Heat Study Solved Example

HEAT Page # 17 SOLVED EXAMPLE Ex.1 Show that the coefficient of area expansions, (A/A)/?T, of a rectangular sheet of the solid is twice its linear expansivity,1 Ans. Consider a rectangular sheet of the solid material of length a and breadth b When the temperature increases by ?T, a increases by a = 1aT and b increases by ? b = 1bT. From Fig. 11.8, the increase in area A = A1 +A2 + A3 A = ab + ba + (a) (b) = a 1bT + b1 a T + (1)2 ab (T)2 = 1 ab T (2 + 1 T) = 1 A T (2 + 1T) Since 1 . 10–5 K–1, the product 1T for fractional temperature is small in comparision with 2 and may be neglected. Hence,  A  1  21    A  A Ex.2 A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27 °C. To what temperature should the ring be heated so as to fit the rim of the wheel? Ans. Given, T1 = 27°C LT1 = 5.231 m LT2 = 5.243 m So, LT2 =LT1 [1+1 (T2 – T1)] 5.243 m = 5.231 m [1 + 1.20 × 10–5 K–1 (T2 – 27°C)] or T2 = 218°C. Ex.3 A sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100°C. It is then immediately transfered to 0.14 kg copper calorimeter containing 0.25 kg of water at 20°C. Thetemperature of water rises and attains a steady state at 23°C. Calculate the specific heat capacity of aluminium. Ans. In solving this example we shall use the fact that at a steady state, heat given by an aluminium sphere will be equal to the heat absorbed by the water and calorimeter. Mass of aluminium sphere (m1) = 0.047 kg Initial temp. of aluminium sphere = 100°C Final temp. = 23°C Change in temp (T ) = (100°C - 23°C) = 77°CLet specific heat capacity of aluminium be SAl. The amount of heat lost by the aluminiumsphere =m1sAlT = 0.047kg × SAl × 77ºC Mass of water (m2) = 0.25 kg Mass of calorimeter (m3) = 0.14 kg Initial temp. of water and calorimeter = 20°C Final temp. of the mixture = 23°C Change in temp. (T2) = 23°C – 20°C = 3°C Specific heat capacity of water (sw) = 4.18 × 103 J kg–1 K–1 Specific heat capacity of copper calorimeter = 0.386 × 103 J kg–1 K–1 The amount of heat gained by water and calorimeter = m2 sw T2 + m3SCu T2 = (m2 + m3) (T2) = 0.25 kg ×4.18 × 103 J kg–1 K–1 + 0.14 kg × 0.386 × 103 J kg–1 K–1) (23°C – 20°C) state heat lost by the aluminiumsphere = heat gained by water + heat gained by calorimeter. So, 0.047 kg × SAl × 77°C = (0.25 kg × 4.18 × 103 J kg–1 K–1+ 0.14 kg × 0.386 × 103 J kg–1 K–1)(3°C) sAl = 0.911 kJ kg–1 K–1 Ex.4 When 0.15 kg of ice of 0°C mixed with 0.30 kg of water at 50°C in a container, the resulting temperature is 6.7°C. Calculate the heat of fusion of ice. ( Swater = 4186 J kg–1 K–1) Ans. Heat lost by water = MSw (f – i)w = (0.30 kg) (4186 J kg–1 K–1) (50.0°C – 6.7°C) = 54376.14 J Heat required to melt ice = m2Lf = (0.15 kg) Lf Heat required to raise temperature of ice water to final temperature = mISw (f – i)I = (0.15 kg) (4186 J kg–1 K–1) (6.7°C – 0°C) = 4206.93 J Heat lost = heat gained 54376.14 J = (0.15 kg ) Lf + 4206.93 J Lf = 3.34 × 105 J kg–1. Ex.5 Calculate the heat required to convert 3 kg of ice at –12 °C kept in a calorimeter to steam at 100 °C at atmospheric pressure. Given specific heat capacity of ice = 2100 J kg–1 K–1, specific heat capacity of water = 4186 J kg– 1 K–1, latent heat of fusion of ice = 3.35 × 105 J kg–1 and latent heat of steam = 2.256 ×106 J kg–1. Ans. We have Mass of the ice, m = 3 kg specific heat capacity of ice, sice = 2100 J kg–1 K–1 specific heat capacity of water, swater = 4186 J kg–1 K–1 HEAT Page # 18 latent heat of fusion of ice, Lf ice = 3.35 × 105 J kg–1 latent heat of steam, Lsteam = 2.256 × 106 J kg–1 Now, Q = heat required to convert 3 kg of ice at –12 °C to steam at 100 °C, Q1 = heat required to convert ice at –12 °C to ice at 0°C. = m sice T1 = (3 kg) (2100 J kg–1.K–1) [0–(–12)]°C = 75600 J Q2 = heat required to melt ice at 0°C to water at 0°C = m Lf ice = (3 kg) (3.35 × 105 J kg) = 1005000 J Q3 = heat required to convert water at 0 °C to water at 100 °C. = mSw T2 = (3kg) (4186J kg–1 K–1) (100°C) = 1255800 J So, Q4 = heat required to convert water at 100 °C to steam at 100°C. = m Lsteam = (3 kg) (2.256 × 106 J/kg–1) = 6768000 J Q = Q1 + Q2 + Q3 + Q4 = 75600J + 1005000 J + 1255800 J + 6768000 J = 9.1×106 J Ex.6 What is the temperature of the steel-copper junction in the steady state of the system shown in Fig. 11.15. Length of the steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace = 300°C, temperature of the other end = 0°C. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel = 50.2 J s–1 m–1K–1; and of copper = 385 J s–1m–1K–1). Ans. The insulating material around the rods reduces heat loss from the sides of the rods. Therefore, heat flows only along the length of the rods. Consider any cross section of the rod. In the steady state, heat flowing into the element must equal the heat flowing out of it; otherwise there would be a net gain or loss of heat by the element and its temperature would not be steady. Thus in the steady state, rate of heat flowing across a cross section of the rod is the same at every point along the length of the combined steelcopper rod. Let T be thetemperature of the steelcopper junction in the steady state. Then, K1 A1 (300 T) L1 K A (T - 0) = 2 2 L2 where 1 and 2 refer to the steel and copper rod respectively. For A1 = 2 A2, L1 = 15.0 cm, L2 = 10.0 cm, K1 = 50.2 J s–1 m–1 K–1, K2 = 385 J s–1 m–1 K–1, we have 50.2  2(300  T) 385T  15 10 which gives T = 44.4°C Ex.7 The density of water is 1000 kg m–3. The density of water vapour at 100 °C and 1 atm pressure is 0.6 kg m–3. The volume of a molecule multiplied by the total number gives ,what is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure. Ans. For a given mass of water molecules, the density is less if volume is large. So the volume of the vapour is 1000/0.6 = /(6 ×10-4 ) times larger. If densities of bulk water and water molecules are same, then the fraction of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same amount, i.e. 6×10-4. ?? Ex.8 Estimate the volume of a water molecule using the data in Example 13.1. Ans. In the liquid (or solid) phase, the molecules of water are quite closely packed. The density of water molecule may therefore, be regarded as roughly equal to the density of bulk water = 1000 kg m–3. To estimate the volume of a water molecule, we need to know the mass of a single water molecule. We know that 1 mole of water has a mass approximately equal to (2 + 16)g = 18 g = 0.018 kg. Since 1 mole contains about 6 × 1023 molecules (Avogadro’s number), the mass of a molecule of water is (0.018)/(6 × 1023) kg = 3 × 10–26 kg. Therefore, a rough estimate of the volume of a water molecule is as follows : Volume of a water molecule = (3 × 10–26 kg)/ (1000 kg m–3) = 3 × 10–29 m3 = (4/3) ? (Radius)3 Hence, Radius ? 2 ×10-10 m = 2 Å ?? Ex.9 What is the average distance between atoms (interatomic distance) in water? Use the data given in Examples 13.1 and 13.2. Ans. A given mass of water in vapour state has 1.67 × 103 times the volume of the same mass of water in liquid state (Ex. 13.1). This is also the increase in the amount of volume available for each molecule of water. When volume increases by 103 times the radius increases by V1/3 or 10 times, i.e., 10 × 2 Å = 20 Å. So the average distance is 2 × 20 = 40 Å. HEAT Page # 19 Ex.10 A vessel contains two nonreactive gases : neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3:2. Estimate the ratio of (i) number of molecules and (ii) mass density of neon and oxygen in the vessel. Atomic mass of Ne = 20.2 u, molecular mass of O2 = 32.0 u. Ans. Partial pressure of a gas in a mixture is the pressure it would have for the same volume and temperature if it alone occupied the vessel. (The total pressure of a mixture of non-reactive gases is the sum of partial pressures due to its constituent gases.) Each gas (assumed ideal) obeys the gas law. Since V and T are common to the two gases, we have P1V = 1 RT and P2V = 2 RT, i.e. (P1/P2) = (1 /2). Here 1 and 2 refer to neon and oxygen respectively. Since (P1/P2) = (3/2) (given), (1/2) = 3/2. (i) By definition 1 = (N1/NA ) and 2 = (N2/NA) where N1 and N2 are the number of molecules of 1 and 2, and NA is the Avogadro’s number. Therefore, (N1/N2) = (1 / 2) = 3/2. (ii) We can also write 1 = (m1/M1) and 2 = (m2/M2) where m1 and m2 are the masses of 1 and 2; and M1 and M2 are their molecular masses. (Both m1 and M1; as well as m2 and M2 should be expressed in the same units). If 1 and 2 are the mass densities of 1 and 2 respectively, we have 1 m1 /V 2 m2 /V = m1 m2 = μ1 μ2 × M1 M2 3 20.2   0.947 2 32.0 Ex.11 A flask contains argon and chlorine in the ratio of 2:1 by mass. The temperature of the mixture is 27 °C. Obtain the ratio of (i) average kinetic energy per molecule, and (ii) root mean square speed vrms of the molecules of the two gases. Atomic mass of argon = 39.9 u; Molecular mass of chlorine = 70.9 u. Ans. The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to (3/2) kBT. It depends only on temperature, and is independent of the nature of the gas. (i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1 : 1. (ii) Now ½ m V2rms2 = average kinetic energy per molecule = (3/2) ) kBT where m is the mass of a molecule of the gas. Therefore, where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides, Ex.12 Uranium has two isotopes of masses 235 and 238 units. If both are present in Uranium hexafluoride gas which would have the larger average speed ? If atomic mass of fluorine is 19 units, estimate the percentage difference in speeds at any temperature. Ans. At a fixed temperature the average energy = ½ m < v 2 > is constant. So smaller the mass of the molecule, faster will be the speed. The ratio of speeds is inversely proportional to the square root of the ratio of the masses. The masses are 349 and 352 units. So v349 / v352 = ( 352/ 349)1/2 = 1.0044 . Hence dv difference = 0.44 %. v Ex.13 (a) When a molecule (or an elastic ball) hits a ( massive) wall, it rebounds with the same speed. When a ball hits a massive bat held firmly, the same thing happens. However, when the bat is moving towards the ball, the ball rebounds with a different speed. Does the ball move faster or slower? (Ch.6 will refresh your memory on elastic collisions.) (b) When gas in a cylinder is compressed by pushing in a piston, its temperature rises. Guess at an explanation of this in terms of kinetic theory using (a) above. (c) What happens when a compressed gas pushes a piston out and expands. What would you observe ? (d) Sachin Tendulkar uses a heavy cricket bat while playing. Does it help him in anyway ? Ans. (a) Let the speed of the ball be u relative to the wicket behind the bat. If the bat is moving towards the ball with a speed V relative to the wicket, then the relative speed of the ball to bat is V + u towards the bat. When the ball rebounds (after hitting the massive bat) its speed, relative to bat, is V + u moving away from the bat. So relative to the wicket the speed of the rebounding ball is V + (V + u) = 2V + u, moving away from the wicket. So the ball speeds up after the collision with the bat. The rebound speed will be less than u if the bat is not massive. For amolecule this would imply an increase in temperature. You should be able to answer (b) (c) and (d) based on the answer to (a). (Hint: Note the correspondence, piston bat, cylinder wicket, molecule ball.)