C. Therefore, the specific heat capacity of aluminum is 0.897 5 Question 6 Question A 50 g piece of aluminum at 100◦ C is dropped into 200 g of water at 20◦ C in a calorimeter. The final temperature of the system is 25◦ C. Assuming all the heat lost by the aluminum is gained by the water and the calorimeter, what is the heat capacity of the calorimeter? The specific heat capacity of aluminum is 0.90 J/g◦ C and that of water is 4.18 J/g◦ C. Solution Step 1: Calculate the heat lost by the aluminum. The heat lost by the aluminum can be calculated using the formula: Qaluminum = maluminum × caluminum × ∆T where: - maluminum is the mass of aluminum (50 g) - caluminum is the specific heat capacity of aluminum (0.90 J/g◦ C) - ∆T is the change in temperature (final temperature - initial temperature) ◦ Qaluminum = 50 g × 0.90 J/g × (25 − 100)◦ CQaluminum = 50 × 0.90 × (−75) Qaluminum = −3375 J Step 2: Calculate the heat gained by the water and the calorimeter. Since the heat lost by the aluminum is equal to the heat gained by the water and the calorimeter, we have: Qaluminum = Qwater + Qcalorimeter Substitute the values we know: ◦ −3375 J = (200 g + Ccalorimeter ) × 4.18 J/g × (25 − 20)◦ C −3375 J = (200 + Ccalorimeter ) × 4.18 × 5 −3375 = (200 + Ccalorimeter ) × 20.9 −3375 = 4180 + 20.9Ccalorimeter 20.9Ccalorimeter = −7555 Ccalorimeter = −361.2 J/◦ CJ/◦ C J/◦ C J/◦ C Therefore, the heat capacity of the calorimeter is -361.2 J/◦ C. Question 7 Question A 50 gram piece of iron at 80°C is placed in 200 grams of water at 20°C. If the final temperature of the system is 25°C, what is the specific heat capacity of iron? Assume no heat is lost to the surroundings. 6 Solution Step 1: Calculate the heat lost by the iron and the heat gained by the water using the formula: qiron = −qwater where q = mc∆T (m = mass, c = specific heat capacity, ∆T = change in temperature). Step 2: Calculate the heat lost by the iron: qiron = miron ciron ∆Tiron qiron = (0.05 kg)(ciron )(80 − 25) Step 3: Calculate the heat gained by the water: qwater = mwater cwater ∆Twater ◦ qwater = (0.2 kg)(4190 J/kg C)(25 − 20) Step 4: Set the two equations equal and solve for the specific heat capacity of iron: ◦ (0.05 kg)(ciron )(55) = (0.2 kg)(4190 J/kg C)(5) ◦ ciron = (0.2 kg)(4190 J/kg C)(5) (0.05 kg)(55) ◦ ciron = 3821 J/kg C Question 8 Question A 50 g aluminum block at 80°C is dropped into 200 g of water at 20°C in a calorimeter. If the final temperature of the system is 25°C, calculate the specific heat capacity of aluminum. Assume that the specific heat capacity of water is 4.18 J/g°C and neglect any heat loss to the surroundings. 7 Solution Step 1: Calculate the heat lost by the aluminum block as it cools down from 80°C to 25°C. The formula for heat transfer is given by: q = mc∆T Where: m = mass of the substance (Aluminum block) in grams = 50 g c = specific heat capacity of the substance (Aluminum) in J/g°C ∆T = change in temperature = 25C − 80C = −55C (negative since it is cooling down) Substitute the values into the formula: qAluminum = 50 × c × (−55) Step 2: Calculate the heat gained by the water as it warms up from 20°C to 25°C. Since the water gains the same amount of heat that the aluminum loses, we can write: qAluminum = qWater The formula for heat transfer in water is similar to that of the aluminum block: qWater = mc∆T Where: m = mass of the substance (Water) in grams = 200 g c = specific heat capacity of the substance (Water) = 4.18 J/g°C ∆T = change in temperature = 25C − 20C = 5C Substitute the values into the formula: 50 × c × (−55) = 200 × 4.18 × 5 Step 3: Solve the equation obtained in Step 2 to find the specific heat capacity of aluminum (c). 50c × (−55) = 200 × 4.18 × 5 200 × 4.18 × 5 50 × 55 4180 c= 55 c ≈ 76.0 J/g°C c= Therefore, the specific heat capacity of aluminum is approximately 76.0 J/g°C. Question 9 Question A 50 g lead bullet at 100°C is fired into a large block of ice at 0°C. If the bullet comes to rest after melting some ice, what is the mass of ice melted? (Specific heat of lead = 0.128 cal/g°C, heat of fusion of ice = 79.7 cal/g, specific heat of ice = 0.5 cal/g°C) 8 Solution Step 1: Calculate the heat lost by the lead bullet as it cools down to the final temperature. The specific heat formula Q = mc∆T can be used where Q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. Given: mlead = 50 g, clead = 0.128 cal/g°C, Ti-lead = 100C, Tf-lead = x Heat lost by lead = −(50 g)(0.128 cal/g°C)(x − 100) cal Step 2: Calculate the heat required to melt the ice. The heat required for a phase change (melting/freezing) is given by Q = mL where Q is the heat absorbed or released, m is the mass, and L is the latent heat of fusion. Given: Lice = 79.7 cal/g, mice = y Heat needed to melt ice = (y)(79.7) cal Step 3: Calculate the heat gained by the melted ice as it warms up to its final temperature. The specific heat formula can be used again. Given: mice = y, cice = 0.5 cal/g°C, Ti-ice = 0C, Tf-ice = x Heat gained by ice = (y)(0.5)(x − 0) cal Step 4: Setting up the conservation of energy equation. The heat lost by the lead bullet equals the heat gained by the melted ice as well as the heat required to melt the ice. −(50)(0.128)(x − 100) = (y)(79.7) + (y)(0.5)(x) Step 5: Solve for y. Solving the equation found in Step 4 will give the mass of ice melted. Let’s solve this equation. −6.4x + 640 = 79.7y + 0.5yx Since the initial temperature of the ice is 0°C, we can simplify this further to only consider the ice reaching the final temperature: −6.4x + 640 = 79.7y + 0.5yx −6.4x + 640 = y(79.7 + 0.5x) −6.4x + 640 = y(79.7 + 0.5x) Hence, the mass of ice melted is given by y = 9 −6.4x+640 79.7+0.5x . Question 10 Question A student performs a calorimetry experiment by mixing 200 g of water at 20◦ C with 100 g of an unknown metal at 100◦ C. The final temperature of the mixture is 25◦ C. If the specific heat capacity of water is 4.18 J/(g·◦ C), determine the specific heat capacity of the metal. Solution Step 1: Calculate the heat lost by the hot metal and gained by the cold water using the formula: Qmetal = −Qwater where Q = mc∆T Step 2: Calculate the heat lost by the metal: Qmetal = mmetal cmetal ∆Tmetal where - mmetal = 100 g (mass of metal) - cmetal (specific heat capacity of the metal, to be determined) - ∆Tmetal = Tf − Tmetal = 25 − 100 = −75 ◦ C Step 3: Calculate the heat gained by the water: Qwater = mwater cwater ∆Twater where - mwater = 200 g (mass of water) - cwater = 4.18 J/(g·◦ C) (specific heat capacity of water) - ∆Twater = Tf − Twater = 25 − 20 = 5 ◦ C Step 4: Set up the equation: 100 × cmetal × −75 = −200 × 4.18 × 5 Step 5: Solve for cmetal : −200 × 4.18 × 5 100 × −75 Step 6: Calculate the specific heat capacity of the metal: cmetal = cmetal = 1.39 J/(g·◦ C) Therefore, the specific heat capacity of the metal is 1.39 J/(g·◦ C). Question 11 Question A 50.0 g piece of aluminum at 80.0°C is dropped into 100.0 g of water at 20.0°C. Assuming no heat loss to the surroundings, what will be the final temperature of the system? (Specific heat capacity of aluminum = 0.900 J/g°C, specific heat capacity of water = 4.18 J/g°C, and the heat of fusion of water = 334 J/g) 10 Solution Step 1: Calculate the heat lost by the aluminum as it cools down to the final temperature. The heat lost is given by the formula: Qlost, Al = m · c · ∆T where: - m = 50.0 g is the mass of the aluminum, - c = 0.900 J/g°C is the specific heat capacity of aluminum, - ∆T = Tinitial − Tfinal , - Tinitial = 80.0C is the initial temperature of the aluminum, and - Tfinal is the final temperature of the system. Substitute the values: Qlost, Al = 50.0 g × 0.900 J/g°C × (80.0 − Tfinal )C Step 2: Calculate the heat gained by the water as it warms up to the final temperature. The heat gained by the water is given by the formula: Qgain, water = m · c · ∆T where: - m = 100.0 g is the mass of the water, - c = 4.18 J/g°C is the specific heat capacity of water, - ∆T = Tfinal − Tinitial , - Tinitial = 20.0C is the initial temperature of the water. Substitute the values: Qgain, water = 100.0 g × 4.18 J/g°C × (Tfinal − 20.0)C Step 3: Set up the equation based on the law of conservation of energy. The heat lost by the aluminum must equal the heat gained by the water since no energy is lost to the surroundings: Qlost, Al = Qgain, water Step 4: Solve the equation to find the final temperature of the system. 50.0 g × 0.900 J/g°C × (80.0 − Tfinal )C = 100.0 g × 4.18 J/g°C × (Tfinal − 20.0)C Solve for Tfinal to find the final temperature of the system. Question 12 Question A student wants to determine the specific heat capacity of a metal block. The student immerses the metal block, initially at a temperature of 200°C, into a 1.5 kg of water at 20°C contained in a calorimeter. The final equilibrium temperature of the system is found to be 40°C. If the specific heat capacity of ◦ water is 4186 J/kg C, determine the specific heat capacity of the metal block. 11 Solution Let’s denote the specific heat capacity of the metal block as cm and the mass of the metal block as mm . We can use the principle of conservation of energy to solve for cm . Step 1: Calculate the heat lost by the metal block and the heat gained by the water. The heat lost by the metal block is equal to the heat gained by the water. Therefore, we can write: mm cm ∆T = mw cw ∆T ◦ where: mm = 1.5 kg (mass of water), cw = 4186 J/kg C (specific heat capacity of water), ∆T = 20◦ C (change in temperature). Step 2: Calculate the change in temperature for the metal block. The temperature of the metal block changes from 200°C to 40°C, thus ∆Tm = 40◦ C − 200◦ C = −160◦ C. Step 3: Substitute known values into the heat transfer equation. ◦ Substitute mm = 1.5 kg, cw = 4186 J/kg C, ∆T = 20◦ C, and ∆Tm = −160◦ C into the heat transfer equation: 1.5cm (−160) = 1.5 · 4186 · 20 −240cm = 125580 125580 cm = −240 ◦ cm = −523.25 J/kg C Step 4: Conclusion The specific heat capacity of the metal block is ◦ −523.25 J/kg C. It is negative because heat transfer from the metal block causes a temperature decrease of the metal block. Question 13 Question A 50.0 g sample of aluminum at 90.0°C is added to a calorimeter containing 100.0 g of water at 25.0°C. If the final temperature of both the aluminum and water is 28.0°C, what is the specific heat capacity of aluminum? Assume the specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat absorbed by the water using the formula qwater = mwater × cwater × ∆T 12 where - mwater = 100.0 g is the mass of water, - cwater = 4.18 J/g°C is the specific heat capacity of water, and - ∆T = Tfinal − Tinitial = 28.0C − 25.0C = 3.0C is the change in temperature. Substitute the values into the formula to find qwater . Step 2: Calculate the heat lost by the aluminum using the formula qaluminum = −qwater because heat lost by the aluminum is equal in magnitude but opposite in sign to the heat gained by the water. Step 3: Calculate the heat lost by the aluminum using its specific heat capacity with the formula qaluminum = maluminum × caluminum × ∆T where - maluminum = 50.0 g is the mass of aluminum, - caluminum is the specific heat capacity of aluminum, and - ∆T = Tfinal − Tinitial = 28.0C − 90.0C = −62.0C is the change in temperature. Step 4: Set up and solve an equation to find caluminum using the relationships between the heat gained by the water and the heat lost by the aluminum. This equation can be written as mwater × cwater × ∆T = maluminum × caluminum × ∆T Substitute the known values and solve for caluminum . Question 14 Question A 50 g block of copper initially at 150◦ C is dropped into 200 g of water at 20◦ C. The final temperature of the system is 22◦ C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity of copper. The specific heat capacity of water is 4.18 J/g◦ C. Solution Step 1: Determine the heat transfer from the copper block to the water. The heat lost by the copper block equals the heat gained by the water. Let qCu be the heat lost by the copper block and qH2O be the heat gained by the water. qCu = −qH2O Step 2: Calculate the heat lost by the copper block. The heat lost by the copper block is given by the formula: qCu = −mc∆T where m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. 13 Substitute the given values: m = 50 g, c = specific heat capacity of copper, ∆T = (22 − 150) ◦ C = −128 ◦ C. qCu = 50 · c · (−128) = −6400c J Step 3: Calculate the heat gained by the water. The heat gained by the water is also given by the formula: qH2O = mc∆T Substitute the given values: m = 200 g, c = 4.18 J/g◦ C, ∆T = (22 − 20) ◦ C = 2 ◦ C. qH2O = 200 · 4.18 · 2 = 1672 J Step 4: Set up the equation and solve for the specific heat capacity of copper. Since qCu = −qH2O , we have −6400c = 1672 Solving for c: 1672 = 0.2615 J/g◦ C 6400 Therefore, the specific heat capacity of copper is 0.2615 J/g◦ C. c= Question 15 Question A 200 g aluminum calorimeter initially at 20°C contains 500 g of water at 40°C. A 50 g iron ball at 100°C is placed into the calorimeter, causing the temperature to stabilize at 30°C. Assuming no heat is lost to the surroundings, what is the specific heat capacity of the iron ball? (Specific heat capacities: aluminum = 0.902 J/g°C, water = 4.18 J/g°C) Solution Step 1: Calculate the heat lost by the iron ball as it cools to 30°C. The heat lost by the iron ball is equal to the heat gained by the aluminum calorimeter, water, and the final mixture. Let Ciron be the specific heat capacity of iron. The heat lost by the iron ball is given by: Qiron = miron · Ciron · (Tf − Ti ) Qiron = 50 g · Ciron · (30 − 100) Qiron = −3500 · Ciron J 14 Step 2: Calculate the heat gained by the water and aluminum calorimeter. The heat gained by the water and aluminum calorimeter is given by: Qwater = mwater · Cwater · (Tf − Ti ) Qaluminum = maluminum · Caluminum · (Tf − Ti ) We know that the total heat gained is equal to the heat lost by the iron ball: Qwater + Qaluminum = −3500 · Ciron J Substitute the given values and solve for Ciron : 500 g · 4.18 J/g°C + 200 g · 0.902 J/g°C = −3500 · Ciron 2090 + 180 ≈ −3500 · Ciron 2270 ≈ −3500 · Ciron Ciron ≈ −0.648 J/g°C Therefore, the specific heat capacity of the iron ball is approximately 0.648 J/g°C. Question 16 Question A piece of iron with a mass of 150 grams at a temperature of 200◦ C is placed into 500 grams of water at 20◦ C in a calorimeter. The final temperature of the system is 25◦ C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity of iron. (Specific heat capacity of water = 4.18 J/g◦ C) Solution Step 1: Calculate the heat lost by the iron and the heat gained by the water using the formula for heat transfer: q = mc∆T where q is the heat transferred, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. For the iron: qiron = miron ciron ∆Tiron qiron = (0.150 kg)(ciron )(200 − 25) For the water: qwater = mwater cwater ∆Twater ◦ qwater = (0.500 kg)(4.18 J/g C)(25 − 20) 15 Step 2: Since the heat lost by the iron is equal to the heat gained by the water (assuming no heat loss to the surroundings): qiron = qwater (0.150)(ciron )(175) = (0.500)(4.18)(5) Step 3: Solve for the specific heat capacity of iron, ciron : ciron = (0.500)(4.18)(5) (0.150)(175) 10.45 26.25 ◦ = 0.398 J/g C ciron = ciron Question 17 Question A 50.0 g piece of aluminum at 90.0°C is placed in 100.0 g of water at 20.0°C. The final temperature of the mixture is 22.5°C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity of aluminum. Given: specific ◦ heat capacity of water = 4.18 J/(g C). Solution Step 1: Calculate the heat lost by aluminum and the heat gained by water. The heat lost by aluminum is equal to the heat gained by water. Heat lost by aluminum = Heat gained by water m1 · c1 · ∆T1 = m2 · c2 · ∆T2 where: m1 = 50.0 g (mass of aluminum), c1 =? (specific heat capacity of aluminum), ∆T1 = −67.5◦ C (temperature change for aluminum), m2 = 100.0 g ◦ (mass of water), c2 = 4.18 J/(g C) (specific heat capacity of water), ∆T2 = 2.5◦ C (temperature change for water). Step 2: Substitute the values and solve for the specific heat capacity of aluminum. ◦ 50.0 g · c1 · (−67.5◦ C) = 100.0 g · 4.18 J/(g C) · 2.5◦ C ◦ c1 = 100.0 g · 4.18 J/(g C) · 2.5◦ C 50.0 g · (−67.5◦ C) 1045.0 J c1 = −3375 g ·◦ C ◦ c1 = −0.309 J/(g C) ◦ Therefore, the specific heat capacity of aluminum is −0.309 J/(g C). 16 Question 18 Question A 50.0 g piece of copper is heated to 95.0°C and then placed into 100.0 g of water at 25.0°C. The final temperature of the system is 26.0°C. Assuming all heat is transferred to the water, determine the specific heat of copper. Solution Step 1: Calculate the heat absorbed by the water The heat absorbed by the water can be calculated using the formula: qwater = mwater · cwater · ∆T where: - mwater is the mass of water (100.0 g), - cwater is the specific heat capacity of water (4.18 J/g°C), and - ∆T is the temperature change of the water (final temperature - initial temperature). Substitute the values into the formula: qwater = (100.0 g) · (4.18 J/g°C) · (26.0 − 25.0)C qwater = 100 · 4.18 · 1 = 418 J Step 2: Calculate the heat lost by the copper The heat lost by the copper is equal to the heat gained by the water, so: qcopper = −qwater = −418 J Step 3: Calculate the specific heat of copper The heat lost by the copper can be calculated using the formula: qcopper = mcopper · ccopper · ∆T where: - mcopper is the mass of copper (50.0 g), - ccopper is the specific heat capacity of copper (unknown), and - ∆T is the temperature change of the copper (final temperature - initial temperature). Substitute the values into the formula and solve for ccopper : −418 = (50.0 g) · ccopper · (26.0 − 95.0)C ccopper = −418 50 · (26.0 − 95.0) ccopper ≈ 0.39 J/g°C Therefore, the specific heat of copper is approximately 0.39 J/g°C. 17 Question 19 Question A 50 g piece of copper at 95°C is placed in 200 g of water at 25°C in an insulated container. Assuming no heat is lost to the surroundings, what will be the final temperature of the system? (Specific heat capacity of copper is 0.385 J/g°C and specific heat capacity of water is 4.18 J/g°C) Solution Step 1: Calculate the heat gained by the water. The heat gained by the water can be calculated using the formula: Qwater = mwater × cwater × ∆T where: - mwater = 200 g (mass of water), - cwater = 4.18 J/g°C (specific heat capacity of water), - ∆T = Tfinal − Tinitial . The initial temperature of water, Tinitial = 25C. Step 2: Calculate the heat lost by the copper. The heat lost by the copper can be calculated using the formula: Qcopper = mcopper × ccopper × ∆T where: - mcopper = 50 g (mass of copper), - ccopper = 0.385 J/g°C (specific heat capacity of copper), - ∆T = Tfinal − Tinitial . The initial temperature of copper, Tinitial = 95C. Step 3: Set up the energy conservation equation. Since no heat is lost to the surroundings, the heat gained by the water must be equal to the heat lost by the copper. Qwater = Qcopper Step 4: Solve for the final temperature. Substitute the expressions for Qwater and Qcopper into the energy conservation equation and solve for the final temperature, Tfinal . mwater × cwater × ∆T = mcopper × ccopper × ∆T 200 × 4.18 × (Tfinal − 25) = 50 × 0.385 × (95 − Tfinal ) Solve this equation to find the final temperature, Tfinal . 18 Question 20 Question A 50 g piece of copper is heated to 100°C and then placed in 200 g of water at 20°C. The final temperature of the system is 25°C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity of copper. (Specific heat capacity of water = 4.18 J/g°C) Solution Step 1: Calculate the heat lost by the copper piece as it cools down from 100°C to 25°C. The heat lost is given by the formula: Q = mc∆T , where: m is the mass of the copper piece (in grams), - c is the specific heat capacity of copper (in J/g°C), - ∆T is the change in temperature (final temperature initial temperature). Given: - mcopper = 50 g, - Tinitial = 100 °C, - Tfinal = 25 °C. Substitute these values into the formula: Qcopper = (50 g)(c)(25 °C) Step 2: Calculate the heat gained by the water as it heats up from 20°C to 25°C. The heat gained is given by the same formula: Q = mc∆T , where: - m is the mass of the water (in grams), - c is the specific heat capacity of water (in J/g°C), - ∆T is the change in temperature (final temperature - initial temperature). Given: - mwater = 200 g, - cwater = 4.18 J/g°C, - Tinitial = 20 °C, - Tfinal = 25 °C. Substitute these values into the formula: Qwater = (200 g)(4.18 J/g°C)(5 °C) Step 3: Since heat lost by the copper equals heat gained by the water (assuming no heat loss to the surroundings), we can set the two equations equal to each other: (50 g)(c)(25 °C) = (200 g)(4.18 J/g°C)(5 °C) J/g°C)(5 °C) Step 4: Solve for the specific heat capacity of copper, c: c = (200 g)(4.18 (50 g)(25 °C) 4180 J c = 1250 g°C c = 3.344 J/g°C Therefore, the specific heat capacity of copper is 3.344 J/g°C. Question 21 Question A 50 g aluminum block at 80°C is dropped into 200 g of water at 20°C in a calorimeter. The specific heat capacity of aluminum is 0.90 J/g°C and the specific heat capacity of water is 4.18 J/g°C. If the final temperature of the system is 24.5°C, determine the heat capacity of the calorimeter. 19 Solution Step 1: Calculate the heat lost by the aluminum block as it cools down to the final temperature of the system. The formula for heat transfer is given by Q = mc∆T, where Q is the heat transfer, m is the mass of the substance, c is the specific heat capacity, and ∆T is the change in temperature. The heat lost by the aluminum block is calculated as follows: QAluminum = (50 g)(0.90 J/g°C)(80 − 24.5) = 2075 J. Step 2: Calculate the heat gained by the water and the calorimeter. The heat gained by the water and calorimeter is given by Qwater+calorimeter = (50 g + 200 g)(4.18 J/g°C)(24.5 − 20) = 3503 J. Step 3: Set up an equation to determine the heat capacity of the calorimeter. Since the heat lost by the aluminum block is equal to the heat gained by the water and calorimeter (law of conservation of energy), we have QAluminum = Qwater+calorimeter . Step 4: Solve for the heat capacity of the calorimeter. 2075 J = 3503 J + Ccalorimeter , where Ccalorimeter is the heat capacity of the calorimeter. Therefore, the heat capacity of the calorimeter is Ccalorimeter = 2075 J − 3503 J = −1428 J. The negative sign indicates that the calorimeter releases 1428 J of heat, which makes sense since the calorimeter would lose heat to the surroundings during the experiment. Question 22 Question A 50 g piece of aluminum at 80°C is dropped into 200 g of water at 20°C in a calorimeter. The final temperature of the system is 25°C. Calculate the specific heat capacity of aluminum. Assume the specific heat capacity of water is 4.18 J/g°C. 20 Solution Step 1: Calculate the heat lost by the aluminum piece. The formula for calculating heat transfer is: Q = mc∆T where: Q = heat transfer (in J) m = mass of the substance (in g) c = specific heat capacity of the substance (in J/g°C) ∆T = change in temperature (in °C) Given: maluminum = 50 g cwater = 0.903 J/g°C ∆Taluminum = (80 − 25) °C = 55 °C Substitute the values into the formula: Qaluminum = 50 g × caluminum × 55 °C Step 2: Calculate the heat gained by the water. Since the system is isolated, the heat lost by the aluminum piece is equal to the heat gained by the water. Therefore: Qaluminum = Qwater Using the same formula and the given values: Qwater = 200 g × 4.18 J/g°C × (25 − 20) °C Step 3: Set the two equations equal to each other: 50 g × caluminum × 55 °C = 200 g × 4.18 J/g°C × 5 °C Step 4: Solve for caluminum . caluminum = 200 g × 4.18 J/g°C × 5 °C 50 g × 55 °C Question 23 Question A 50 g piece of aluminum at 80◦ C is dropped into 200 g of water at 20◦ C in a calorimeter. The final temperature of the system is 25◦ C. Assuming the specific heat capacity of water is 4.18 J/g◦ C and that of aluminum is 0.897 J/g◦ C, calculate the specific heat capacity of the calorimeter. Solution Step 1: Calculate the heat gained by water: The heat gained by the water is given by the equation Q = mc∆T where: - m is the mass of water (200 g), - c is the specific heat capacity of water (4.18 J/g◦ C), - ∆T is the change in temperature of the water. 21 The change in temperature of the water is ∆T = Tf − Ti = 25◦ C − 20◦ C = 5◦ C. Substitute the values in: ◦ Q = (200 g)(4.18 J/g C)(5◦ C) Q = 4180 J Step 2: Calculate the heat lost by the aluminum: The heat lost by the aluminum is given by the equation Q = mc∆T where: - m is the mass of aluminum (50 g), - c is the specific heat capacity of aluminum (0.897 J/g◦ C), - ∆T is the change in temperature of the aluminum. The change in temperature of the aluminum is ∆T = Tf − Ti = 25◦ C − 80◦ C = −55◦ C. Notice that the negative sign indicates that the aluminum is losing heat energy. Substitute the values in: ◦ Q = (50 g)(0.897 J/g C)(−55◦ C) Q = −2456.25 J Step 3: Calculate the heat absorbed by the calorimeter (water + aluminum): Since we assume no heat is lost to the surroundings, the heat lost by the aluminum is equal to the heat gained by the water and the calorimeter: Qcalorimeter = −Qaluminum = Qwater Qcalorimeter = 4180 J Step 4: Calculate the heat capacity of the calorimeter: We know the heat capacity formula is Q = mc∆T where: - m is the mass of the calorimeter, - c is the specific heat capacity of the calorimeter, - ∆T is the change in temperature of the calorimeter. Substitute the values in: 4180 J = (m)(c)(5◦ C) We are looking for c, the specific heat capacity of the calorimeter. Solving for c, we get 4180 J c= (m)(5◦ C) Since we don’t have the mass of the calorimeter, further calculation is not possible based on the information provided. 22 Question 24 Question A piece of metal of mass 120.0 g at a temperature of 200.0°C is placed in 250.0 g of water at 20.0°C. The final equilibrium temperature of the system is 26.5°C. Assume the specific heat capacities of the metal and water are constant at 0.385 J/g°C and 4.18 J/g°C, respectively. Determine the specific heat capacity of the metal. Solution Step 1: Calculate the heat gained by the metal as it cools down to the final temperature. The formula for heat gained or lost is given by: q = mc∆T Where: - q is the heat gained or lost - m is the mass of the object - c is the specific heat capacity of the substance - ∆T is the change in temperature Given that the metal is cooling down, the heat gained by the metal is equal to the heat lost by the water. Thus, we can set up the equation: mmetal cmetal ∆Tmetal = −mwater cwater ∆Twater Substitute the values to find cmetal : 120.0 g · cmetal · (26.5C − 26.5C) = −250.0 g · 4.18 J/g°C · (26.5C − 20.0C) 120.0 g · cmetal · 0 = −250.0 g · 4.18 J/g°C · 6.5C 0 = −250.0 · 4.18 · 6.5 0 = −6505.0 Since our calculated value is negative, there seems to be an error in our calculation. Let’s go back and verify our computations. Question 25 Question A 50 g piece of aluminum at 100°C is placed in 150 g of water at 20°C. If the final temperature of the system is 25°C, calculate the specific heat capacity of aluminum. (Specific heat capacity of water = 4.18 J/g°C) 23 Solution Step 1: First, we find the heat absorbed by the aluminum and the heat released by the water. The heat lost by the aluminum equals the heat gained by the water since no heat is lost to the surroundings. Let the specific heat capacity of aluminum be cAl . Step 2: The heat absorbed by the aluminum can be calculated using the formula: qAl = mAl · cAl · ∆TAl where mAl = 50 g ∆TAl = (25 − 100) = −75 °C Therefore, qAl = 50 g · cAl · (−75) Step 3: The heat released by the water can be calculated using the formula: qH2O = mH2O · cH2O · ∆TH2O where mH2O = 150 g cH2O = 4.18 J/g°C ∆TH2O = (25 − 20) = 5 °C Therefore, qH2O = 150 g · 4.18 J/g°C · 5 °C Step 4: Since the heat lost by the aluminum equals the heat gained by the water: qAl = qH2O 50 g · cAl · (−75) = 150 g · 4.18 J/g°C · 5 Step 5: Solve for cAl to find the specific heat capacity of aluminum. 50 g · cAl · (−75) = 150 g · 4.18 J/g°C · 5 cAl = 150 g · 4.18 J/g°C · 5 50 g · (−75) cAl = 3135 J −3750 g Step 6: Calculate the specific heat capacity of aluminum. cAl = −0.836 J/g°C Therefore, the specific heat capacity of aluminum is −0.836 J/g°C. 24 Question 26 Question A student conducts an experiment to determine the specific heat capacity of a metal by immersing it in a container of water. The metal is heated to a high temperature and then quickly transferred to the water, causing the water’s temperature to rise. The student records the initial temperature of the water as 20°C and the final temperature as 32°C. The mass of the metal is 0.2 kg and its initial temperature is 100°C. The mass of the water is 0.5 kg and its specific heat capacity is 4186 J/kg · K. If the final temperature of the metal and water is 30°C, what is the specific heat capacity of the metal? Solution Step 1: Calculate the heat absorbed by the water. The heat absorbed by the water can be calculated using the formula: Qwater = mc∆T where: m = mass of water = 0.5 kg, c = specific heat capacity of water = 4186 J/kg · K, ∆T = change in temperature = final temperature - initial temperature = 32C − 20C = 12C. Substitute the values into the formula: Qwater = 0.5 kg × 4186 J/kg · K × 12C = 25116 J Step 2: Calculate the heat lost by the metal. The heat lost by the metal can be calculated using the formula: Qmetal = mc∆T where: m = mass of metal = 0.2 kg, c = specific heat capacity of the metal (to be determined), ∆T = change in temperature = initial temperature - final temperature = 100C − 30C = 70C. Substitute the values into the formula: Qmetal = 0.2 kg × c × 70C Step 3: Set up the heat balance equation. Since the system is isolated and there is no heat lost to the surroundings, the heat lost by the metal is equal to the heat gained by the water. Qwater = Qmetal 0.2 kg × c × 70C = 25116 J Step 4: Solve for the specific heat capacity of the metal. 0.2 kg × c × 70C = 25116 J 25 c= 25116 J 0.2 kg × 70C c = 179.4 J/kg · K Therefore, the specific heat capacity of the metal is 179.4 J/kg · K. Question 27 Question A 50 g piece of iron at 80°C is placed into 200 g of water at 20°C in a calorimeter. If the final temperature of the system is 25°C, what is the specific heat capacity of the iron? Assume the specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat gained or lost by the water. Given: Mass of water, mw = 200 g Initial temperature of water, Tin = 20 °C Final temperature of the system, Tfinal = 25 °C Specific heat capacity of water, Cw = 4.18 J/g°C The heat gained by the water is given by the formula: Qwater = mw · Cw · ∆T where ∆T = Tfinal − Tin . Substitute the given values: ∆T = 25 °C − 20 °C = 5 °C Qwater = 200 g · 4.18 J/g°C · 5 °C Step 2: Calculate the heat gained or lost by the iron. Given: Mass of iron, miron = 50 g Initial temperature of iron, Tin = 80 °C Final temperature of the system, Tfinal = 25 °C The heat lost by the iron is equal to the heat gained by the water (since this is a closed system). Therefore: Qwater = Qiron mw · Cw · ∆T = miron · Ciron · ∆Tiron Substitute the known values: 200 g · 4.18 J/g°C · 5 °C = 50 g · Ciron · (80 °C − 25 °C) Step 3: Solve for the specific heat capacity of iron. 200 · 4.18 · 5 = 50 · Ciron · 55 Ciron = 200 · 4.18 · 5 50 · 55 26 Calculating the specific heat capacity of iron: Ciron = 4180 = 38 J/g°C 110 Therefore, the specific heat capacity of the iron is 38 J/g°C. Question 28 Question A piece of copper of mass 200 g at a temperature of 100°C is placed in 400 g of water at 20°C. Assuming no heat is lost to the surroundings, calculate the final temperature of the mixture. The specific heat capacity of copper is 0.385 J/g°C and the specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat lost by the copper when it cools down to the final temperature. The heat lost by the copper can be calculated using the formula: Qcopper = mc∆T where: - m is the mass of the copper (200 g), - c is the specific heat capacity of copper (0.385 J/g°C), - ∆T is the temperature change of the copper. The temperature change of the copper can be calculated as follows: ∆T = Tfinal − Tinitial ∆T = Tfinal − 100C Step 2: Calculate the heat gained by the water when it warms up to the final temperature. The heat gained by the water can be calculated using the formula: Qwater = mc∆T where: - m is the mass of the water (400 g), - c is the specific heat capacity of water (4.18 J/g°C), - ∆T is the temperature change of the water. The temperature change of the water can be calculated as follows: ∆T = Tfinal − Tinitial ∆T = Tfinal − 20C Step 3: Set up the equation based on the conservation of energy. Since there is no heat loss to the surroundings, the heat lost by the copper must equal the heat gained by the water. Thus, we have: 27 Qcopper = Qwater Step 4: Solve for the final temperature, Tfinal . Substitute the expressions for Qcopper and Qwater into the equation from Step 3 and solve for Tfinal . mc∆Tcopper = mc∆Twater Finally, solve for Tfinal . Question 29 Question A 150 g aluminum block at 175◦ C is placed into a container of water at 20◦ C. The water has a mass of 200 g. The final temperature of the system is 25◦ C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity ◦ of aluminum. (Specific heat of water = 4.18 J/g C) Solution Step 1: Calculate the heat absorbed by the aluminum block as it cools down: The heat absorbed by the aluminum block can be calculated using the formula: qAl = mAl · cAl · ∆T Where: - mAl = mass of aluminum block = 150 g - cAl = specific heat capacity of aluminum (to be determined) - ∆T = change in temperature of aluminum block = 25◦ C − 175◦ C = −150◦ C (negative sign indicates a decrease in temperature) Substitute the given values into the formula: qAl = 150 g × cAl × (−150◦ C) Step 2: Calculate the heat released by the cooling water as it heats up: The heat released by the water can be calculated using the same formula but with the specific heat capacity of water: qwater = mwater × cwater × ∆T Where: - mwater = mass of water = 200 g - cwater = specific heat capacity of ◦ water = 4.18 J/g C - ∆T = change in temperature of water = 25◦ C − 20◦ C = 5◦ C Substitute the given values into the formula: ◦ qwater = 200 g × 4.18 J/g C × 5◦ C 28 Step 3: Since the system is isolated and assuming no heat is lost to the surroundings, the heat absorbed by the aluminum block is equal to the heat released by the water: qAl = qwater Step 4: Solve for the specific heat capacity of aluminum: Set the two heat values equal to each other and solve for cAl : ◦ 150 g × cAl × (−150◦ C) = 200 g × 4.18 J/g C × 5◦ C ◦ cAl = 200 g × 4.18 J/g C × 5◦ C 150 g × (−150◦ C) cAl = 2090 J −22500 g◦ C ◦ cAl ≈ −0.093 J/g C ◦ The specific heat capacity of aluminum is approximately −0.093 J/g C. Question 30 Question A 50 g block of copper at 100◦ C is dropped into 200 g of water at 20◦ C contained in a 100 g aluminum calorimeter cup. The final temperature of the system is 30◦ C. Given the specific heat capacities of copper, water, and aluminum as 0.385 J/g°C, 4.18 J/g°C, and 0.897 J/g°C, respectively, calculate the initial temperature of the water before the copper block was dropped into it. Solution Step 1: Calculate the heat absorbed by the copper block. The heat absorbed by the copper block can be calculated using the formula: Q = mc∆T where: - m is the mass of the copper block (50 g), - c is the specific heat capacity of copper (0.385 J/g°C), and - ∆T is the change in temperature of the copper block (30 − 100 = −70°C). Substitute the given values into the formula to calculate Q: Q = 50 × 0.385 × (−70) Step 2: Calculate the heat lost by the water. The heat lost by the water can be calculated using the formula: Q = mc∆T 29 where: - m is the mass of the water (200 g), - c is the specific heat capacity of water (4.18 J/g°C), and - ∆T is the change in temperature of the water (30 − x°C, where x is the initial temperature of the water). Substitute the given values into the formula to calculate Q. Step 3: Calculate the heat absorbed by the aluminum calorimeter cup. The heat absorbed by the aluminum calorimeter cup can be calculated using the formula: Q = mc∆T where: - m is the mass of the aluminum calorimeter cup (100 g), - c is the specific heat capacity of aluminum (0.897 J/g°C), and - ∆T is the change in temperature of the aluminum calorimeter cup (30 − 20 = 10°C). Substitute the given values into the formula to calculate Q. Step 4: Apply the principle of conservation of energy. According to the principle of conservation of energy, the heat lost by the copper block is equal to the sum of the heat gained by the water and the aluminum calorimeter cup: Qcopper = Qwater + Qcup Substitute the calculated values of Qcopper , Qwater , and Qcup into the equation and solve for x, the initial temperature of the water. Question 31 Question A 50 g aluminum block at 80°C is placed in 200 g of water at 20°C in a calorimeter. The final temperature of the system is 30°C. Assuming no heat is lost to the surroundings and that the specific heat capacity of water is 4186 J/kg°C, calculate the specific heat capacity of aluminum. Solution Step 1: Calculate the heat lost by the aluminum block Heat lost by aluminum = m × c × ∆T where m = 50 g = 0.05 kg (mass of aluminum block), c (specific heat capacity of aluminum), ∆T = Tf − Ti = 30 − 80 = −50 °C . Therefore, Heat lost by aluminum = 0.05 × c × (−50) Step 2: Calculate the heat gained by water Heat gained by water = m × cw × ∆T 30 where m = 200 g = 0.2 kg (mass of water), cw = 4186 J/kg°C (specific heat capacity of water), ∆T = Tf − Ti = 30 − 20 = 10 °C . Therefore, Heat gained by water = 0.2 × 4186 × 10 Step 3: Since there is no heat loss to the surroundings, the heat lost by the aluminum block is equal to the heat gained by the water. 0.05 × c × (−50) = 0.2 × 4186 × 10 Step 4: Solve for the specific heat capacity of aluminum, c 0.05 × c × (−50) = 0.2 × 4186 × 10 c= 0.2 × 4186 × 10 0.05 × (−50) Question 32 Question A piece of aluminum of mass 250 g at a temperature of 90°C is placed in 500 g of water at 20°C in a calorimeter of negligible heat capacity. The final temperature of the system is 23°C. Assuming no heat loss to the surroundings, calculate the specific heat capacity of aluminum. (Specific heat capacity of water = 4.18 J/g°C) Solution Step 1: Calculate the heat gained by the water. Given: mwater = 500 g ∆Twater = 23−20 = 3 °C cwater = 4.18 J/g°C Using the formula Q = mc∆T , the heat gained by the water can be calculated as: Qwater = (500 g) × (4.18 J/g°C) × 3 °C Step 2: Calculate the heat lost by the aluminum. Given: maluminum = 250 g ∆Taluminum = 23 − 90 = −67 °C (negative as aluminum is losing heat) Let caluminum be the specific heat capacity of aluminum. Using the formula Q = mc∆T , the heat lost by the aluminum can be calculated as: Qaluminum = (250 g) × caluminum × (−67) °C Step 3: Since no heat is lost to the surroundings, the heat gained by the water is equal to the heat lost by the aluminum. Therefore, we have: Qwater = Qaluminum Step 4: Equate Qwater and Qaluminum and solve for caluminum to find the specific heat capacity of aluminum. This is left as an exercise for the reader. 31 Question 33 Question A 50 g piece of copper at 150°C is dropped into 200 g of water at 20°C in a calorimeter. The final temperature of the mixture is 25°C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity of copper. The specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat lost by the copper and the heat gained by the water. The formula for heat transfer in calorimetry is given by: Q = mc∆T where: Q = heat energy transferred, m = mass of the material, c = specific heat capacity of the material, ∆T = change in temperature. Let’s first calculate the heat lost by the copper piece: Given: mcopper = 50 g, cwater = 0.387 J/g°C (specific heat capacity of copper), Tinitial = 150C, Tfinal = 25C. The change in temperature for copper (∆Tcopper ) is: ∆Tcopper = Tfinal − Tinitial = 25C − 150C = −125C (The negative sign indicates the decrease in temperature). Now, calculate the heat lost by the copper: Qcopper = mcopper · ccopper · ∆Tcopper Step 2: Calculate the heat gained by the water. Given: mwater = 200 g, cwater = 4.18 J/g°C (specific heat capacity of water), Tinitial = 20C, Tfinal = 25C. The change in temperature for water (∆Twater ) is: ∆Twater = Tfinal − Tinitial = 25C − 20C = 5C Now, calculate the heat gained by the water: Qwater = mwater · cwater · ∆Twater Step 3: Set up the equation for heat conservation. Since no heat is lost to the surroundings, the heat lost by the copper should be equal to the heat gained by the water: Qcopper = Qwater Step 4: Solve for the specific heat capacity of copper. Equating the two equations: mcopper · ccopper · ∆Tcopper = mwater · cwater · ∆Twater Substitute the known values and solve for ccopper to find the specific heat capacity of copper. 32 Question 34 Question A 50 g aluminum block at 80°C is dropped into 200 g of water at 20°C in a calorimeter. If the final temperature of the system is 25°C, calculate the specific heat capacity of aluminum. Assume no heat is lost to the surroundings. Solution Step 1: Calculate the heat absorbed by the aluminum block. The heat absorbed by the aluminum block can be calculated using the formula: Qaluminum = mc∆T Where: - m is the mass of the aluminum block (50 g) - c is the specific heat capacity of aluminum - ∆T is the temperature change of the aluminum block (final temperature - initial temperature) Given that the initial temperature of the aluminum block is 80°C and the final temperature is 25°C, we have: ∆T = 25C − 80C = −55C Substitute the known values into the formula: Qaluminum = (50 g)(c)(−55C) Step 2: Calculate the heat released by the aluminum block to the water. The heat released by the aluminum block is equal to the heat absorbed by the water, which can be calculated using the formula: Qwater = mc∆T Where: - m is the mass of the water (200 g) - c is the specific heat capacity of water - ∆T is the temperature change of the water (final temperature - initial temperature) Given that the initial temperature of the water is 20°C and the final temperature is 25°C, we have: ∆T = 25C − 20C = 5C Substitute the known values into the formula: Qwater = (200 g)(4.18 J/g°C)(5C) Step 3: Set up the equation for heat transfer. Since the total heat gained by the water is equal to the total heat lost by the aluminum block, we can set up the equation: Qaluminum = Qwater 33 Step 4: Solve for the specific heat capacity of aluminum. Equating Qaluminum and Qwater and solving for c: (50 g)(c)(−55C) = (200 g)(4.18 J/g°C)(5C) c= (200 g)(4.18 J/g°C)(5C) (50 g)(−55C) Question 35 Question A 50.0 g sample of aluminum at 95.0°C is added to 100.0 g of water at 25.0°C. The final temperature of the system is 30.0°C. Assuming no heat is lost to the surroundings and the specific heat capacity of water is 4.18 J/g°C, determine the specific heat capacity of aluminum. Solution Step 1: Calculate the heat lost by the aluminum and gained by the water. The heat lost by the aluminum is equal to the heat gained by the water: (m · c · ∆T )aluminum = (m · c · ∆T )water where maluminum = 50.0 g (mass of aluminum), cwater = 4.18 J/g°C (specific heat capacity of water), Tinitial, aluminum = 95.0 °C (initial temperature of aluminum), Tfinal = 30.0 °C (final temperature of the system), mwater = 100.0 g (mass of water), Tinitial, water = 25.0 °C (initial temperature of water). Therefore, the equation becomes: (50.0 g · caluminum · (30.0 °C − 95.0 °C)) = (100.0 g · 4.18 J/g°C · (30.0 °C − 25.0 °C)) Step 2: Solve for caluminum . Begin by simplifying both sides of the equation: 6750 · caluminum = 20.9 · 100 20.9 · 100 = 0.310 J/g°C 6750 Therefore, the specific heat capacity of aluminum is 0.310 J/g°C. caluminum = 34