CHEM 107: Question Bank - Set 3

CHEM 107 - ESSENTIALS OF GENERAL AND ORGANIC CHEMISTRY - Calorimetry Question Bank - Set 3 Liberty University Question 1 Question A student wants to determine the specific heat capacity of a metal sample using a calorimeter. The student places the metal sample, initially at a temperature of 100◦ C, into a calorimeter containing 200 g of water at 20◦ C. After thermal equilibrium is reached, the final temperature of the system is 25◦ C. If the metal sample has a mass of 50 g, what is the specific heat capacity of the metal? Solution Step 1: Calculate the heat gained by the water. The heat gained by the water can be calculated using the formula: Qwater = mc∆T, where: - m = 200 g is the mass of the water, - c is the specific heat capacity of water (4.18 J/g◦ C), - ∆T = Tfinal − Tinitial = 25 − 20 = 5 ◦ C. Substitute these values into the formula to find Qwater : Qwater = 200 × 4.18 × 5 = 4180J. 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 = 50 g is the mass of the metal, - c is the specific heat capacity of the metal (to be determined), - ∆T = Tfinal − Tinitial = 25 − 100 = −75 ◦ C (negative because the metal is losing heat). Substitute these values into the formula to find Qmetal . Qmetal = 50 × c × (−75) = −3750cJ. Step 3: Since energy is conserved, the heat gained by the water is equal to the heat lost by the metal. Qwater = Qmetal 4180 = −3750c 4180 = −1.115 J/g◦ C c= −3750 Therefore, the specific heat capacity of the metal is 1.115 J/g◦ C. Question 2 Question A 100 g piece of aluminum at 100°C is dropped into a calorimeter containing 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 the calorimeter (assume it is the same as water, 4.18 J/g°C). Solution Step 1: Calculate the heat lost by the aluminum piece. The specific heat capacity of aluminum is 0.903 J/g°C. The formula for heat transfer is: q = mcT , where - q is the heat transferred, - m is the mass of the substance, - c is the specific heat capacity of the substance, and - T is the change in temperature. The heat lost by the aluminum piece is given by: qAl = mAl cAl TAl Plugging in the values: qAl = (100 g)(0.903 J/g°C)(100 − 25)C qAl = 64725 J Step 2: Calculate the heat gained by the water and the calorimeter. The formula for heat transfer is: q = mcT , where now m is the total mass of water and the calorimeter. The specific heat capacity of water is 4.18 J/g°C. The total mass of water and calorimeter is 200 g +mcal , where mcal is the mass of the calorimeter (water equivalent). The heat gained by the water and calorimeter is given by: qw+cal = (200 g + mcal )(4.18 J/g°C)(25 − 20)C 2 qw+cal = (200 g + mcal )(20.9 J) Step 3: Since no heat is lost to the surroundings, the heat lost by aluminum must be equal to the heat gained by the water and calorimeter. Therefore, 64725 J = (200 g + mcal )(20.9 J) Step 4: Solve for mcal to determine the mass of the calorimeter. 64725 J − 200 g 20.9 J = 3085.65 g − 200 g = 2885.65 g mcal = mcal Step 5: Calculate the specific heat capacity of the calorimeter. Since the specific heat capacity of the calorimeter is assumed to be the same as water (4.18 J/g°C), the specific heat capacity of the calorimeter is: ccal = 4.18 J/g°C Question 3 Question A 50 g piece of iron at 100◦ C is placed in 100 g of water at 20◦ C. If the final temperature of the system is 25◦ C, what is the specific heat capacity of the iron? (Specific heat capacity of water = 4.18 J/g◦ C, specific heat capacity of iron = 0.449 J/g◦ C) Solution Let the specific heat capacity of iron be denoted by cFe . Step 1: Calculate the heat lost by the iron and the heat gained by the water. The heat lost by the iron is equal to the heat gained by the water: mFe cFe ∆TFe = −mw cw ∆Tw where: mFe = 50 g (mass of iron), cw = 4.18 J/g◦ C (specific heat capacity of water), cFe =? (specific heat capacity of iron), ∆TFe = (25 − 100) ◦ C, ∆Tw = (25 − 20) ◦ C. Substitute the given values into the equation: 50 × cFe × (25 − 100) = −100 × 4.18 × (25 − 20) Step 2: Solve for cFe . 50 × cFe × (−75) = −100 × 4.18 × 5 cFe = cFe −100 × 4.18 × 5 50 × (−75) = 0.558 J/g◦ C Therefore, the specific heat capacity of iron is 0.558 J/g◦ C. 3 Question 4 Question A 50 g piece of aluminum at 150◦ C is placed in a container of water at 25◦ C. If the final temperature of the system is 30◦ C, calculate the mass of water in ◦ the container. (Specific heat capacity of aluminum = 0.90 J/g C, 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. Step 2: Set the two heat quantities equal and then solve for the mass of water. Step 1: Calculate the heat lost by aluminum and the heat gained by water. The heat lost by the aluminum can be calculated using the formula: qAl = mc∆T where: m = mass of aluminum = 50 g, c = specific heat capacity of aluminum ◦ = 0.90 J/g C, ∆T = change in temperature of aluminum = (30 − 150)◦ C = −120◦ C. Substitute the values into the formula: ◦ qAl = 50 g × 0.90 J/g C × −120◦ C qAl = −5400 J The heat gained by water can be calculated using the same formula: qH2O = mc∆T ◦ where: m = mass of water, c = specific heat capacity of water = 4.18 J/g C, ∆T = change in temperature of water = (30 − 25)◦ C = 5◦ C. Step 2: Set the two heat quantities equal and then solve for the mass of water. Since energy is conserved: qAl = qH2O ◦ −5400 J = m × 4.18 J/g C × 5◦ C Solving for m: −5400 J = 20.9m m= −5400 J ≈ −258.3 g 20.9 J/g 4 Since mass cannot be negative, the mass of water in the container is approximately 258.3 g . Question 5 Question A student is performing a calorimetry experiment by mixing 100 g of water at 20°C with 50 g of water at 60°C. Assuming no heat is lost to the surroundings, calculate the final temperature of the system. Solution Step 1: Calculate the heat lost by the hot water and gained by the cold water using the formula: q = mc∆T For the hot water: qhot = mhot cwater (Tfinal − Thot ) qhot = 50 g × 4.18 J/g°C × (Tfinal − 60C) For the cold water: qcold = mcold cwater (Tfinal − Tcold ) qcold = 100 g × 4.18 J/g°C × (Tfinal − 20C) Since energy is conserved in the system when the two waters combine: qhot = −qcold Step 2: Substitute the expressions for qhot and qcold into the conservation of energy equation. 50 · 4.18 · (Tfinal − 60) = −100 · 4.18 · (Tfinal − 20) Step 3: Solve the equation for the final temperature, Tfinal . 50 · 4.18 · Tfinal − 50 · 4.18 · 60 = −100 · 4.18 · Tfinal + 100 · 4.18 · 20 184.6 · Tfinal − 12516 = −418 · Tfinal + 8360 602 · Tfinal = 20876 Tfinal = 20876 602 Tfinal ≈ 34.7C Therefore, the final temperature of the system is approximately 34.7°C. 5 Question 6 Question A 100 g piece of copper at 100◦ C is placed in 200 g of water at 20◦ C in a perfectly insulated container. Assuming no heat is lost to the surroundings, what will be the final temperature of the system? The specific heat capacity of copper is 0.385 J/g◦ C and that of water is 4.18 J/g◦ C. Solution Step 1: Calculate the heat lost by the copper piece: The heat lost by the copper can be calculated using the formula q = mc∆T , where m is the mass of the copper, c is the specific heat capacity of copper, and ∆T is the change in temperature. Given: mCu = 100 g, cCu = 0.385 J/g◦ C, Tinitial = 100◦ C, and Tfinal = Teq (final equilibrium temperature). The heat lost by the copper is: qCu = mCu · cCu · (Tfinal − Tinitial ) Step 2: Calculate the heat gained by the water: The heat gained by the water can also be calculated using the formula q = mc∆T , with the mass and specific heat capacity of water. Given: mH2O = 200 g, cH2O = 4.18 J/g◦ C, Tinitial = 20◦ C, and Tfinal = Teq . The heat gained by the water is: qH2O = mH2O · cH2O · (Tfinal − Tinitial ) Step 3: Since the total heat lost by the copper equals the total heat gained by the water (assuming no heat is lost to the surroundings), we have: qCu = qH2O mCu · cCu · (Tfinal − Tinitial ) = mH2O · cH2O · (Tfinal − Tinitial ) Step 4: Solve for the final temperature Tfinal : Substitute the given values into the equation and solve for Tfinal : 100 · 0.385 · (Tfinal − 100) = 200 · 4.18 · (Tfinal − 20) Solving for Tfinal will give us the final equilibrium temperature of the system. Question 7 Question A piece of copper of mass 150 g at 200°C is placed in a calorimeter containing 500 g of water at 20°C. If the final temperature of the system is 25°C, calculate the specific heat capacity of copper. Assume no heat is lost to the surroundings. 6 Solution Step 1: Calculate the heat lost by the copper and gained by the water using the formula of heat transfer: Q = mc∆T where: - Q is the heat transfer - m is the mass - c is the specific heat capacity - ∆T is the change in temperature For copper: Qcopper = mc∆T Qcopper = (0.15 kg)(387 J/kg · °C)(200 − 25) °C Qcopper = 6007.5 J For water: Qwater = mc∆T Qwater = (0.5 kg)(4186 J/kg · °C)(25 − 20) °C Qwater = 10465 J Step 2: Since no heat is lost to the surroundings, the heat lost by the copper is equal to the heat gained by the water: Qcopper = Qwater 6007.5 = 10465 Qwater ccopper = mcopper ∆Tcopper ccopper = 10465 0.15 kg · 175 °C 10465 26.25 ≈ 398.86 J/kg · °C ccopper = ccopper Therefore, the specific heat capacity of copper is approximately 398.86 J/kg°C. Question 8 Question A 50 g aluminum block at 100°C is dropped into 200 g of water at 20°C. The final temperature of the mixture is 25°C. Assuming no heat is lost to the surroundings and the specific heat capacity of water is 4186 J/kg°C, calculate the specific heat capacity of aluminum. 7 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. Let Ca be the specific heat capacity of aluminum. We have: ma Ca ∆Ta = mw Cw ∆Tw Where: ma = 0.05 kg (mass of aluminum), Cw = 4186 J/kg°C (specific heat capacity of water), mw = 0.2 kg (mass of water), ∆Ta = Tf − Ta,i = 25 − 100 = −75°C, and ∆Tw = Tf − Tw,i = 25 − 20 = 5°C. Step 2: Substitute the known values into the equation and solve for Ca . 0.05Ca (−75) = 0.2 × 4186 × 5 −3.75Ca = 4186 4186 Ca = −3.75 Ca ≈ −1116.27 J/kg°C Therefore, the specific heat capacity of aluminum is approximately 1116.27 J/kg°C. Question 9 Question A 50 g block of copper at 80°C is placed in 200 g of water at 20°C in an insulated container. 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 and the specific heat capacity of copper is 0.385 J/g°C. Solution Step 1: Calculate the heat lost by the copper block and gained by the water using the formula: qcopper = −qwater Step 2: The heat lost by the copper block is given by: qcopper = mcopper × ccopper × ∆T where: - mcopper = mass of copper = 50 g - ccopper = specific heat capacity of copper = 0.385 J/g°C - ∆T = change in temperature of copper = final temperature - initial temperature - Initial temperature of copper = 80°C - Final temperature of the mixture = 25°C 8 Step 3: Substitute the values into the formula to find qcopper : qcopper = 50 g × 0.385 J/g°C × (25 − 80) °C Step 4: Calculate the heat gained by the water using the formula: 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 = final temperature - initial temperature - Initial temperature of water = 20°C - Final temperature of the mixture = 25°C Step 5: Substitute the values into the formula to find qwater : qwater = 200 g × 4.18 J/g°C × (25 − 20) °C Step 6: Use the fact that qcopper = −qwater to solve for the specific heat capacity of copper: mcopper × ccopper × (Tf − Ti ) = −mwater × cwater × (Tf − Ti ) Step 7: Plug in the values and solve for ccopper : 50 g × ccopper × (25 − 80) = −200 g × 4.18 J/g°C × (25 − 20) Question 10 Question A 50.0 g piece of copper at 150.0°C is placed in 100.0 g of water at 20.0°C. Assuming no heat is lost to the surroundings, what is the final equilibrium temperature of the system? Given: Specific heat capacity of copper = 0.385 J/g°C Specific heat capacity of water = 4.184 J/g°C Solution Step 1: Calculate the heat lost by copper and gained by water using the equation Q = mc∆T , where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. For copper: Qcopper = (50.0 g)(0.385 J/g°C)(Tfinal − 150.0) Step 2: For water: Qwater = (100.0 g)(4.184 J/g°C)(Tfinal − 20.0) Step 3: Since heat lost by copper equals the heat gained by water (assuming no heat loss to the surroundings), we have: Qcopper = Qwater Step 4: Set the two equations equal to each other: (50.0)(0.385)(Tfinal − 150.0) = (100.0)(4.184)(Tfinal − 20.0) Step 5: Simplify and solve for Tfinal : 19.25(Tfinal −150.0) = 418.4(Tfinal −20.0) Step 6: Expand and solve for Tfinal : 19.25Tfinal − 2887.5 = 418.4Tfinal − 8368 9 Step 7: Rearrange the equation: 418.4Tfinal − 19.25Tfinal = 8368 − 2887.5 Step 8: Solve for Tfinal : 399.15Tfinal = 5480.5 5480.5 Step 9: Calculate the final equilibrium temperature: Tfinal = 399.15 ≈ 13.72C Therefore, the final equilibrium temperature of the system is approximately 13.72°C. Question 11 Question A 50 g piece of copper metal is heated to 100◦ C and then 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 the copper metal. The specific heat capacity of water is 4.18 J/(g·◦ C) and the specific heat capacity of the calorimeter is negligible. Solution Step 1: Calculate the heat transfer from the copper piece to the water using the formula: Qcopper = −Qwater Where Qcopper is the heat lost by the copper piece and Qwater is the heat gained by the water. The formula for heat transfer is: Q = mc∆T For the copper piece, the heat lost is: Qcopper = mcopper ccopper ∆T Substitute the values mcopper = 50 g, cwater = 0.387 J/(g·◦ C) (specific heat capacity of copper), ∆T = 100 − 25 = 75: Qcopper = 50 × ccopper × 75 For the water, the heat gained is: Qwater = mwater cwater ∆T Substitute the values mwater = 200 g, cwater = 4.18 J/(g·◦ C) (specific heat capacity of water), ∆T = 25 − 20 = 5: Qwater = 200 × 4.18 × 5 Set Qcopper equal to −Qwater and solve for ccopper . 50 × ccopper × 75 = −200 × 4.18 × 5 10 ccopper = −200 × 4.18 × 5 50 × 75 ccopper = −0.557 J/(g ·◦ C) So, the specific heat capacity of copper is 0.557 J/(g·◦ C). Question 12 Question A piece of copper weighing 200 g at a temperature of 100◦ C is placed in a calorimeter containing 500 g of water at 20◦ C. The final temperature of the system is 25◦ C. Assuming no heat loss to the surroundings, what is the specific heat capacity of copper? (Specific heat capacities: copper = 0.385 J/g◦ C, water = 4.18 J/g◦ C) Solution Step 1: Calculate the heat absorbed by the water: The heat absorbed by the water can be calculated using the formula: Q = mc∆T where: - m is the mass of the water (500 g), - c is the specific heat capacity of water (4.18 J/g◦ C), - ∆T is the temperature change of the water (25◦ C−20◦ C = 5◦ C). Substitute these values into the formula: ◦ Q = (500 g)(4.18 J/g C)(5◦ C) = 10450 J Step 2: Calculate the heat lost by the copper: The heat lost by the copper is equal to the heat absorbed by the water (by the principle of conservation of energy), so Qcopper = −Qwater . Substitute the value of Qwater into the formula: Qcopper = −10450 J Step 3: Calculate the specific heat capacity of copper: 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 (25◦ C − 100◦ C = −75◦ C). 11 Substitute the known values into the formula and solve for c: ◦ −10450 = (200 g)(0.385 J/g C)(−75◦ C) −10450 = −5782.5 Jg × (−75◦ C) −10450 = 433875 J −10450 c= 433875 ◦ c ≈ −0.024 J/g C Therefore, the specific heat capacity of copper is approximately 0.024 J/g◦ C. Question 13 Question A 50 g piece of copper at 100◦ C is placed in 200 g of water at 20◦ C in a calorimeter. The final temperature of the system is 22◦ C. Assuming no heat is lost to the surroundings, determine the specific heat capacity of copper. (Specific heat capacity of water = 4186 J/kg·K) 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 · ∆Twater where: - mwater = 200 g = 0.2 kg (mass of water), - cwater = 4186 J/kg·K (specific heat capacity of water), and - ∆Twater = Tf − Ti = 22 − 20 = 2 K (change in temperature of water). Therefore, Qwater = 0.2 kg × 4186 J/kg · K × 2 K = 1674.4 J Step 2: Calculate the heat lost by the copper. The heat lost by the copper can be calculated using the formula: Qcopper = −mcopper · ccopper · ∆Tcopper where: - mcopper = 50 g = 0.05 kg (mass of copper), - ccopper (specific heat capacity of copper), and - ∆Tcopper = Tf − Ti = 22 − 100 = −78 K (change in temperature of copper). We need to find ccopper . Step 3: Set up the heat balance equation. Since no heat is lost to the surroundings, the total heat lost by the copper must equal the total heat gained by the water: Qcopper = Qwater 12 −0.05 kg × ccopper × −78 K = 1674.4 J Step 4: Solve for the specific heat capacity of copper. 0.05 kg × ccopper × 78 K = 1674.4 J 1674.4 J = 429.69 J/kg · K 0.05 kg × 78 K Therefore, the specific heat capacity of copper is approximately 429.69 J/kg·K. ccopper = Question 14 Question A piece of metal weighing 150 g at a temperature of 180°C is placed in 300 g of water at 20°C. The final temperature of the system is 30°C. Assuming no heat loss to the surroundings, calculate the specific heat capacity of the metal. The specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat absorbed or lost by the water using the formula: qwater = m · c · ∆T where: - m = 300 g (mass of water) - c = 4.18 J/g°C (specific heat capacity of water) - ∆T = 30 − 20 = 10 °C (change in temperature) Calculating: qwater = 300 g × 4.18 J/g°C × 10 °C qwater = 12540 J Step 2: Calculate the heat absorbed or lost by the metal using the formula: qmetal = m · cm · ∆T where: - m = 150 g (mass of metal) - cm is the specific heat capacity of the metal (unknown) - ∆T = 30 − 180 = −150 °C (change in temperature) Step 3: Since the total heat gained by the metal is equal to the total heat lost by the water (assuming no heat loss to the surroundings), we have: qwater = −qmetal Substitute the known values and solve for the specific heat capacity of the metal: 12540 J = 150 g × cm × −150 °C 12540 J cm = 150 g × −150 °C cm = −0.56 J/g°C Therefore, the specific heat capacity of the metal is −0.56 J/g°C. 13 Question 15 Question A 200 g aluminum block at 100◦ C is dropped into 400 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, calculate the specific heat capacity of aluminum. The specific heat capacity of water is 4.18 J/(g·◦ C). Solution Step 1: Find the heat absorbed by the water: The heat absorbed by the water can be calculated using the formula: Q = mc∆T where: - m = 400 g is the mass of water, - c = 4.18 J/(g·◦ C) is the specific heat capacity of water, - ∆T = 25 − 20 = 5 ◦ C is the change in temperature. Substitute the values into the formula: Qwater = 400 g × 4.18 J/(g·◦ C) × 5 ◦ C = 400 × 4.18 × 5 = 8360 J Step 2: Find the heat lost by the aluminum block: The heat lost by the aluminum block will be equal to the heat absorbed by the water, since no heat is lost to the surroundings. Therefore, Qaluminum = −Qwater = −8360 J Step 3: Find the specific heat capacity of aluminum: The heat lost by the aluminum block can be calculated using the formula: Q = mc∆T where: - m = 200 g is the mass of aluminum, - c is the specific heat capacity of aluminum (to be determined), - ∆T = 25 − 100 = −75 ◦ C is the change in temperature. Substitute the values into the formula and solve for c: −8360 J = 200 g × c × −75 ◦ C 8360 = 15000c 8360 = 0.557 J/(g·◦ C) c= 15000 Therefore, the specific heat capacity of aluminum is 0.557 J/(g·◦ C). Question 16 Question A piece of aluminum (specific heat, c = 0.897 J/g°C) at 100°C is placed in 200 g of water at 20°C. Assuming no heat is lost to the surroundings, what is the final temperature of the system? (specific heat of water, c = 4.18 J/g°C) 14 Solution Step 1: First, calculate the heat gained or lost by the aluminum as it cools down to the final temperature. Given: - Mass of aluminum, mAl = 200 g - Initial temperature of aluminum, TAl,i = 100 °C - Final temperature of the system, Tfinal The heat lost by the aluminum can be calculated using the formula: qAl = −mAl cAl (TAl,i − Tfinal ) qAl = −200 × 0.897 × (100 − Tfinal ) Step 2: Next, calculate the heat gained or lost by the water as it warms up to the final temperature. Given: - Mass of water, mw = 200 g - Initial temperature of water, Tw,i = 20 °C - Final temperature of the system, Tfinal The heat gained by the water can be calculated using the formula: qw = mw cw (Tfinal − Tw,i ) qw = 200 × 4.18 × (Tfinal − 20) Step 3: Since no heat is lost to the surroundings, the heat lost by the aluminum is equal to the heat gained by the water. Set qAl = qw . −200 × 0.897 × (100 − Tfinal ) = 200 × 4.18 × (Tfinal − 20) Step 4: Solve the equation for Tfinal to find the final temperature of the system. This involves algebraic manipulation and solving for Tfinal in °C. Question 17 Question A 50 g piece of iron at 80°C is placed 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, 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 gained by the water. The heat lost by the iron is equal to the heat gained by the water. Using the formula Q = mc∆T , where - Q is the heat energy absorbed or released, - m is the mass of the substance, - c is the specific heat capacity of the substance, - ∆T is the change in temperature. Let’s denote the specific heat capacity of iron as cFe . The heat lost by the iron is QFe = mFe · cFe · ∆TFe , where - mFe = 50 g, - ∆TFe = Tfinal − Tinitial = 25 − 80 = −55°C. The heat gained by the water is Qwater = mwater · cwater · ∆Twater , where mwater = 200 g, - cwater = 4.18 J/g°C, - ∆Twater = Tfinal −Tinitial = 25−20 = 5°C. 15 Since the heat lost by the iron is equal to the heat gained by the water, we have: QFe = Qwater Step 2: Set up the equation and solve for cFe . mFe · cFe · ∆TFe = mwater · cwater · ∆Twater Substitute the given values: 50 · cFe · (−55) = 200 · 4.18 · 5 cFe = 0.381 J/g°C Solving for cFe : cFe = 200·4.18·5 50·55 Therefore, the specific heat capacity of iron is 0.381 J/g°C. Question 18 Question A 50.0 g piece of iron at 150.0°C is placed in a calorimeter containing 200.0 g of water at 20.0°C. The final temperature of the iron and water mixture is 25.0°C. Assuming no heat is lost to the surroundings, what is the specific heat capacity of the iron? Solution Step 1: First, calculate the heat lost by the iron and the heat gained by the water using the formula: Q = mc∆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 the iron: Qiron = miron ciron ∆Tiron Qiron = (0.050 kg)(ciron )(25 − 150) Qiron = (0.050 kg)(ciron )(−125) For the water: Qwater = mwater cwater ∆Twater Qwater = (0.200 kg)(4190 J/kg · °C)(25 − 20) Qwater = (0.200 kg)(4190 J/kg · °C)(5) Step 2: Since the system is isolated: Qiron = −Qwater (0.050 kg)(ciron )(−125) = (0.200 kg)(4190 J/kg · °C)(5) Step 3: Solve for ciron : ciron = (0.200 kg)(4190 J/kg · °C)(5) 0.050 kg(−125) 16 ciron = ciron = ((0.200)(4190)(5)) (0.050)(−125) 4190 = −167.6 J/kg · °C −25 Therefore, the specific heat capacity of iron is −167.6 J/kg· °C. The negative sign indicates that heat is lost by the iron. Question 19 Question A 50.0 g piece of iron at 95.0°C is placed in a calorimeter containing 100.0 g of water at 20.0°C. The final temperature of the system is 25.0°C. Assuming no heat is transferred to the surroundings, calculate the specific heat capacity of ◦ the iron. (Specific heat capacity of water = 4.18 J/g C) Solution Step 1: Calculate the heat lost by the iron piece. The heat lost by the iron piece can be calculated using the formula: Qlost = m · c · ∆T Where: - m = 50.0 g (mass of iron) - c is the specific heat capacity of iron (unknown) - ∆T = 95.0 − 25.0 = 70.0 °C (change in temperature of the iron) Plugging in the values, we get: Qlost = 50.0 g × c × 70.0 °C Step 2: Calculate the heat gained by the water. The heat gained by the water can be calculated using the formula: Qgain = m · c · ∆T ◦ Where: - m = 100.0 g (mass of water) - c = 4.18 J/g C (specific heat capacity of water) - ∆T = 25.0 − 20.0 = 5.0 °C (change in temperature of the water) Plugging in the values, we get: ◦ Qgain = 100.0 g × 4.18 J/g C × 5.0 °C Step 3: Set up the equation based on the law of conservation of energy. According to the law of conservation of energy, the heat lost by the iron must be equal to the heat gained by the water. Therefore: Qlost = Qgain 17 Step 4: Solve for the specific heat capacity of iron. Equating Qlost and Qgain , we have: ◦ 50.0 g · c · 70.0 °C = 100.0 g · 4.18 J/g C · 5.0 °C ◦ c= 100.0 g · 4.18 J/g C · 5.0 °C 50.0 g · 70.0 °C Question 20 Question A piece of metal of mass 0.2 kg is heated to 100◦ C and then dropped into 0.5 kg of water at 20◦ C. The final temperature of the mixture is 30◦ C. Given that the specific heat capacity of the metal is 500 J/kg◦ C, calculate the specific heat capacity of the water. Solution Step 1: First, we need to calculate the heat absorbed by the metal as it cools down to the final temperature. The heat absorbed by the metal is given by the formula: Qmetal = mcmetal ∆T ◦ where: - m = 0.2 kg is the mass of the metal, - cmetal = 500 J/kg C is the specific heat capacity of the metal, and - ∆T = Tf −Ti = 30◦ C−100◦ C = −70◦ C is the change in temperature of the metal. Therefore, Qmetal = 0.2 × 500 × (−70) Qmetal = −7000 J Step 2: Next, calculate the heat lost by the metal is gained by the water in order for the system to reach thermal equilibrium. Let’s assume the specific heat capacity of water is cwater (in J/kg◦ C). The heat lost by the metal is equal to the heat gained by the water: mcmetal ∆T = mcwater ∆T Substitute the known values: (−7000) = (0.5)(cwater )(10) Step 3: Solve for the specific heat capacity of water: cwater = −7000 0.5 × 10 ◦ cwater = −1400 J/kg C Therefore, the specific heat capacity of the water is 1400 J/kg◦ C. 18 Question 21 Question A piece of metal with mass 200 g is heated to 100°C and then transferred to a calorimeter of mass 150 g containing 400 g of water at 20°C. The final temperature of the system is 25°C. If the specific heat capacity of the metal is 0.2 J/g°C, calculate the specific heat capacity of the calorimeter. Assume no heat is lost to the surroundings. Solution Step 1: Calculate the heat transfer from the metal to the water in the calorimeter. The heat lost by the metal is equal to the heat gained by the water and calorimeter: mmetal · cmetal · ∆T = (mwater + mcalorimeter ) · cwater · ∆T Substitute the given values: 200 g · 0.2 J/g°C · (100 − 25)C = (400 + 150) g · cwater · (25 − 20)C 3500 J = 550 g · cwater · 5C 3500 ≈ 1.27 J/g°C cwater = 2750 Step 2: Calculate the specific heat capacity of the calorimeter. To find the specific heat capacity of the calorimeter, we can rewrite the heat transfer equation in terms of the calorimeter: mcalorimeter · ccalorimeter · ∆T = (mwater + mmetal ) · cwater · ∆T Substitute the known values: 150 g · ccalorimeter · (25 − 20)C = 550 g · 1.27 J/g°C · 5C 30 g · ccalorimeter = 3183.5 3183.5 ccalorimeter = ≈ 106.12 J/g°C 30 Therefore, the specific heat capacity of the calorimeter is approximately 106.12 J/g°C. Question 22 Question A student carries out an experiment to determine the specific heat capacity of a metal using a calorimeter. The student places 200 g of the metal at an initial temperature of 120°C into 400 g of water at an initial temperature of 20°C. The final temperature of the mixture is 30°C. If the specific heat capacity of water is 4.18 J/g · °C, determine the specific heat capacity of the metal. 19 Solution Step 1: Calculate the heat gained by the water: The heat gained by the water can be calculated using the equation: Qwater = mwater · cwater · ∆T where: - mwater = 400 g is the mass of the water, - cwater = 4.18 J/g · °C is the specific heat capacity of water, and - T = Tfinal − Tinitial = 30 °C − 20 °C = 10 °C is the temperature change. Therefore: Qwater = 400 g × 4.18 J/g · °C × 10 °C = 16720 J Step 2: Calculate the heat lost by the metal: The heat lost by the metal is equal in magnitude but opposite in sign to the heat gained by the water. Therefore, the heat lost by the metal can be calculated as: Qmetal = −Qwater = −16720 J Step 3: Calculate the specific heat capacity of the metal: The heat lost by the metal can be calculated using the equation: Qmetal = mmetal · cmetal · ∆T where: - mmetal = 200 g is the mass of the metal, - cmetal is the specific heat capacity of the metal, and - T = Tfinal − Tinitial = 30 °C − 120 °C = −90 °C is the temperature change. Substitute in the values: −16720 J = 200 g × cmetal × −90 °C Solve for cmetal : cmetal = −16720 J = 9.29 J/g · °C 200 g × −90 °C Therefore, the specific heat capacity of the metal is 9.29 J/g · °C. Question 23 Question A 50 g aluminum sample at 80°C is placed in a calorimeter containing 200 g of water at 20°C. If the final temperature of the system is 25°C, determine the specific heat capacity of the material of the calorimeter. Assume no heat is ◦ lost to the surroundings. (Specific heat capacity of aluminum is 0.9 J/g C and ◦ specific heat capacity of water is 4.18 J/g C.) 20 Solution Step 1: Calculate the heat lost by the aluminum sample. The heat lost by the aluminum sample can be calculated using the formula: qaluminum = m · c · ∆T where: - m is the mass of the aluminum sample (50 g) - c is the specific heat ◦ capacity of aluminum (0.9 J/g C) - ∆T is the change in temperature of the aluminum sample (80◦ C − 25◦ C = 55◦ C) Substitute the values into the formula: ◦ qaluminum = 50 g · 0.9 J/g C · 55◦ C qaluminum = 2475 J Step 2: Calculate the heat gained by the water in the calorimeter. The heat gained by the water in the calorimeter can be calculated using the formula: qwater = m · c · ∆T where: - m is the mass of the water in the calorimeter (200 g) - c is the specific ◦ heat capacity of water (4.18 J/g C) - ∆T is the change in temperature of the water (25◦ C − 20◦ C = 5◦ C) Substitute the values into the formula: ◦ qwater = 200 g · 4.18 J/g C · 5◦ C qwater = 4180 J Step 3: Determine the specific heat capacity of the calorimeter material. Since there is no heat lost to the surroundings, the heat lost by the aluminum sample is equal to the heat gained by the water: qaluminum = qwater 2475 J = 4180 J + m · ccalorimeter · ∆T m · ccalorimeter = 2475 J − 4180 J m · ccalorimeter = −1705 J Since the mass of the calorimeter material is not given, we can solve for the specific heat capacity of the calorimeter material: ccalorimeter = −1705 J m · ∆T Therefore, the specific heat capacity of the material of the calorimeter is J ccalorimeter = −1705 m·∆T . 21 Question 24 Question A 50 g piece of copper at 200°C is placed in 100 g of water at 20°C. Assuming no heat is lost to the surroundings, what will be the final temperature of the system? The specific heat capacity of copper is 0.385 J/g°C and that of water is 4.18 J/g°C. Solution Step 1: Calculate the heat lost by the copper piece as it cools down to the final temperature. The formula for calculating heat is Q = m · c · ∆T , where: - Q is the heat energy, - m is the mass of the substance, - c is the specific heat capacity of the substance, - ∆T is the change in temperature. Given that the initial temperature of the copper piece is 200°C and the final temperature is T (in °C), the change in temperature is ∆T = 200 − T °C. Therefore, the heat lost by the copper piece is: Qcopper = 50 g · 0.385 J/g°C · (200 − T ) °C Step 2: Calculate the heat gained by the water as it warms up to the final temperature. Following the same formula as above and with the same reasoning, the heat gained by the water is: Qwater = 100 g · 4.18 J/g°C · (T − 20) °C Step 3: Since the system is isolated and assuming no heat is lost to the surroundings, the heat lost by the copper will be equal to the heat gained by the water. Therefore, Qcopper = Qwater Step 4: Set the two expressions for heat equal to each other and solve for T. 50 · 0.385 · (200 − T ) = 100 · 4.18 · (T − 20) Step 5: Solve the equation for T to find the final temperature of the system. Question 25 Question A 50 g gold coin at a temperature of 85°C is dropped into 200 g of water at 15°C. If the final temperature of the mixture is 25°C, calculate the specific heat capacity of gold. Assume no heat is lost to the surroundings. 22 Solution Step 1: Calculate the heat lost by the gold coin as it cools down to the final temperature of the mixture. The heat lost by the gold coin can be calculated using the formula: Qlost, gold = mgold · cgold · (Tf − Ti ) where: - mgold = 50 g = 0.05 kg is the mass of the gold coin, - cgold is the specific heat capacity of gold (to be determined), - Tf = 25◦ C = 25 + 273 = 298 K is the final temperature of the mixture, - Ti = 85◦ C = 85 + 273 = 358 K is the initial temperature of the gold coin. Substitute the given values into the formula: Qlost, gold = 0.05 kg · cgold · (298 K − 358 K) Qlost, gold = −15 J · cgold Step 2: Calculate the heat gained by the water as it heats up to the final temperature of the mixture. The heat gained by the water can be calculated using the formula: Qgained, water = mwater · cwater · (Tf − Ti ) where: - mwater = 200 g = 0.2 kg is the mass of the water, - cwater = 4186 J/kg·K is the specific heat capacity of water, - Tf = 298 K is the final temperature of the mixture, - Ti = 15◦ C = 15 + 273 = 288 K is the initial temperature of the water. Substitute the given values into the formula: Qgained, water = 0.2 kg · 4186 J/kg · K · (298 K − 288 K) Qgained, water = 4186 J Step 3: Since no heat is lost to the surroundings, the heat lost by the gold coin must equal the heat gained by the water. Set the expressions for heat lost and heat gained equal to each other: −15 J · cgold = 1846 J Step 4: Solve for the specific heat capacity of gold. cgold = 4186 J 0.05 kg · (358 K − 298 K) 4186 J 0.05 kg · 60 K 4186 J = 3 J/kg · K cgold = cgold cgold = 1395.33 J/kg · K Therefore, the specific heat capacity of gold is approximately 1395.33 J/kg·K. 23 Question 26 Question A 100 g piece of aluminum at 80°C is placed in 200 g of water at 20°C in an insulated container. The final equilibrium temperature of the system is 30°C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity ◦ of aluminum. (Specific heat capacity of water is 4.18 J/g C) Solution Step 1: First, we need to calculate the heat lost by the aluminum and the heat gained by the water. The heat lost by the aluminum can be calculated using the formula: QAluminum = m × c × ∆T where: - m is the mass of aluminum (100 g) - c is the specific heat capacity of aluminum (to be calculated) - ∆T is the change in temperature of aluminum Given that the initial temperature of aluminum (Tinitial, Al ) is 80°C and the final temperature of the system is 30°C, the change in temperature of aluminum is: ∆TAluminum = Tfinal − Tinitial, Al = 30C − 80C = −50C (We use a negative sign because the temperature of aluminum is decreasing) Step 2: Now, we calculate the heat gained by the water. This can be calculated using the formula: QWater = m × c × ∆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 water Given that the initial temperature of water (Tinitial, Water ) is 20°C and the final temperature of the system is 30°C, the change in temperature of water is: ∆TWater = Tfinal − Tinitial, Water = 30C − 20C = 10C Step 3: Since the system is insulated and no heat is lost to the surroundings, we can assume that the heat lost by the aluminum is equal to the heat gained by the water: QAluminum = QWater mAl × cAl × ∆TAl = mWater × cWater × ∆TWater Substitute the given values: 100 × cAl × (−50) = 200 × 4.18 × 10 Step 4: Solve the equation for the specific heat capacity of aluminum: cAl = 200 × 4.18 × 10 100 × (−50) 24 cAl = 8360 −5000 ◦ cAl = −1.672 J/g C ◦ Therefore, the specific heat capacity of aluminum is −1.672 J/g C. Question 27 Question A 50 g piece of iron at 80°C is placed into 200 g of water at 20°C. Assuming no heat is lost to the surroundings, calculate the final temperature of the system. The specific heat capacity of iron is 0.45 J/g°C and the specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat gained by the water. The formula for heat gained or lost is given by: Q = mc∆T , where Q is the heat, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. Given: mwater = 200 g, cwater = 4.18 J/g°C, Tinitial, water = 20°C, Tfinal, water = x (unknown). Qwater = mwater × cwater × ∆Twater Qwater = 200 × 4.18 × (x − 20) Step 2: Calculate the heat lost by the iron. Given: miron = 50 g, ciron = 0.45 J/g°C, Tinitial, iron = 80°C, Tfinal, iron = x (unknown). Qiron = miron × ciron × ∆Tiron Qiron = 50 × 0.45 × (x − 80) Step 3: Since the total heat gained by the water should be equal to the total heat lost by the iron (neglecting any heat loss to the surroundings), we can set Qwater equal to Qiron and solve for the final temperature, x. 200 × 4.18 × (x − 20) = 50 × 0.45 × (x − 80) 25 Question 28 Question A 50 g piece of aluminum at 80°C is placed in a calorimeter containing 200 g of water at 20°C. If the final temperature of the system is 30°C, what is the heat capacity of the calorimeter? Assume no heat is lost to the surroundings. Solution Step 1: Calculate the heat lost by the aluminum and gained by the water. The heat lost by the aluminum can be calculated using the formula: QAl = mAl · cAl · ∆TAl where: - mAl is the mass of aluminum (50 g), - cAl is the specific heat capacity of aluminum (0.903 J/g°C), - ∆TAl is the change in temperature of aluminum (final temperature - initial temperature). Substitute the given values: QAl = 50 g × 0.903 J/g°C × (30 − 80)°C QAl = 50 × 0.903 × (−50) Step 2: Calculate the heat gained by the water. The heat gained by the water can be calculated using the formula: QH2O = mH2O · cH2O · ∆TH2O where: - mH2O is the mass of water (200 g), - cH2O is the specific heat capacity of water (4.18 J/g°C), - ∆TH2O is the change in temperature of water. Substitute the given values: QH2O = 200 g × 4.18 J/g°C × (30 − 20)°C Step 3: Recall that in an isolated system, heat lost equals heat gained. Set up an equation: QAl = QH2O and solve for the heat capacity of the calorimeter, denoted as Ccal . 50 × 0.903 × (−50) = 200 × 4.18 × (30 − 20) + Ccal × (30 − 20) This equation can be solved to find the heat capacity of the calorimeter. Question 29 Question A student wants to determine the specific heat capacity of an unknown metal. To do this, the student heats a 150 g metal sample to 100°C and then drops it into 200 g of water initially at 20°C in a calorimeter. The final temperature of the system is 25°C. Assuming no heat is lost to the surroundings, determine the specific heat capacity of the metal. 26 Solution Step 1: Calculate the heat gained by the water. The heat gained by the water can be calculated using the formula: Q = mc∆T where: - m = 200 g is the mass of water, - c = 4.18 J/g°C is the specific heat capacity of water, - ∆T = (Tf − Ti ) = (25C − 20C) = 5C is the change in temperature. Substitute the values into the formula: Qwater = (200 g)(4.18 J/g°C)(5C) Qwater = 4180 J Therefore, the heat gained by the water is 4180 J. Step 2: Calculate the heat lost by the metal. The heat lost by the metal is equal to the heat gained by the water, as no heat is lost to the surroundings (assuming an insulated calorimeter). Therefore, the heat lost by the metal can be calculated as: Qmetal = 4180 J Step 3: Determine the specific heat capacity of the metal. The heat lost by the metal can be calculated using the formula: Qmetal = mc∆T where: - m = 150 g is the mass of the metal, - c is the specific heat capacity of the metal, - ∆T = 100C − 25C = 75C is the change in temperature. Substitute the values into the formula and solve for c: 4180 J = (150 g)c(75C) c= 4180 J (150 g)(75C) c ≈ 37.07 J/g°C Therefore, the specific heat capacity of the metal is approximately 37.07 J/g°C. 27 Question 30 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 no heat is lost to the surroundings and the specific heat capacity of water is 4.18 J/g°C, calculate the specific heat capacity of aluminum. Solution Step 1: Calculate the heat lost by aluminum. The formula for calculating heat is: Q = mc∆T where: - Q is the heat absorbed or lost, - m is the mass of the material, - c is the specific heat capacity of the material, - T is the change in temperature. Substitute the given values: QAl = (50 g)(cAl )(25 − 100) QAl = −50cAl Step 2: Calculate the heat gained by water. QH2 O = (200 g)(4.18 J/g°C)(25 − 20) QH2 O = 4180 J Step 3: Since the system is isolated, the heat lost by aluminum is equal to the heat gained by water. −50cAl = 4180 4180 cAl = − 50 cAl = −83.6 J/g°C Therefore, the specific heat capacity of aluminum is 83.6 J/g°C. Question 31 Question A 50.0 g piece of aluminum at 90.0◦ C is added to 100.0 g of water at 20.0◦ C in a calorimeter. The final temperature of the mixture is 27.0◦ C. Assume all the heat lost by the aluminum is gained by the water and the calorimeter itself. 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. What is the heat capacity (in J◦ C) of the calorimeter? 28 Solution Step 1: Calculate the heat gained by the water. The heat gained by the water can be calculated using the formula: qwater = mc∆T where: m = mass of water in grams c = specific heat capacity of water in J/g◦ C ∆T = change in temperature of the water Given that m = 100.0 g, c = 4.184 J/g◦ C, and ∆T = 27.0 − 20.0 = 7.0 ◦ C, we can calculate: ◦ qwater = 100.0 g × 4.184 J/g C × 7.0 ◦ C qwater = 2928.8 J Step 2: Calculate the heat lost by the aluminum. The heat lost by the aluminum can be calculated using the same formula: qaluminum = mc∆T where: m = mass of aluminum in grams c = specific heat capacity of aluminum in J/g◦ C ∆T = change in temperature of the aluminum Given that m = 50.0 g, c = 0.900 J/g◦ C, and ∆T = 27.0 − 90.0 = −63.0 ◦ C (negative because the aluminum is losing heat), we can calculate: ◦ qaluminum = 50.0 g × 0.900 J/g C × −63.0 ◦ C qaluminum = −2835.0 J Step 3: Calculate the heat capacity of the calorimeter. Since the heat lost by the aluminum is equal to the heat gained by the water and the calorimeter, we can write: qaluminum = qwater + qcalorimeter Substitute the calculated values for qaluminum and qwater : −2835.0 = 2928.8 + qcalorimeter qcalorimeter = −2835.0 − 2928.8 qcalorimeter = −5763.8 J Therefore, the heat capacity of the calorimeter is 5763.8 J◦ C . 29 Question 32 Question A 100 g piece of copper at 200°C is dropped into 200 g of water at 20°C in a calorimeter. If the final temperature of the system is 22.5°C, find the specific heat capacity of copper. Assume no heat is lost to the surroundings. The specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat lost by copper and the heat gained by water. The heat lost by copper is equal to the heat gained by water: (mcopper )(ccopper )(∆Tcopper ) = (mwater )(cwater )(∆Twater ) where: - mcopper is the mass of copper (100 g) - ccopper is the specific heat capacity of copper (to be found) - ∆Tcopper is the change in temperature of copper (final temp - initial temp) - mwater is the mass of water (200 g) - cwater is the specific heat capacity of water (4.18 J/g°C) - ∆Twater is the change in temperature of water (final temp - initial temp) Step 2: Calculate the change in temperature for copper and water. ∆Tcopper = final temp − initial temp = 22.5 − 200 = −177.5C ∆Twater = final temp − initial temp = 22.5 − 20 = 2.5C Step 3: Substitute the known values into the heat equation and solve for ccopper . (100)(ccopper )(−177.5) = (200)(4.18)(2.5) −17750ccopper = 2095 ccopper = 2095 = −0.118 J/g°C −17750 Therefore, the specific heat capacity of copper is −0.118 J/g°C . Question 33 Question A 50 g block of copper at 95°C is dropped into 200 g of water at 25°C in a calorimeter. If the final temperature of the copper-water mixture is 30°C, what is the specific heat capacity of the copper block? The specific heat capacity of water is 4.18 J/g°C. 30 Solution Step 1: Calculate the heat absorbed by the water. Given: Mass of water, mw = 200 g Initial temperature of water, Twi = 25°C Final temperature of water and copper mixture, Tf = 30°C Specific heat capacity of water, cw = 4.18 J/g°C Using the formula for heat absorbed or released: Q = mc∆T where Q is the heat, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. The heat absorbed by the water can be calculated as: Qw = mw cw ∆T Qw = (200 g)(4.18 J/g°C)(30 − 25) °C Qw = 4180 J Step 2: Calculate the heat lost by the copper block. Given: Mass of copper block, mc = 50 g Initial temperature of copper block, Tci = 95°C Final temperature of copper-water mixture, Tf = 30°C Specific heat capacity of copper, cc (to be determined) The heat lost by the copper block can also be calculated using the formula for heat: Qc = mc∆T Step 3: Set up the conservation of energy equation. According to the principle of conservation of energy, the heat lost by the copper block is equal to the heat gained by the water: Qc = Qw Step 4: Solve for the specific heat capacity of the copper block. Substitute the known values and solve for cc : mc cc ∆T = mw cw ∆T (50 g)cc (30 − 95) °C = 4180 J cc (−65) = 4180 cc = 4180 −65 cc ≈ −64.31 J/g°C Therefore, the specific heat capacity of the copper block is approximately −64.31 J/g°C. Note the negative sign indicates that the direction of heat transfer for the copper block is opposite to that of the water. 31 Question 34 Question A 50 g piece of metal at 150°C is placed into a calorimeter containing 200 g of water at 20°C. The final temperature of the system is 25°C. If the specific heat capacity of the metal is 0.5 J/g°C and the specific heat capacity of water is 4.18 J/g°C, calculate the initial temperature of the metal before it was added to the calorimeter. Solution Step 1: Calculate the heat gained by the water Let the initial temperature of the metal be Tm °C. The heat lost by the metal = heat gained by the water mm · cm · (Tf − Tm ) = mw · cw · (Tf − Tw ) 50 · 0.5 · (25 − Tm ) = 200 · 4.18 · (25 − 20) 25(25 − Tm ) = 836 625 − 25Tm = 836 25Tm = −211 Tm = −8.44 Therefore, the initial temperature of the metal was -8.44°C before it was added to the calorimeter. Question 35 Question A 50 g cube of iron at 80°C is placed in 200 g of water at 20°C in a perfectly insulated container. Assuming no heat is lost to the surroundings, what will be the final temperature of the system? (Specific heat capacity of iron = 450 J/kg · K, specific heat capacity of water = 4186 J/kg · K) Solution Step 1: Calculate the heat gained by the water: The heat gained by the water can be expressed as Q = mc∆T , where m is the mass of the water, c is the specific heat capacity of water, and ∆T is the change in temperature. Given: m = 200 g = 0.2 kg, c = 4186 J/kg · K, and ∆T = Tf − 20 (since the water is originally at 20°C). Therefore, Q = (0.2)(4186)(Tf − 20). Step 2: Calculate the heat lost by the iron: Similarly, the heat lost by the iron can be calculated as Q = mc∆T , where m is the mass of the iron, c is the specific heat capacity of iron, and ∆T is the change in temperature. Given: 32 Question 4 Question A 50 g piece of aluminum at 150◦ C is placed in a container of water at 25◦ C. If the final temperature of the system is 30◦ C, calculate the mass of water in ◦ the container. (Specific heat capacity of aluminum = 0.90 J/g C, 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. Step 2: Set the two heat quantities equal and then solve for the mass of water. Step 1: Calculate the heat lost by aluminum and the heat gained by water. The heat lost by the aluminum can be calculated using the formula: qAl = mc∆T where: m = mass of aluminum = 50 g, c = specific heat capacity of aluminum ◦ = 0.90 J/g C, ∆T = change in temperature of aluminum = (30 − 150)◦ C = −120◦ C. Substitute the values into the formula: ◦ qAl = 50 g × 0.90 J/g C × −120◦ C qAl = −5400 J The heat gained by water can be calculated using the same formula: qH2O = mc∆T ◦ where: m = mass of water, c = specific heat capacity of water = 4.18 J/g C, ∆T = change in temperature of water = (30 − 25)◦ C = 5◦ C. Step 2: Set the two heat quantities equal and then solve for the mass of water. Since energy is conserved: qAl = qH2O ◦ −5400 J = m × 4.18 J/g C × 5◦ C Solving for m: −5400 J = 20.9m m= −5400 J ≈ −258.3 g 20.9 J/g 4 Since mass cannot be negative, the mass of water in the container is approximately 258.3 g . Question 5 Question A student is performing a calorimetry experiment by mixing 100 g of water at 20°C with 50 g of water at 60°C. Assuming no heat is lost to the surroundings, calculate the final temperature of the system. Solution Step 1: Calculate the heat lost by the hot water and gained by the cold water using the formula: q = mc∆T For the hot water: qhot = mhot cwater (Tfinal − Thot ) qhot = 50 g × 4.18 J/g°C × (Tfinal − 60C) For the cold water: qcold = mcold cwater (Tfinal − Tcold ) qcold = 100 g × 4.18 J/g°C × (Tfinal − 20C) Since energy is conserved in the system when the two waters combine: qhot = −qcold Step 2: Substitute the expressions for qhot and qcold into the conservation of energy equation. 50 · 4.18 · (Tfinal − 60) = −100 · 4.18 · (Tfinal − 20) Step 3: Solve the equation for the final temperature, Tfinal . 50 · 4.18 · Tfinal − 50 · 4.18 · 60 = −100 · 4.18 · Tfinal + 100 · 4.18 · 20 184.6 · Tfinal − 12516 = −418 · Tfinal + 8360 602 · Tfinal = 20876 Tfinal = 20876 602 Tfinal ≈ 34.7C Therefore, the final temperature of the system is approximately 34.7°C. 5 Question 6 Question A 100 g piece of copper at 100◦ C is placed in 200 g of water at 20◦ C in a perfectly insulated container. Assuming no heat is lost to the surroundings, what will be the final temperature of the system? The specific heat capacity of copper is 0.385 J/g◦ C and that of water is 4.18 J/g◦ C. Solution Step 1: Calculate the heat lost by the copper piece: The heat lost by the copper can be calculated using the formula q = mc∆T , where m is the mass of the copper, c is the specific heat capacity of copper, and ∆T is the change in temperature. Given: mCu = 100 g, cCu = 0.385 J/g◦ C, Tinitial = 100◦ C, and Tfinal = Teq (final equilibrium temperature). The heat lost by the copper is: qCu = mCu · cCu · (Tfinal − Tinitial ) Step 2: Calculate the heat gained by the water: The heat gained by the water can also be calculated using the formula q = mc∆T , with the mass and specific heat capacity of water. Given: mH2O = 200 g, cH2O = 4.18 J/g◦ C, Tinitial = 20◦ C, and Tfinal = Teq . The heat gained by the water is: qH2O = mH2O · cH2O · (Tfinal − Tinitial ) Step 3: Since the total heat lost by the copper equals the total heat gained by the water (assuming no heat is lost to the surroundings), we have: qCu = qH2O mCu · cCu · (Tfinal − Tinitial ) = mH2O · cH2O · (Tfinal − Tinitial ) Step 4: Solve for the final temperature Tfinal : Substitute the given values into the equation and solve for Tfinal : 100 · 0.385 · (Tfinal − 100) = 200 · 4.18 · (Tfinal − 20) Solving for Tfinal will give us the final equilibrium temperature of the system. Question 7 Question A piece of copper of mass 150 g at 200°C is placed in a calorimeter containing 500 g of water at 20°C. If the final temperature of the system is 25°C, calculate the specific heat capacity of copper. Assume no heat is lost to the surroundings. 6 Solution Step 1: Calculate the heat lost by the copper and gained by the water using the formula of heat transfer: Q = mc∆T where: - Q is the heat transfer - m is the mass - c is the specific heat capacity - ∆T is the change in temperature For copper: Qcopper = mc∆T Qcopper = (0.15 kg)(387 J/kg · °C)(200 − 25) °C Qcopper = 6007.5 J For water: Qwater = mc∆T Qwater = (0.5 kg)(4186 J/kg · °C)(25 − 20) °C Qwater = 10465 J Step 2: Since no heat is lost to the surroundings, the heat lost by the copper is equal to the heat gained by the water: Qcopper = Qwater 6007.5 = 10465 Qwater ccopper = mcopper ∆Tcopper ccopper = 10465 0.15 kg · 175 °C 10465 26.25 ≈ 398.86 J/kg · °C ccopper = ccopper Therefore, the specific heat capacity of copper is approximately 398.86 J/kg°C. Question 8 Question A 50 g aluminum block at 100°C is dropped into 200 g of water at 20°C. The final temperature of the mixture is 25°C. Assuming no heat is lost to the surroundings and the specific heat capacity of water is 4186 J/kg°C, calculate the specific heat capacity of aluminum. 7 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. Let Ca be the specific heat capacity of aluminum. We have: ma Ca ∆Ta = mw Cw ∆Tw Where: ma = 0.05 kg (mass of aluminum), Cw = 4186 J/kg°C (specific heat capacity of water), mw = 0.2 kg (mass of water), ∆Ta = Tf − Ta,i = 25 − 100 = −75°C, and ∆Tw = Tf − Tw,i = 25 − 20 = 5°C. Step 2: Substitute the known values into the equation and solve for Ca . 0.05Ca (−75) = 0.2 × 4186 × 5 −3.75Ca = 4186 4186 Ca = −3.75 Ca ≈ −1116.27 J/kg°C Therefore, the specific heat capacity of aluminum is approximately 1116.27 J/kg°C. Question 9 Question A 50 g block of copper at 80°C is placed in 200 g of water at 20°C in an insulated container. 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 and the specific heat capacity of copper is 0.385 J/g°C. Solution Step 1: Calculate the heat lost by the copper block and gained by the water using the formula: qcopper = −qwater Step 2: The heat lost by the copper block is given by: qcopper = mcopper × ccopper × ∆T where: - mcopper = mass of copper = 50 g - ccopper = specific heat capacity of copper = 0.385 J/g°C - ∆T = change in temperature of copper = final temperature - initial temperature - Initial temperature of copper = 80°C - Final temperature of the mixture = 25°C 8 Step 3: Substitute the values into the formula to find qcopper : qcopper = 50 g × 0.385 J/g°C × (25 − 80) °C Step 4: Calculate the heat gained by the water using the formula: 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 = final temperature - initial temperature - Initial temperature of water = 20°C - Final temperature of the mixture = 25°C Step 5: Substitute the values into the formula to find qwater : qwater = 200 g × 4.18 J/g°C × (25 − 20) °C Step 6: Use the fact that qcopper = −qwater to solve for the specific heat capacity of copper: mcopper × ccopper × (Tf − Ti ) = −mwater × cwater × (Tf − Ti ) Step 7: Plug in the values and solve for ccopper : 50 g × ccopper × (25 − 80) = −200 g × 4.18 J/g°C × (25 − 20) Question 10 Question A 50.0 g piece of copper at 150.0°C is placed in 100.0 g of water at 20.0°C. Assuming no heat is lost to the surroundings, what is the final equilibrium temperature of the system? Given: Specific heat capacity of copper = 0.385 J/g°C Specific heat capacity of water = 4.184 J/g°C Solution Step 1: Calculate the heat lost by copper and gained by water using the equation Q = mc∆T , where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. For copper: Qcopper = (50.0 g)(0.385 J/g°C)(Tfinal − 150.0) Step 2: For water: Qwater = (100.0 g)(4.184 J/g°C)(Tfinal − 20.0) Step 3: Since heat lost by copper equals the heat gained by water (assuming no heat loss to the surroundings), we have: Qcopper = Qwater Step 4: Set the two equations equal to each other: (50.0)(0.385)(Tfinal − 150.0) = (100.0)(4.184)(Tfinal − 20.0) Step 5: Simplify and solve for Tfinal : 19.25(Tfinal −150.0) = 418.4(Tfinal −20.0) Step 6: Expand and solve for Tfinal : 19.25Tfinal − 2887.5 = 418.4Tfinal − 8368 9 Step 7: Rearrange the equation: 418.4Tfinal − 19.25Tfinal = 8368 − 2887.5 Step 8: Solve for Tfinal : 399.15Tfinal = 5480.5 5480.5 Step 9: Calculate the final equilibrium temperature: Tfinal = 399.15 ≈ 13.72C Therefore, the final equilibrium temperature of the system is approximately 13.72°C. Question 11 Question A 50 g piece of copper metal is heated to 100◦ C and then 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 the copper metal. The specific heat capacity of water is 4.18 J/(g·◦ C) and the specific heat capacity of the calorimeter is negligible. Solution Step 1: Calculate the heat transfer from the copper piece to the water using the formula: Qcopper = −Qwater Where Qcopper is the heat lost by the copper piece and Qwater is the heat gained by the water. The formula for heat transfer is: Q = mc∆T For the copper piece, the heat lost is: Qcopper = mcopper ccopper ∆T Substitute the values mcopper = 50 g, cwater = 0.387 J/(g·◦ C) (specific heat capacity of copper), ∆T = 100 − 25 = 75: Qcopper = 50 × ccopper × 75 For the water, the heat gained is: Qwater = mwater cwater ∆T Substitute the values mwater = 200 g, cwater = 4.18 J/(g·◦ C) (specific heat capacity of water), ∆T = 25 − 20 = 5: Qwater = 200 × 4.18 × 5 Set Qcopper equal to −Qwater and solve for ccopper . 50 × ccopper × 75 = −200 × 4.18 × 5 10 ccopper = −200 × 4.18 × 5 50 × 75 ccopper = −0.557 J/(g ·◦ C) So, the specific heat capacity of copper is 0.557 J/(g·◦ C). Question 12 Question A piece of copper weighing 200 g at a temperature of 100◦ C is placed in a calorimeter containing 500 g of water at 20◦ C. The final temperature of the system is 25◦ C. Assuming no heat loss to the surroundings, what is the specific heat capacity of copper? (Specific heat capacities: copper = 0.385 J/g◦ C, water = 4.18 J/g◦ C) Solution Step 1: Calculate the heat absorbed by the water: The heat absorbed by the water can be calculated using the formula: Q = mc∆T where: - m is the mass of the water (500 g), - c is the specific heat capacity of water (4.18 J/g◦ C), - ∆T is the temperature change of the water (25◦ C−20◦ C = 5◦ C). Substitute these values into the formula: ◦ Q = (500 g)(4.18 J/g C)(5◦ C) = 10450 J Step 2: Calculate the heat lost by the copper: The heat lost by the copper is equal to the heat absorbed by the water (by the principle of conservation of energy), so Qcopper = −Qwater . Substitute the value of Qwater into the formula: Qcopper = −10450 J Step 3: Calculate the specific heat capacity of copper: 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 (25◦ C − 100◦ C = −75◦ C). 11 Substitute the known values into the formula and solve for c: ◦ −10450 = (200 g)(0.385 J/g C)(−75◦ C) −10450 = −5782.5 Jg × (−75◦ C) −10450 = 433875 J −10450 c= 433875 ◦ c ≈ −0.024 J/g C Therefore, the specific heat capacity of copper is approximately 0.024 J/g◦ C. Question 13 Question A 50 g piece of copper at 100◦ C is placed in 200 g of water at 20◦ C in a calorimeter. The final temperature of the system is 22◦ C. Assuming no heat is lost to the surroundings, determine the specific heat capacity of copper. (Specific heat capacity of water = 4186 J/kg·K) 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 · ∆Twater where: - mwater = 200 g = 0.2 kg (mass of water), - cwater = 4186 J/kg·K (specific heat capacity of water), and - ∆Twater = Tf − Ti = 22 − 20 = 2 K (change in temperature of water). Therefore, Qwater = 0.2 kg × 4186 J/kg · K × 2 K = 1674.4 J Step 2: Calculate the heat lost by the copper. The heat lost by the copper can be calculated using the formula: Qcopper = −mcopper · ccopper · ∆Tcopper where: - mcopper = 50 g = 0.05 kg (mass of copper), - ccopper (specific heat capacity of copper), and - ∆Tcopper = Tf − Ti = 22 − 100 = −78 K (change in temperature of copper). We need to find ccopper . Step 3: Set up the heat balance equation. Since no heat is lost to the surroundings, the total heat lost by the copper must equal the total heat gained by the water: Qcopper = Qwater 12 −0.05 kg × ccopper × −78 K = 1674.4 J Step 4: Solve for the specific heat capacity of copper. 0.05 kg × ccopper × 78 K = 1674.4 J 1674.4 J = 429.69 J/kg · K 0.05 kg × 78 K Therefore, the specific heat capacity of copper is approximately 429.69 J/kg·K. ccopper = Question 14 Question A piece of metal weighing 150 g at a temperature of 180°C is placed in 300 g of water at 20°C. The final temperature of the system is 30°C. Assuming no heat loss to the surroundings, calculate the specific heat capacity of the metal. The specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat absorbed or lost by the water using the formula: qwater = m · c · ∆T where: - m = 300 g (mass of water) - c = 4.18 J/g°C (specific heat capacity of water) - ∆T = 30 − 20 = 10 °C (change in temperature) Calculating: qwater = 300 g × 4.18 J/g°C × 10 °C qwater = 12540 J Step 2: Calculate the heat absorbed or lost by the metal using the formula: qmetal = m · cm · ∆T where: - m = 150 g (mass of metal) - cm is the specific heat capacity of the metal (unknown) - ∆T = 30 − 180 = −150 °C (change in temperature) Step 3: Since the total heat gained by the metal is equal to the total heat lost by the water (assuming no heat loss to the surroundings), we have: qwater = −qmetal Substitute the known values and solve for the specific heat capacity of the metal: 12540 J = 150 g × cm × −150 °C 12540 J cm = 150 g × −150 °C cm = −0.56 J/g°C Therefore, the specific heat capacity of the metal is −0.56 J/g°C. 13 Question 15 Question A 200 g aluminum block at 100◦ C is dropped into 400 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, calculate the specific heat capacity of aluminum. The specific heat capacity of water is 4.18 J/(g·◦ C). Solution Step 1: Find the heat absorbed by the water: The heat absorbed by the water can be calculated using the formula: Q = mc∆T where: - m = 400 g is the mass of water, - c = 4.18 J/(g·◦ C) is the specific heat capacity of water, - ∆T = 25 − 20 = 5 ◦ C is the change in temperature. Substitute the values into the formula: Qwater = 400 g × 4.18 J/(g·◦ C) × 5 ◦ C = 400 × 4.18 × 5 = 8360 J Step 2: Find the heat lost by the aluminum block: The heat lost by the aluminum block will be equal to the heat absorbed by the water, since no heat is lost to the surroundings. Therefore, Qaluminum = −Qwater = −8360 J Step 3: Find the specific heat capacity of aluminum: The heat lost by the aluminum block can be calculated using the formula: Q = mc∆T where: - m = 200 g is the mass of aluminum, - c is the specific heat capacity of aluminum (to be determined), - ∆T = 25 − 100 = −75 ◦ C is the change in temperature. Substitute the values into the formula and solve for c: −8360 J = 200 g × c × −75 ◦ C 8360 = 15000c 8360 = 0.557 J/(g·◦ C) c= 15000 Therefore, the specific heat capacity of aluminum is 0.557 J/(g·◦ C). Question 16 Question A piece of aluminum (specific heat, c = 0.897 J/g°C) at 100°C is placed in 200 g of water at 20°C. Assuming no heat is lost to the surroundings, what is the final temperature of the system? (specific heat of water, c = 4.18 J/g°C) 14 Solution Step 1: First, calculate the heat gained or lost by the aluminum as it cools down to the final temperature. Given: - Mass of aluminum, mAl = 200 g - Initial temperature of aluminum, TAl,i = 100 °C - Final temperature of the system, Tfinal The heat lost by the aluminum can be calculated using the formula: qAl = −mAl cAl (TAl,i − Tfinal ) qAl = −200 × 0.897 × (100 − Tfinal ) Step 2: Next, calculate the heat gained or lost by the water as it warms up to the final temperature. Given: - Mass of water, mw = 200 g - Initial temperature of water, Tw,i = 20 °C - Final temperature of the system, Tfinal The heat gained by the water can be calculated using the formula: qw = mw cw (Tfinal − Tw,i ) qw = 200 × 4.18 × (Tfinal − 20) Step 3: Since no heat is lost to the surroundings, the heat lost by the aluminum is equal to the heat gained by the water. Set qAl = qw . −200 × 0.897 × (100 − Tfinal ) = 200 × 4.18 × (Tfinal − 20) Step 4: Solve the equation for Tfinal to find the final temperature of the system. This involves algebraic manipulation and solving for Tfinal in °C. Question 17 Question A 50 g piece of iron at 80°C is placed 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, 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 gained by the water. The heat lost by the iron is equal to the heat gained by the water. Using the formula Q = mc∆T , where - Q is the heat energy absorbed or released, - m is the mass of the substance, - c is the specific heat capacity of the substance, - ∆T is the change in temperature. Let’s denote the specific heat capacity of iron as cFe . The heat lost by the iron is QFe = mFe · cFe · ∆TFe , where - mFe = 50 g, - ∆TFe = Tfinal − Tinitial = 25 − 80 = −55°C. The heat gained by the water is Qwater = mwater · cwater · ∆Twater , where mwater = 200 g, - cwater = 4.18 J/g°C, - ∆Twater = Tfinal −Tinitial = 25−20 = 5°C. 15 Since the heat lost by the iron is equal to the heat gained by the water, we have: QFe = Qwater Step 2: Set up the equation and solve for cFe . mFe · cFe · ∆TFe = mwater · cwater · ∆Twater Substitute the given values: 50 · cFe · (−55) = 200 · 4.18 · 5 cFe = 0.381 J/g°C Solving for cFe : cFe = 200·4.18·5 50·55 Therefore, the specific heat capacity of iron is 0.381 J/g°C. Question 18 Question A 50.0 g piece of iron at 150.0°C is placed in a calorimeter containing 200.0 g of water at 20.0°C. The final temperature of the iron and water mixture is 25.0°C. Assuming no heat is lost to the surroundings, what is the specific heat capacity of the iron? Solution Step 1: First, calculate the heat lost by the iron and the heat gained by the water using the formula: Q = mc∆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 the iron: Qiron = miron ciron ∆Tiron Qiron = (0.050 kg)(ciron )(25 − 150) Qiron = (0.050 kg)(ciron )(−125) For the water: Qwater = mwater cwater ∆Twater Qwater = (0.200 kg)(4190 J/kg · °C)(25 − 20) Qwater = (0.200 kg)(4190 J/kg · °C)(5) Step 2: Since the system is isolated: Qiron = −Qwater (0.050 kg)(ciron )(−125) = (0.200 kg)(4190 J/kg · °C)(5) Step 3: Solve for ciron : ciron = (0.200 kg)(4190 J/kg · °C)(5) 0.050 kg(−125) 16 ciron = ciron = ((0.200)(4190)(5)) (0.050)(−125) 4190 = −167.6 J/kg · °C −25 Therefore, the specific heat capacity of iron is −167.6 J/kg· °C. The negative sign indicates that heat is lost by the iron. Question 19 Question A 50.0 g piece of iron at 95.0°C is placed in a calorimeter containing 100.0 g of water at 20.0°C. The final temperature of the system is 25.0°C. Assuming no heat is transferred to the surroundings, calculate the specific heat capacity of ◦ the iron. (Specific heat capacity of water = 4.18 J/g C) Solution Step 1: Calculate the heat lost by the iron piece. The heat lost by the iron piece can be calculated using the formula: Qlost = m · c · ∆T Where: - m = 50.0 g (mass of iron) - c is the specific heat capacity of iron (unknown) - ∆T = 95.0 − 25.0 = 70.0 °C (change in temperature of the iron) Plugging in the values, we get: Qlost = 50.0 g × c × 70.0 °C Step 2: Calculate the heat gained by the water. The heat gained by the water can be calculated using the formula: Qgain = m · c · ∆T ◦ Where: - m = 100.0 g (mass of water) - c = 4.18 J/g C (specific heat capacity of water) - ∆T = 25.0 − 20.0 = 5.0 °C (change in temperature of the water) Plugging in the values, we get: ◦ Qgain = 100.0 g × 4.18 J/g C × 5.0 °C Step 3: Set up the equation based on the law of conservation of energy. According to the law of conservation of energy, the heat lost by the iron must be equal to the heat gained by the water. Therefore: Qlost = Qgain 17 Step 4: Solve for the specific heat capacity of iron. Equating Qlost and Qgain , we have: ◦ 50.0 g · c · 70.0 °C = 100.0 g · 4.18 J/g C · 5.0 °C ◦ c= 100.0 g · 4.18 J/g C · 5.0 °C 50.0 g · 70.0 °C Question 20 Question A piece of metal of mass 0.2 kg is heated to 100◦ C and then dropped into 0.5 kg of water at 20◦ C. The final temperature of the mixture is 30◦ C. Given that the specific heat capacity of the metal is 500 J/kg◦ C, calculate the specific heat capacity of the water. Solution Step 1: First, we need to calculate the heat absorbed by the metal as it cools down to the final temperature. The heat absorbed by the metal is given by the formula: Qmetal = mcmetal ∆T ◦ where: - m = 0.2 kg is the mass of the metal, - cmetal = 500 J/kg C is the specific heat capacity of the metal, and - ∆T = Tf −Ti = 30◦ C−100◦ C = −70◦ C is the change in temperature of the metal. Therefore, Qmetal = 0.2 × 500 × (−70) Qmetal = −7000 J Step 2: Next, calculate the heat lost by the metal is gained by the water in order for the system to reach thermal equilibrium. Let’s assume the specific heat capacity of water is cwater (in J/kg◦ C). The heat lost by the metal is equal to the heat gained by the water: mcmetal ∆T = mcwater ∆T Substitute the known values: (−7000) = (0.5)(cwater )(10) Step 3: Solve for the specific heat capacity of water: cwater = −7000 0.5 × 10 ◦ cwater = −1400 J/kg C Therefore, the specific heat capacity of the water is 1400 J/kg◦ C. 18 Question 21 Question A piece of metal with mass 200 g is heated to 100°C and then transferred to a calorimeter of mass 150 g containing 400 g of water at 20°C. The final temperature of the system is 25°C. If the specific heat capacity of the metal is 0.2 J/g°C, calculate the specific heat capacity of the calorimeter. Assume no heat is lost to the surroundings. Solution Step 1: Calculate the heat transfer from the metal to the water in the calorimeter. The heat lost by the metal is equal to the heat gained by the water and calorimeter: mmetal · cmetal · ∆T = (mwater + mcalorimeter ) · cwater · ∆T Substitute the given values: 200 g · 0.2 J/g°C · (100 − 25)C = (400 + 150) g · cwater · (25 − 20)C 3500 J = 550 g · cwater · 5C 3500 ≈ 1.27 J/g°C cwater = 2750 Step 2: Calculate the specific heat capacity of the calorimeter. To find the specific heat capacity of the calorimeter, we can rewrite the heat transfer equation in terms of the calorimeter: mcalorimeter · ccalorimeter · ∆T = (mwater + mmetal ) · cwater · ∆T Substitute the known values: 150 g · ccalorimeter · (25 − 20)C = 550 g · 1.27 J/g°C · 5C 30 g · ccalorimeter = 3183.5 3183.5 ccalorimeter = ≈ 106.12 J/g°C 30 Therefore, the specific heat capacity of the calorimeter is approximately 106.12 J/g°C. Question 22 Question A student carries out an experiment to determine the specific heat capacity of a metal using a calorimeter. The student places 200 g of the metal at an initial temperature of 120°C into 400 g of water at an initial temperature of 20°C. The final temperature of the mixture is 30°C. If the specific heat capacity of water is 4.18 J/g · °C, determine the specific heat capacity of the metal. 19 Solution Step 1: Calculate the heat gained by the water: The heat gained by the water can be calculated using the equation: Qwater = mwater · cwater · ∆T where: - mwater = 400 g is the mass of the water, - cwater = 4.18 J/g · °C is the specific heat capacity of water, and - T = Tfinal − Tinitial = 30 °C − 20 °C = 10 °C is the temperature change. Therefore: Qwater = 400 g × 4.18 J/g · °C × 10 °C = 16720 J Step 2: Calculate the heat lost by the metal: The heat lost by the metal is equal in magnitude but opposite in sign to the heat gained by the water. Therefore, the heat lost by the metal can be calculated as: Qmetal = −Qwater = −16720 J Step 3: Calculate the specific heat capacity of the metal: The heat lost by the metal can be calculated using the equation: Qmetal = mmetal · cmetal · ∆T where: - mmetal = 200 g is the mass of the metal, - cmetal is the specific heat capacity of the metal, and - T = Tfinal − Tinitial = 30 °C − 120 °C = −90 °C is the temperature change. Substitute in the values: −16720 J = 200 g × cmetal × −90 °C Solve for cmetal : cmetal = −16720 J = 9.29 J/g · °C 200 g × −90 °C Therefore, the specific heat capacity of the metal is 9.29 J/g · °C. Question 23 Question A 50 g aluminum sample at 80°C is placed in a calorimeter containing 200 g of water at 20°C. If the final temperature of the system is 25°C, determine the specific heat capacity of the material of the calorimeter. Assume no heat is ◦ lost to the surroundings. (Specific heat capacity of aluminum is 0.9 J/g C and ◦ specific heat capacity of water is 4.18 J/g C.) 20 Solution Step 1: Calculate the heat lost by the aluminum sample. The heat lost by the aluminum sample can be calculated using the formula: qaluminum = m · c · ∆T where: - m is the mass of the aluminum sample (50 g) - c is the specific heat ◦ capacity of aluminum (0.9 J/g C) - ∆T is the change in temperature of the aluminum sample (80◦ C − 25◦ C = 55◦ C) Substitute the values into the formula: ◦ qaluminum = 50 g · 0.9 J/g C · 55◦ C qaluminum = 2475 J Step 2: Calculate the heat gained by the water in the calorimeter. The heat gained by the water in the calorimeter can be calculated using the formula: qwater = m · c · ∆T where: - m is the mass of the water in the calorimeter (200 g) - c is the specific ◦ heat capacity of water (4.18 J/g C) - ∆T is the change in temperature of the water (25◦ C − 20◦ C = 5◦ C) Substitute the values into the formula: ◦ qwater = 200 g · 4.18 J/g C · 5◦ C qwater = 4180 J Step 3: Determine the specific heat capacity of the calorimeter material. Since there is no heat lost to the surroundings, the heat lost by the aluminum sample is equal to the heat gained by the water: qaluminum = qwater 2475 J = 4180 J + m · ccalorimeter · ∆T m · ccalorimeter = 2475 J − 4180 J m · ccalorimeter = −1705 J Since the mass of the calorimeter material is not given, we can solve for the specific heat capacity of the calorimeter material: ccalorimeter = −1705 J m · ∆T Therefore, the specific heat capacity of the material of the calorimeter is J ccalorimeter = −1705 m·∆T . 21 Question 24 Question A 50 g piece of copper at 200°C is placed in 100 g of water at 20°C. Assuming no heat is lost to the surroundings, what will be the final temperature of the system? The specific heat capacity of copper is 0.385 J/g°C and that of water is 4.18 J/g°C. Solution Step 1: Calculate the heat lost by the copper piece as it cools down to the final temperature. The formula for calculating heat is Q = m · c · ∆T , where: - Q is the heat energy, - m is the mass of the substance, - c is the specific heat capacity of the substance, - ∆T is the change in temperature. Given that the initial temperature of the copper piece is 200°C and the final temperature is T (in °C), the change in temperature is ∆T = 200 − T °C. Therefore, the heat lost by the copper piece is: Qcopper = 50 g · 0.385 J/g°C · (200 − T ) °C Step 2: Calculate the heat gained by the water as it warms up to the final temperature. Following the same formula as above and with the same reasoning, the heat gained by the water is: Qwater = 100 g · 4.18 J/g°C · (T − 20) °C Step 3: Since the system is isolated and assuming no heat is lost to the surroundings, the heat lost by the copper will be equal to the heat gained by the water. Therefore, Qcopper = Qwater Step 4: Set the two expressions for heat equal to each other and solve for T. 50 · 0.385 · (200 − T ) = 100 · 4.18 · (T − 20) Step 5: Solve the equation for T to find the final temperature of the system. Question 25 Question A 50 g gold coin at a temperature of 85°C is dropped into 200 g of water at 15°C. If the final temperature of the mixture is 25°C, calculate the specific heat capacity of gold. Assume no heat is lost to the surroundings. 22 Solution Step 1: Calculate the heat lost by the gold coin as it cools down to the final temperature of the mixture. The heat lost by the gold coin can be calculated using the formula: Qlost, gold = mgold · cgold · (Tf − Ti ) where: - mgold = 50 g = 0.05 kg is the mass of the gold coin, - cgold is the specific heat capacity of gold (to be determined), - Tf = 25◦ C = 25 + 273 = 298 K is the final temperature of the mixture, - Ti = 85◦ C = 85 + 273 = 358 K is the initial temperature of the gold coin. Substitute the given values into the formula: Qlost, gold = 0.05 kg · cgold · (298 K − 358 K) Qlost, gold = −15 J · cgold Step 2: Calculate the heat gained by the water as it heats up to the final temperature of the mixture. The heat gained by the water can be calculated using the formula: Qgained, water = mwater · cwater · (Tf − Ti ) where: - mwater = 200 g = 0.2 kg is the mass of the water, - cwater = 4186 J/kg·K is the specific heat capacity of water, - Tf = 298 K is the final temperature of the mixture, - Ti = 15◦ C = 15 + 273 = 288 K is the initial temperature of the water. Substitute the given values into the formula: Qgained, water = 0.2 kg · 4186 J/kg · K · (298 K − 288 K) Qgained, water = 4186 J Step 3: Since no heat is lost to the surroundings, the heat lost by the gold coin must equal the heat gained by the water. Set the expressions for heat lost and heat gained equal to each other: −15 J · cgold = 1846 J Step 4: Solve for the specific heat capacity of gold. cgold = 4186 J 0.05 kg · (358 K − 298 K) 4186 J 0.05 kg · 60 K 4186 J = 3 J/kg · K cgold = cgold cgold = 1395.33 J/kg · K Therefore, the specific heat capacity of gold is approximately 1395.33 J/kg·K. 23 Question 26 Question A 100 g piece of aluminum at 80°C is placed in 200 g of water at 20°C in an insulated container. The final equilibrium temperature of the system is 30°C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity ◦ of aluminum. (Specific heat capacity of water is 4.18 J/g C) Solution Step 1: First, we need to calculate the heat lost by the aluminum and the heat gained by the water. The heat lost by the aluminum can be calculated using the formula: QAluminum = m × c × ∆T where: - m is the mass of aluminum (100 g) - c is the specific heat capacity of aluminum (to be calculated) - ∆T is the change in temperature of aluminum Given that the initial temperature of aluminum (Tinitial, Al ) is 80°C and the final temperature of the system is 30°C, the change in temperature of aluminum is: ∆TAluminum = Tfinal − Tinitial, Al = 30C − 80C = −50C (We use a negative sign because the temperature of aluminum is decreasing) Step 2: Now, we calculate the heat gained by the water. This can be calculated using the formula: QWater = m × c × ∆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 water Given that the initial temperature of water (Tinitial, Water ) is 20°C and the final temperature of the system is 30°C, the change in temperature of water is: ∆TWater = Tfinal − Tinitial, Water = 30C − 20C = 10C Step 3: Since the system is insulated and no heat is lost to the surroundings, we can assume that the heat lost by the aluminum is equal to the heat gained by the water: QAluminum = QWater mAl × cAl × ∆TAl = mWater × cWater × ∆TWater Substitute the given values: 100 × cAl × (−50) = 200 × 4.18 × 10 Step 4: Solve the equation for the specific heat capacity of aluminum: cAl = 200 × 4.18 × 10 100 × (−50) 24 cAl = 8360 −5000 ◦ cAl = −1.672 J/g C ◦ Therefore, the specific heat capacity of aluminum is −1.672 J/g C. Question 27 Question A 50 g piece of iron at 80°C is placed into 200 g of water at 20°C. Assuming no heat is lost to the surroundings, calculate the final temperature of the system. The specific heat capacity of iron is 0.45 J/g°C and the specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat gained by the water. The formula for heat gained or lost is given by: Q = mc∆T , where Q is the heat, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. Given: mwater = 200 g, cwater = 4.18 J/g°C, Tinitial, water = 20°C, Tfinal, water = x (unknown). Qwater = mwater × cwater × ∆Twater Qwater = 200 × 4.18 × (x − 20) Step 2: Calculate the heat lost by the iron. Given: miron = 50 g, ciron = 0.45 J/g°C, Tinitial, iron = 80°C, Tfinal, iron = x (unknown). Qiron = miron × ciron × ∆Tiron Qiron = 50 × 0.45 × (x − 80) Step 3: Since the total heat gained by the water should be equal to the total heat lost by the iron (neglecting any heat loss to the surroundings), we can set Qwater equal to Qiron and solve for the final temperature, x. 200 × 4.18 × (x − 20) = 50 × 0.45 × (x − 80) 25 Question 28 Question A 50 g piece of aluminum at 80°C is placed in a calorimeter containing 200 g of water at 20°C. If the final temperature of the system is 30°C, what is the heat capacity of the calorimeter? Assume no heat is lost to the surroundings. Solution Step 1: Calculate the heat lost by the aluminum and gained by the water. The heat lost by the aluminum can be calculated using the formula: QAl = mAl · cAl · ∆TAl where: - mAl is the mass of aluminum (50 g), - cAl is the specific heat capacity of aluminum (0.903 J/g°C), - ∆TAl is the change in temperature of aluminum (final temperature - initial temperature). Substitute the given values: QAl = 50 g × 0.903 J/g°C × (30 − 80)°C QAl = 50 × 0.903 × (−50) Step 2: Calculate the heat gained by the water. The heat gained by the water can be calculated using the formula: QH2O = mH2O · cH2O · ∆TH2O where: - mH2O is the mass of water (200 g), - cH2O is the specific heat capacity of water (4.18 J/g°C), - ∆TH2O is the change in temperature of water. Substitute the given values: QH2O = 200 g × 4.18 J/g°C × (30 − 20)°C Step 3: Recall that in an isolated system, heat lost equals heat gained. Set up an equation: QAl = QH2O and solve for the heat capacity of the calorimeter, denoted as Ccal . 50 × 0.903 × (−50) = 200 × 4.18 × (30 − 20) + Ccal × (30 − 20) This equation can be solved to find the heat capacity of the calorimeter. Question 29 Question A student wants to determine the specific heat capacity of an unknown metal. To do this, the student heats a 150 g metal sample to 100°C and then drops it into 200 g of water initially at 20°C in a calorimeter. The final temperature of the system is 25°C. Assuming no heat is lost to the surroundings, determine the specific heat capacity of the metal. 26 Solution Step 1: Calculate the heat gained by the water. The heat gained by the water can be calculated using the formula: Q = mc∆T where: - m = 200 g is the mass of water, - c = 4.18 J/g°C is the specific heat capacity of water, - ∆T = (Tf − Ti ) = (25C − 20C) = 5C is the change in temperature. Substitute the values into the formula: Qwater = (200 g)(4.18 J/g°C)(5C) Qwater = 4180 J Therefore, the heat gained by the water is 4180 J. Step 2: Calculate the heat lost by the metal. The heat lost by the metal is equal to the heat gained by the water, as no heat is lost to the surroundings (assuming an insulated calorimeter). Therefore, the heat lost by the metal can be calculated as: Qmetal = 4180 J Step 3: Determine the specific heat capacity of the metal. The heat lost by the metal can be calculated using the formula: Qmetal = mc∆T where: - m = 150 g is the mass of the metal, - c is the specific heat capacity of the metal, - ∆T = 100C − 25C = 75C is the change in temperature. Substitute the values into the formula and solve for c: 4180 J = (150 g)c(75C) c= 4180 J (150 g)(75C) c ≈ 37.07 J/g°C Therefore, the specific heat capacity of the metal is approximately 37.07 J/g°C. 27 Question 30 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 no heat is lost to the surroundings and the specific heat capacity of water is 4.18 J/g°C, calculate the specific heat capacity of aluminum. Solution Step 1: Calculate the heat lost by aluminum. The formula for calculating heat is: Q = mc∆T where: - Q is the heat absorbed or lost, - m is the mass of the material, - c is the specific heat capacity of the material, - T is the change in temperature. Substitute the given values: QAl = (50 g)(cAl )(25 − 100) QAl = −50cAl Step 2: Calculate the heat gained by water. QH2 O = (200 g)(4.18 J/g°C)(25 − 20) QH2 O = 4180 J Step 3: Since the system is isolated, the heat lost by aluminum is equal to the heat gained by water. −50cAl = 4180 4180 cAl = − 50 cAl = −83.6 J/g°C Therefore, the specific heat capacity of aluminum is 83.6 J/g°C. Question 31 Question A 50.0 g piece of aluminum at 90.0◦ C is added to 100.0 g of water at 20.0◦ C in a calorimeter. The final temperature of the mixture is 27.0◦ C. Assume all the heat lost by the aluminum is gained by the water and the calorimeter itself. 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. What is the heat capacity (in J◦ C) of the calorimeter? 28 Solution Step 1: Calculate the heat gained by the water. The heat gained by the water can be calculated using the formula: qwater = mc∆T where: m = mass of water in grams c = specific heat capacity of water in J/g◦ C ∆T = change in temperature of the water Given that m = 100.0 g, c = 4.184 J/g◦ C, and ∆T = 27.0 − 20.0 = 7.0 ◦ C, we can calculate: ◦ qwater = 100.0 g × 4.184 J/g C × 7.0 ◦ C qwater = 2928.8 J Step 2: Calculate the heat lost by the aluminum. The heat lost by the aluminum can be calculated using the same formula: qaluminum = mc∆T where: m = mass of aluminum in grams c = specific heat capacity of aluminum in J/g◦ C ∆T = change in temperature of the aluminum Given that m = 50.0 g, c = 0.900 J/g◦ C, and ∆T = 27.0 − 90.0 = −63.0 ◦ C (negative because the aluminum is losing heat), we can calculate: ◦ qaluminum = 50.0 g × 0.900 J/g C × −63.0 ◦ C qaluminum = −2835.0 J Step 3: Calculate the heat capacity of the calorimeter. Since the heat lost by the aluminum is equal to the heat gained by the water and the calorimeter, we can write: qaluminum = qwater + qcalorimeter Substitute the calculated values for qaluminum and qwater : −2835.0 = 2928.8 + qcalorimeter qcalorimeter = −2835.0 − 2928.8 qcalorimeter = −5763.8 J Therefore, the heat capacity of the calorimeter is 5763.8 J◦ C . 29 Question 32 Question A 100 g piece of copper at 200°C is dropped into 200 g of water at 20°C in a calorimeter. If the final temperature of the system is 22.5°C, find the specific heat capacity of copper. Assume no heat is lost to the surroundings. The specific heat capacity of water is 4.18 J/g°C. Solution Step 1: Calculate the heat lost by copper and the heat gained by water. The heat lost by copper is equal to the heat gained by water: (mcopper )(ccopper )(∆Tcopper ) = (mwater )(cwater )(∆Twater ) where: - mcopper is the mass of copper (100 g) - ccopper is the specific heat capacity of copper (to be found) - ∆Tcopper is the change in temperature of copper (final temp - initial temp) - mwater is the mass of water (200 g) - cwater is the specific heat capacity of water (4.18 J/g°C) - ∆Twater is the change in temperature of water (final temp - initial temp) Step 2: Calculate the change in temperature for copper and water. ∆Tcopper = final temp − initial temp = 22.5 − 200 = −177.5C ∆Twater = final temp − initial temp = 22.5 − 20 = 2.5C Step 3: Substitute the known values into the heat equation and solve for ccopper . (100)(ccopper )(−177.5) = (200)(4.18)(2.5) −17750ccopper = 2095 ccopper = 2095 = −0.118 J/g°C −17750 Therefore, the specific heat capacity of copper is −0.118 J/g°C . Question 33 Question A 50 g block of copper at 95°C is dropped into 200 g of water at 25°C in a calorimeter. If the final temperature of the copper-water mixture is 30°C, what is the specific heat capacity of the copper block? The specific heat capacity of water is 4.18 J/g°C. 30 Solution Step 1: Calculate the heat absorbed by the water. Given: Mass of water, mw = 200 g Initial temperature of water, Twi = 25°C Final temperature of water and copper mixture, Tf = 30°C Specific heat capacity of water, cw = 4.18 J/g°C Using the formula for heat absorbed or released: Q = mc∆T where Q is the heat, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. The heat absorbed by the water can be calculated as: Qw = mw cw ∆T Qw = (200 g)(4.18 J/g°C)(30 − 25) °C Qw = 4180 J Step 2: Calculate the heat lost by the copper block. Given: Mass of copper block, mc = 50 g Initial temperature of copper block, Tci = 95°C Final temperature of copper-water mixture, Tf = 30°C Specific heat capacity of copper, cc (to be determined) The heat lost by the copper block can also be calculated using the formula for heat: Qc = mc∆T Step 3: Set up the conservation of energy equation. According to the principle of conservation of energy, the heat lost by the copper block is equal to the heat gained by the water: Qc = Qw Step 4: Solve for the specific heat capacity of the copper block. Substitute the known values and solve for cc : mc cc ∆T = mw cw ∆T (50 g)cc (30 − 95) °C = 4180 J cc (−65) = 4180 cc = 4180 −65 cc ≈ −64.31 J/g°C Therefore, the specific heat capacity of the copper block is approximately −64.31 J/g°C. Note the negative sign indicates that the direction of heat transfer for the copper block is opposite to that of the water. 31 Question 34 Question A 50 g piece of metal at 150°C is placed into a calorimeter containing 200 g of water at 20°C. The final temperature of the system is 25°C. If the specific heat capacity of the metal is 0.5 J/g°C and the specific heat capacity of water is 4.18 J/g°C, calculate the initial temperature of the metal before it was added to the calorimeter. Solution Step 1: Calculate the heat gained by the water Let the initial temperature of the metal be Tm °C. The heat lost by the metal = heat gained by the water mm · cm · (Tf − Tm ) = mw · cw · (Tf − Tw ) 50 · 0.5 · (25 − Tm ) = 200 · 4.18 · (25 − 20) 25(25 − Tm ) = 836 625 − 25Tm = 836 25Tm = −211 Tm = −8.44 Therefore, the initial temperature of the metal was -8.44°C before it was added to the calorimeter. Question 35 Question A 50 g cube of iron at 80°C is placed in 200 g of water at 20°C in a perfectly insulated container. Assuming no heat is lost to the surroundings, what will be the final temperature of the system? (Specific heat capacity of iron = 450 J/kg · K, specific heat capacity of water = 4186 J/kg · K) Solution Step 1: Calculate the heat gained by the water: The heat gained by the water can be expressed as Q = mc∆T , where m is the mass of the water, c is the specific heat capacity of water, and ∆T is the change in temperature. Given: m = 200 g = 0.2 kg, c = 4186 J/kg · K, and ∆T = Tf − 20 (since the water is originally at 20°C). Therefore, Q = (0.2)(4186)(Tf − 20). Step 2: Calculate the heat lost by the iron: Similarly, the heat lost by the iron can be calculated as Q = mc∆T , where m is the mass of the iron, c is the specific heat capacity of iron, and ∆T is the change in temperature. Given: 32 m = 50 g = 0.05 kg, c = 450 J/kg · K, and ∆T = 80 − Tf (since the iron is originally at 80°C). Therefore, Q = (0.05)(450)(80 − Tf ). Step 3: Set up the energy balance equation: Since the system is perfectly insulated, the heat gained by the water must be equal to the heat lost by the iron: (0.2)(4186)(Tf − 20) = (0.05)(450)(80 − Tf ) Step 4: Solve for the final temperature (Tf ): Expanding and simplifying the equation from step 3: 837.2Tf − 8372 = 2025 − 22.5Tf 837.2Tf + 22.5Tf = 8372 + 2025 859.7Tf = 10397 Tf ≈ 12.1C Therefore, the final temperature of the system will be approximately 12.1C. 33