Problem Set 9 Due: see website for due date Chapter 12: Thermal Properties of Matter Q12.1: Solution Q12.4. Reason: Q12.20: Solution Q12.20.

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1 Problem Set 9 Due: see website for due date Chapter 1: Thermal Properties of Matter Questions: 4, 0, 1, 5 Exercises & Problems: 9, 16, 5, 54, 57, 6, 80, 8, 98 Q1.1: If you buy a sealed bag of potato chips in Miami and drive with it to Denver, where the atmospheric pressure is lower, you will find that the bag gets very "puffy." Explain why. Q1.4. Reason: From the ideal gas law, the lower the pressure of a container of gas, the higher the volume. The pressure inside the bag equals the pressure outside. So when the atmospheric pressure goes down, the pressure in the bag goes down and the volume goes up. Q1.0: A student is heating chocolate in a pan on the stove. He uses a cooking thermometer to measure the temperature of the chocolate and sees it varies as shown. Describe what is happening to the chocolate in each of the three portions of the graph. Q1.0. Reason: The chocolate starts at some temperature. As it s heated the temperature rises to a point where the chocolate changes phase. Assuming the chocolate started as a solid, the chocolate melts during the second portion of the graph. During melting, the temperature is constant. After all the chocolate melts, the temperature rises as the liquid chocolate is heated. Assess: Compare to Figure 1.1. Q1.1: If you bake a cake at high elevation, where atmospheric pressure is lower than at sea level, you will need to adjust the recipe. You will need to cook the cake for a longer time, and you will need to add less baking powder. (Baking powder is a leavening agent. As it heats up, it releases gas bubbles that cause the cake to rise.) Explain why those adjustments are necessary. Q1.1. Reason: At higher elevations the air pressure is lower. The amount of leavening agent should be reduced, since bubbles will form more readily in the lower pressure. Water will also vaporize at a lower temperature, so the baking temperature should be decreased. Decreasing the temperature requires an increase in the baking times. Assess: Most non-physicist cooks will suggest increasing the amount of water in a recipe instead of increasing baking time. Q1.5: If you live somewhere with cold, clear nights, you may have noticed some mornings when there was frost on open patches of ground but not under trees. This is because the ground under trees does not get as cold as open ground. Explain how tree cover keeps the ground under trees warmer. 1

2 Q1.5. Reason: The trees help prevent the energy from being radiated out into space on a cold clear night; the trees reflect back down some of the infrared radiation and keep the ground under them warmer. In contrast, the open ground radiates its thermal energy into space without the blanket of the trees or clouds to keep the energy in. Assess: Gardeners in northern climes know to cover their plants on clear fall nights to keep the radiation in and keep the plants from freezing. These early first frosts in the fall are even called radiation frosts. They take place on clear nights with calm winds. Another type, called advective freeze, occurs when very cold air moves in (by convection); advective freezes can take place with winds and clouds present, and are much harder to protect plants against. P1.9: Many cultures around the world still use a simple weapon called a blowgun, a tube with a dart that fits tightly inside. A sharp breath into the end of the tube launches the dart. When exhaling forcefully, a healthy person can supply air at a gauge pressure of 6.0 kpa. What force does this pressure exert on a dart in a 1.5-cm-diameter tube? P1.9. Prepare: Equation 1.9 gives the force due to a pressure applied over an area. The preliminary calculation is to compute the cross section area of the tube. We are given p 6.0 kpa m 4 A R m 4 F pa (6.0 kpa)( m ) 1.1 N Assess: 1.1 N is not a large force, but it is pushing a light dart, so the dart achieves a respectable acceleration. P1.16: A cylinder contains.0 L of oxygen at 00 K and.4 atom. The gas is heated, causing a piston in the cylinder to move outward. The heating causes the temperature to rise to 600 K and the volume of the cylinder to increase to 9.0 L. What is the final gas pressure? P1.16. Prepare: The gas is assumed to be ideal. As a general rule, we must convert all quantities into SI units. In the present case, however, we will be dealing with the ratio of the final and the initial value of V, so we do not have to convert L into m. The before-and-after relationship of an ideal gas is pv 1 1 pv V1 T.0 L 600 K p p1 (.4 atm) 1.6 atm T1 T V T1 9.0 L 00 K P1.5: The maximum amount of water an adult in temperature climates can perspire in one hour is typically 1.8 L. However, after several weeks in a tropical climate the body can adapt, increasing the maximum perspiration rate to.5 L/h. At what rate, in watts, is energy being removed when perspiring that rapidly? Assume all the perspired water evaporates. At body temperature, the heat of vaporization of water is L v = J/kg. P1.5. Prepare: We ll compute the energy necessary to evaporate.5 L and then divide by an hour to get the rate. One liter of water has a mass of one kilogram. 5 6 Q ML f (.5 kg)(410 J/K) J The rate is the energy divided by the time. 6 Q J 1h rate 00 W t 1h 600 s Assess: That is an impressive power output, but necessary to keep cool in tropical climates. Notice that the value given for L f at body temperature is different from the one given in Table 1.5 for standard temperature (0C).

3 P1.54: A copper-bottomed kettle, its bottom 4 cm in diameter and.0 mm thick, sits on a burner. The kettle holds boiling water, and energy flows into the water from the kettle bottom at 800 W. What is the temperature of the bottom surface of the kettle? P1.54. Prepare: The bottom of the interior of the kettle is the same temperature of the boiling water, 100C. Equation 1.1 can be used. Solving Equation 1.1 for the temperature difference, Q L.010 m 1 T (800 W) K t ka (400 W/(m K)) (0. 1m) The bottom of the kettle is only 0.1 K hotter than the interior. Assess: This result makes sense. P1.57: Seals may cool themselves by using thermal windows, patches on their bodies with much higher than average surface temperature. Suppose a seal has a 0.00 m thermal window at a temperature of 0 o C. If the seal's surroundings are a frosty 10 C, what is the net rate of energy loss by radiation? Assume an emissivity equal to that of a human (e = 0.97). P1.57. Prepare: The rate of net energy loss by radiation is given by Equation 1.4. Qnet 4 4 e A( T T 0) t where T 0 is the termperature of the surroundings. We are given T 0C 0 K, T 0 10C 6 K, and A 0.00 m. We are told to assume the emissivity of seal skin is the same as human skin; the text gives this vaule as e The textbook gives Stefan s constant as W/(m K ). Qnet e AT T 0 (0.97)( W/(m K ))(0.00 m )[(0 K) (6 K) ] 6.0 W t Assess: 6 W isn t a lot, but it is sufficient to cool the seal when the surroundings are very cool. If there were no thermal windows the seal would have difficulty regulating its temperature. P1.6: A 15-cm-diameter compressed-air tank is 50 cm tall. The pressure at 0 o C is 150 atm. a. How many moles of air are in the tank? b. What volume would this air occupy at STP? P1.6. Prepare: Treat the air in the compressed-air tank as an ideal gas and use Equation 1.1 to find n. We will, however, need to convert pressure and temperature to SI units using 1 atm Pa and T (in K) T (in C) 7. Also note that r h is the volume of a cylinder. (a) From the ideal-gas law pv nrt, the number of moles n is 5 pv p( r h) (150 atm)( Pa/1atm)[ (0.075 m) (0.50 m)] n 55.1 mol RT RT (8.1J/(mol K))[(7 0) K] which will be reported as 55 mol. (b) At STP, the ideal-gas law yields nrt (55.1 mol)(8.1 J/(mol K))(7 K) V 1. m 5 p Pa Assess: The volume of the compressed air tank is (r )h m. The volume at STP is 140 times the volume of the tank. That is, the air is compressed 140 times compared to STP values, which does not look unreasonable. P1.80: Susan, whose mass is 68 kg, climbs 59 m to the top of the Cape Hatteras lighthouse.

4 a. During the climb, by how much does her potential energy increase? b. For a typical efficiency of 5%, what metabolic energy does she require to complete the climb? c. When exercising, the body must perspire and use other mechanisms to cool itself to avoid potentially dangerous increases on body temperature. If we assume that Susan doesn't perspire or otherwise cool herself and that all of the "lost" energy goes into increasing her body temperature, by how much would her body temperature increase during this climb? P1.80. Prepare: We will need the formula for gravitational potential energy, Ug mgh, as well as a formula which relates a change in temperature to a change in thermal energy. Equation 1.4 gives the heat needed to change the temperature of an object. In this case, the heat produced by the body s metabolism is turned into thermal energy so we write Eth mc T. We will use the mammalian specific heat from Table 1.4. (a) Susan s change in potential energy is given by the following equation: Ug mgh (68 kg)(9.8 m/s )(59 m) 900 J 9000 J (b) If this increase in potential energy represents 5% of the total energy used by her body, then she has used four times this much energy, or 157 kj 160 kj. (c) 75% of the energy used is wasted and goes to thermal energy so Eth (0.75)(157 kj) 118 kj. We can solve the equation E mc T for T to obtain th Eth J T mc (68 kg)(400 J/(kg K)) 0.51 K Assess: This increase is about 0.5 C which can be converted to 0.9 F. This doesn t seem like much, but on the other hand, this exercise was less strenuous than it might seem. The energy spent, 160 kj can be converted to about 8 dietary Calories. The same amount of energy would be spent in walking about two fifths of a mile. P1.8: A 100 kg car traveling at 60 mph quickly brakes to a halt. The kinetic energy of the car is converted to thermal energy of the disk brakes. The brake disks (one per wheel) are iron disks with a mass of 4.0 kg. Estimate the temperature rise in each disk as the car stops. (c iron = 449 J/(kg K)) P1.8. Prepare: The entire kinetic energy of the car goes into heating the four brakes. Equation 1.1 applies. Table 1.4 gives the specific heat of iron, c 449 J/(kg K). We are given the mass of each brake disk, m 4.0 kg and the mass of the car M 100 kg. The speed of the car is v 60 mph, which we convert to SI units (keeping one extra significant figure in the intermediate calculation) m/s v 60 mph 6.8 m/s 1mph Setting the kinetic energy equal to the heat going into the four brakes gives 1 Q 4mcT Mv Now solve for the temperature increase. 4

5 1 Mv 1 (100 kg)(6.8 m/s) T 60 K 8 mc 8 (4.0 kg)(449 J/(kg K)) Assess: This is a significant rise in temperature, but brakes really do heat up quite a bit. The kinetic energy of the car must go somewhere. One other option is to let the translational kinetic energy of the car be transformed into rotational kinetic energy of a flywheel, so that it can be recovered after the light turns green. Careful examination of the units shows they cancel properly. P1.98: Suppose you go outside in your fiber-filled jacket on a windless but very cold day. The thickness of the jacket is.5 cm, and it covers 1.1 m of your body. The purpose of fiber- or down-filled jackets is to trap a layer of air, and it's really the air layer that provides the insulation. If your skin temperature is 4 o C while the air temperature is 0 o C, at what rate is heat being conducted through the jacket and away from your body? P1.98. Prepare: Assume all the insulation comes from the air layer. Assume the value for the thermal conductivity of air in Table 1.7 is a good approximation for these conditions. Equation 1.1 applies. Using Equation 1.1, Q ka (0.06 W/ (m K)(1.1m ) T (4 C ( 0 C)) 6 W t L 0.05 m Assess: This is a large amount of power, even though air has the lowest thermal conductivity of all the materials in Table