Chapter 18: Temperature, Heat & the First Law of Thermodynamics Example Questions and Problems

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1 Chapter 18: Temperature, Heat & the First Law of Thermodynamics Example Questions and Problems E Q W W pdv Q mct Q ml Q ml int f f V V Q T T Q H L 4 L LT P ka P AT rad t L t Example 18.1 a. The desert sand is very hot in the day and very cool at night. What does this tell you about its specific heat? How about Space Shuttle tiles? b. In some household air itioners used in dry climates, air is cooled by blowing it through a water-soaked filter, evaporating some of the water. How does this cool the air? Would such a system work well in a high-humidity climate? Why or why not? c. Desert travelers sometimes keep water in a canvas bag. Some water seeps through the bag and evaporates. How does this cool the water inside? d. Explain the wind-chill effect that meteorologists mention during cold weather. e. In Alaska, a lack of snow allowed the ground to freeze down to a depth of about one meter, causing buried water pipers to freeze and burst. Why did a lack of snow lead to this situation? f. When you first step out of the shower, you feel cold. But as soon as you are dry you feel warmer, even though the room temperature does not change. Why? g. Why does a black charcoal glow bright red in a fire? Example 18.1 A thermodynamic system is taken from an initial state A to another state B and back again to A, via state C, as shown by path ABCA in the pvdiagram. a. (i) Complete the table filling in either ± for the sign of each thermodynamic quantity associated with each step of the cycle. (ii) Draw an energy bar diagram for each process. b. Calculate the numerical value of the work done by the system for the complete cycle ABCA. Example 18.2 A sample of gas expands from 1.0 m 3 to 4.0 m 3 while its pressure decreases from 40 Pa to 10 Pa. Assume E int,f > E int,0. (a) (i) Rank the works, heats, and internal energies for processes A, B, and C, greatest first. (ii) Draw an energy bar diagram for each process. (b) (iii) How much work is done by the gas if its pressure changes with volume via path A, B, and C? Does this answer agree with part (a)? 18-1

2 Example 18.3 A 50.0 g aluminum (c Al = 900 J/kg K) disk at 300 o C is placed in 200 cm 3 (or kg) of ethyl alcohol (c ethyl = 2400 J/kg K) at 10.0 o C, then quickly removed. The aluminum temperature is found to have dropped to 120 o C. What is the new temperature of the ethyl alcohol? Example 18.4 Two 50 g ice cubes are dropped into 200 g of water in a thermally insulated container. The water is initially at 25 o C and the ice comes directly from a freezer at 15 o C. a. (i) Physically explain if only part of the ice melts or all of the ice melts? Now draw a (ii) T vs. Q diagram and an (iii) energy bar diagram for the situation. b. What is the final temperature of the water? If only part of the ice melts, how much of it is left? Example 18.5 a. The radius of the sun is m. At the distance of the earth, m, the intensity of solar radiation (measured by satellites above the atmosphere) is 1370 W/m 2. What is the temperature of the sun s surface? b. The sun s intensity at the distance of the earth has this energy is reflected by water and clouds by 30%; 70% is absorbed. What would be the earth s average temperature (in o C) if the earth had no atmosphere? The emissivity of the surface is very close to

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4 Example A A steel rod is cm in diameter at 25 o C. A brass ring has an interior diameter of cm at 25 o C. At what common temperature will the ring just slide onto the rod? Solution After the change in temperature the diameter of the steel rod is d s = d s0 + sd s0 T and the diameter of the brass ring is d b = d b0 + bd b0 T, where d s0 and d b0 are the original diameters, s and b are the coefficients of linear expansion, and T is the change in temperature. The rod just fits through the ring if d s = d b. This means Therefore, d s0 + sd s0 T = d b0 + bd b0 T. Example B a. What is the rate of energy loss in watts per square meter through a glass window 3.0 mm thick I the outside temperature is 20 o F and the inside temperature is +72 o F? b. A storm window having the same thickness of glass is installed parallel to the first window, with an air gap of 7.5 cm between the two windows. What now is the rate of energy loss if uction is the only important energy-loss mechanism? Solution a. The power radiated through the glass window per are is given by T T P T T H C H C P ka k L A L where the uctivity of glass I looked up and is given by 1.0 W/m K. Converting the temperature difference into the Celsius scale, we get H C T T 72F 20F 92 F (92) 51.1C 51.1K converting to celsius 5 9 So the power radiated via uction per square meter is then P T T H C 51.1C k 1.0 W m K W/m 3 A L m P A 4 2 b. The energy now passes in succession through 3 layers, one of air and two of glass. The power transfer is the same in each layer and given by AT T H C P Lk where the sum in the denominator is over the layers. If L g is the thickness of a glass layer, L a is the thickness of the air layer, k g is the thermal uctivity of glass, and k a is the thermal uctivity of air, then the denominator is L 2L L 2L k L k k k k k k g a g a a g g a a g Therefore, the heat ucted per unit area occurs at the following rate:. 18-4

5 P T T k k A 2L k L k H C a g g a a g C W m K 1.0 W m K P 18 W m m W m K 0.075m 1.0 W m K A