## Chapter 17, example problems:

Chapter 17, example problems: (17.12) Gas thermometer. Absolute pressure = 325 mm Hg, when in contact with water at triple point. What pressure does i...
Author: Rosa Young
Chapter 17, example problems: (17.12) Gas thermometer. Absolute pressure = 325 mm Hg, when in contact with water at triple point. What pressure does it read when in contact with water at normal boiling point? Normal boiling point of water means boiling point at 1 atm pressure. That is, it means 100 °C, or 373.15K. Triple point of water is @ 0.01°C = 273.16 K. The pressure of the gas in the gas thermometer is proportional to its absolute temperature (because its volume is held fixed). Hence the new gas pressure in the gas thermometer is 325 mm Hg × (373.15 K / 273.16 K) = 444 mm Hg. (The pressure should become higher since its temperature is raised.) (17.18) Aluminum Rivets made slightly larger than the rivet holes. Cooled by dried ice (solid CO2) to −78.0 °C. Diameter of hole = 4.500 mm. Diameter of rivet @ 23.0 °C = D (unknown). Two diameters become equal when rivets cooled to −78.0 °C. Find D. We first set up an equation. From Table 17.1, αAl = 2.4 ×10-5 C°-1. Hence 4.500 mm = D × {1 − 2.4 ×10-5 C°-1 × [23.0 °C − (−78.0 °C)]} It gives D = 4.500 mm / {1 − 2.4 ×10-5 C°-1 × 101.0 C°} = 4.511 mm. (Note: In this problem, the material containing the hole is not cooled. So its diameter does not change. One simply cooled the rivet to fit into the hole. If both are cooled, then both sides of the equation must involve thermal expansion or contraction.) (17.40) 25,000 kg subway train. Initially traveling at 15.5 m/s. Slows to a stop in a station, and wait until brake is cooled. Station dimensions 65.0 m long, 20.0 m wide, 12.0 m high. All work done by the brakes transformed to heat to heat up the air in the station. Find the rise in temperature of the air. Given: Density of air = 1.20 kg / m3. Specific heat of air cair = 1020 J/kg·K. In the problem the kinetic energy of the train is converted by the brake to heat to raise the temperature of the air in the station. K.E. of train = (1/2) × 25,000 kg × (15.5 m/s) 2 = 3003125 J. Hence 3003125 J = 1020 J/kg·K × (1.20 kg / m3 × 65.0 m × 20.0 m × 12.0 m) × ΔT giving ΔT = 3003125 J / 19094400 J/K = 0.157 K. That is, the air temperature in the station will rise by 0.157 C°. (17.50) Open container holds 0.550 kg ice @ -15.0 °C. Mass of container can be ignored. Heat supplied at rate 800.0 J/min for 500.0 min. (a) When the ice begins to melt? From Table 17.3, the specific heat of ice is 2100 J/kg·C°. Hence 2100 J/kg·C° × 0.550 kg × [0 °C - (-15.0 °C)] = 800.0 J/min × x, giving x = 21.66 min. (b) When does the temperature rise above 0 °C? It happens when all ice melts. From Table 17.4, the latent heat of fusion of ice is 334 × 103 J/kg. Hence 334 × 103 J/kg × 0.550 kg = 800.0 J/min × y, giving y = 229.6 min. It should be added to 21.66 min to give a total time of 251.3 min.

(c) The heater will run for another 500.0 min − 251.3 min = 248.7 min. It will further raise the temperature to t, where, using the specific heat of water, 4190 J/kg·C°, to get 4190 J/kg·C° × 0.550 kg × (t − 0 °C) = 800.0 J/min × 248.7 min, giving t = 86.34 °C. Hence the temperature plot is t (°C) 100

At the end of 500 min t = 86.34 °C .

melting begins

melting ends

0 -15

time (min) 0

500

(Note: We use the notation t to denote temperature in °C, and T to denote temperature in K.) (17.62) Copper calorimeter can. Mass 0.100 kg. Contains 0.160 kg water and 0.0180 kg ice in thermal equilibrium @ 1 atm. 0.750 kg lead @ 255 °C is dropped in. Find final temperature t final. Assume no heat loss. From Table 17.3, ccopper = 390 J/kg·K. clead = 130 J/kg·K. Hence, 130 J/kg·K × 0.750 kg × (255 °C − t final) = 390 J/kg·K × 0.100 kg × (t final − 0 °C) + 0.0180 kg × 334 ×103 J/kg + 4190 J/kg·K × (0.0180 kg + 0.160 kg) × (t final − 0 °C), or 24862.5 J − 97.5 J/K × t final = 6012 J + (39 J/K + 745.82 J/K) × t final Hence t final = 18850.5 J / 882.32 J/K = 21.36 °C. (Note: Here all temperatures are in °C. J/kg·K is the same as J/kg·C°, and C° just mean change of temperature which is in units of °C. J/kg·K means Joule per kilogram per degree change in K, so it is the same as J/kg·C°. t final is not a change of temperature, but is an actual temperature, so it is not in C°, but is in °C.)

(17.70) Long rod. No heat loss along side. In perfect thermal contact with boiling water (at 1 atm) at one end, and with ice-water mixture at the other end. Made of 1.00 m of Cu joined end to end with L2 m of steel. 65 °C

100 °C (boiling water)

Cu

Steel

0 °C (ice-water mixture)

Cross-sectional area A = 4.00 cm2. Steady state: t at junction = 65 °C. (a) Heat flow formula: H = dQ/dt = k A (TH - TC) / L. Applying it to the Cu part: H = 385.0 W/m·C° × 4.00×10−4 m2 × (100 °C − 65 °C) / 1.00 m = 5.39 W. (W = J/s) (b) The same heat current also flows through the steel part. So 5.39 W = 50.2 W/m·C° × 4.00×10−4 m2 × (65 °C − 0 °C) / L2 , giving L2 = 50.2 W/m·C° × 4.00×10−4 m2 × 65 C° / 5.39 W = 0.242 m. (17.86) 108 cm3 ethanol @ −10.0 °C is poured into a graduated glass cylinder @ 20.0 °C, filling it to the top. Specific heat of glass = 840 J/kg·K. Its coefficient of expansion = 1.2 ×10-5 K-1 and Its mass = 0.110 kg. The mass of ethanol = 0.0873 kg. (a) Final temperature of ethanol? From Table 17.3, the specific heat of ethanol = 2428 J/kg·C°. Hence: 2428 J/kg·C° × 0.0873 kg × [tfinal − (−10.0 °C)] = 840 J/kg·C° × 0.110 kg × (20.0 °C − tfinal), or 211.96 J/C° × t final + 2119.6 J = 1848 J − 92.4 J/C° × tfinal , or 1848 J − 2119.64 J = (211.96 J/C° + 92.4 J/C°) × tfinal . Hence t final = − 271.64 J / 304.36 J/C° = − 0.892 °C. Note: (i) The absolute temperature unit, K, is not the same as the centigrade or Celsius temperature unit, °C, but “per K” is the same as “per C°”, since only change of either temperature is involved. Thus the specific heat unit in Table 17.3, viz., J/kg·K, can be changed to J/kg·C°. Alternatively, you can change every temperature in the problem to absolute temperature by adding 273.15 to its value. The unit would then be “K”. After obtaining the final absolute temperature in K, you must still subtract 273.15 to get the final temperature in °C. (ii) No latent heats are involved in this problem since neither ethanol nor glass melts or evaporates in this temperature range. (iii) To get three significant digits for the final answer, you need to keep more than three significant digits in the calculation, whenever there are possible cancellations of the leading digit(s). The example here is the calculation: 1848 J − 2119.64 J = −271.64 J, where the 0.6 part has been promoted from hte fifth digit to the fourth digit due to the cancllation of the leading digit. Whenever you are not sure whether there might be cancelation of leading digit(s), keep more significant digits in the calculation. (b) How much overflow? Assuming that the heat exchange needed to reach

equilibrium happens before overflow happens (which is not true in practice), then we need the thermal expansion coefficient of ethanol. From Table 17.2, it is βethanol = 75 ×10-5 C°-1. That of glass is given in the problem as being βglass = 1.2 ×10-5 C°-1. Thus the volume of ethanol will increase by 108 cm3 × 75 ×10-5 C°-1 × [(− 0.892 °C) − (−10.0 °C)] = 0.7377 cm3, and the volume of glass will shrink (because its temperature has decreased) by 108 cm3 × 1.2 ×10-5 C°-1 ×[ 20.0 °C − (− 0.892 °C)] = 0.0271 cm3. So the overflow amount of ethanol is 0.7377 cm3 + 0.0271 cm3 = 0.7648 cm3. (Actually the overflow amount will be less than this because some ethanol will overflow before it reaches the final temperature. So this problem is, strictly speaking, much more difficult than what we have done above, and its true answer requires calculus to obtain.) (17.100) Given: Molar heat capacity Cunknown substance = 29.5 J/mol·K + (8.20 ×10-3 J/mol·K2) T . Since this heat capacity is not a constant, but varies with T, we must now use dQ/dt = dQ/dT = nC, and perform integration: 273.15 + 227

Q = Û273.15 + 27 (3 mol) × [29.5 J/mol·K + (8.20 ×10-3 J/mol·K2) T] dT = 88.5 J / K × [(273.15 + 227) K - (273.15 + 27) K] + (1/2) × (24.60 ×10-3 J / K2) × [(273.15 + 227)2 K2 - (273.15 + 27) 2 K2] = 17700 J + 1968.74 J = 19669 J (Note that since C is given as a function of T, we must integrate with respect to T, and not t. That is, we must convert all temperatures to absolute temperatures before we do the integration.) (17.102) Hot water vs steam heating. Water delivered to radiator @ 70.0 °C (158.0 °F), and leaves @ 28.0 °C (82.4 °F). The system is to be replaced by a steam system in which steam at atmospheric pressure condenses in the radiators and the condensed steam leaves the radiators @ 35.0 °C (95.0 °F). How many kg of steam will aupply the same amount of heat as was delivered by 1.00 kg of hot water in the first system? Let it be x kg of steam, and it is @ 100.0 °C (212 °F), and not super-heated to higher temperatures. Then, using the data in Table 17.4, x kg ×[2256 ×103 J / kg + 4190 J / kg·C° ×(100 °C − 35 °C)] = 1 kg ×[4190 J / kg·C° ×(70 °C − 28 °C)] , giving x kg = 175980 J / 2528350 J/kg = 0.0696 kg. This we see that 0.0696 kg of steam can already replace 1 kg of water! This is of course because of the very large heat of vaporization of water. (17.116) Radiation from the sun I = 1.50 kW/m2. Distance between the sun and the earth = 1.50 ×1011 m. Radius of the sun = 6.96 ×108 m. (a) Rate of radiation of energy per unit area from the sun’s surface = 1.50 kW/m2 × [4π (1.50 ×1011 m) 2 / 4π (6.96 ×108 m) 2] = 6.97 ×104 kW/m2. (or 69.7 MW/m2.)

(b) If the sun radiates as an ideal blackbody, find the surface temperature of the sun. 6.97 ×107 W/m2 = 5.6704 ×10-8 W / m2 ·K2 × T 4, giving T = (6.97 ×107 W/m2 / 5.6704 ×10-8 W / m2 ·K2 ) 1/4 = 5921 K. (This is the same as 5648 °C, or 10198.4 °F.)