Chapter 17
University Physics with Modern Physics · 77 exercises
Problem 76
A surveyor's 30.0-m steel tape is correct at 20.0\(^\circ\)C. The distance between two points, as measured by this tape on a day when its temperature is 5.00\(^\circ\)C, is 25.970 m. What is the true distance between the points?
5 step solution
Problem 77
A metal rod that is 30.0 cm long expands by 0.0650 cm when its temperature is raised from 0.0\(^\circ\)C to 100.0\(^\circ\)C. A rod of a different metal and of the same length expands by 0.0350 cm for the same rise in temperature. A third rod, also 30.0 cm long, is made up of pieces of each of the above metals placed end to end and expands 0.0580 cm between 0.0\(^\circ\)C and 100.0\(^\circ\)C. Find the length of each portion of the composite rod.
6 step solution
Problem 78
On a cool (4.0\(^\circ\)C) Saturday morning, a pilot fills the fuel tanks of her Pitts S-2C (a two-seat aerobatic airplane) to their full capacity of 106.0 L. Before flying on Sunday morning, when the temperature is again 4.0\(^\circ\)C, she checks the fuel level and finds only 103.4 L of gasoline in the aluminum tanks. She realizes that it was hot on Saturday afternoon and that thermal expansion of the gasoline caused the missing fuel to empty out of the tank's vent. (a) What was the maximum temperature (in \(^\circ\)C) of the fuel and the tank on Saturday afternoon? The coefficient of volume expansion of gasoline is \(9.5 \times 10{^-}{^4} K{^-}{^1}\). (b) To have the maximum amount of fuel available for flight, when should the pilot have filled the fuel tanks?
6 step solution
Problem 82
A typical doughnut contains 2.0 g of protein, 17.0 g of carbohydrates, and 7.0 g of fat. Average food energy values are 4.0 kcal/g for protein and carbohydrates and 9.0 kcal/g for fat. (a) During heavy exercise, an average person uses energy at a rate of 510 kcal/h. How long would you have to exercise to 'work off' one doughnut? (b) If the energy in the doughnut could somehow be converted into the kinetic energy of your body as a whole, how fast could you move after eating the doughnut? Take your mass to be 60 kg, and express your answer in m/s and in km/h.
8 step solution
Problem 83
BIO Shivering. Shivering is your body's way of generating heat to restore its internal temperature to the normal 37\(^\circ\)C, and it produces approximately 290 W of heat power per square meter of body area. A 68-kg, 1.78-m-tall woman has approximately 1.8 m\(^2\) of surface area. How long would this woman have to shiver to raise her body temperature by 1.0 C\(^\circ\), assuming that the body loses none of this heat? The body’s specific heat capacity is about 3500 J/kg \(\cdot\) K.
3 step solution
Problem 86
CP A person of mass 70.0 kg is sitting in the bathtub. The bathtub is 190.0 cm by 80.0 cm; before the person got in, the water was 24.0 cm deep. The water is at 37.0\(^\circ\)C. Suppose that the water were to cool down spontaneously to form ice at 0.0\(^\circ\)C, and that all the energy released was used to launch the hapless bather vertically into the air. How high would the bather go? (As you will see in Chapter 20, this event is allowed by energy conservation but is prohibited by the second law of thermodynamics.)
1 step solution
Problem 87
(a) A typical student listening attentively to a physics lecture has a heat output of 100 W. How much heat energy does a class of 140 physics students release into a lecture hall over the course of a 50-min lecture? (b) Assume that all the heat energy in part (a) is transferred to the 3200 m\(^3\) of air in the room. The air has specific heat 1020 J/kg \(\cdot\) K and density 1.20 kg/m\(^3\). If none of the heat escapes and the air conditioning system is off, how much will the temperature of the air in the room rise during the 50-min lecture? (c) If the class is taking an exam, the heat output per student rises to 280 W. What is the temperature rise during 50 min in this case?
4 step solution
Problem 88
The molar heat capacity of a certain substance varies with temperature according to the empirical equation $$C = 29.5 J/mol \cdot K + (8.20 \times 10{^-}{^3} J/mol \cdot K{^2})T$$ How much heat is necessary to change the temperature of 3.00 mol of this substance from 27\(^\circ\)C to 227\(^\circ\)C? (Hint: Use Eq. (17.18) in the form d\(Q\) = n\(C\) d\(T\) and integrate.)
7 step solution
Problem 89
If the air temperature is the same as the temperature of your skin (about 30\(^\circ\)C), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical 70-kg person's body produces energy at a rate of about 500 W due to metabolism, 80% of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is \(2.42 \times 10{^6} J/kg\). (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many 750-mL bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is 1.0 kg.)
3 step solution
Problem 91
The African bombardier beetle (Stenaptinus insignis) can emit a jet of defensive spray from the movable tip of its abdomen (Fig. P17.91). The beetle's body has reservoirs containing two chemicals; when the beetle is disturbed, these chemicals combine in a reaction chamber, producing a compound that is warmed from 20\(^\circ\)C to 100\(^\circ\)C by the heat of reaction. The high pressure produced allows the compound to be sprayed out at speeds up to 19 m/s 168 km/h2, scaring away predators of all kinds. (The beetle shown in Fig. P17.91 is 2 cm long.) Calculate the heat of reaction of the two chemicals (in J/kg). Assume that the specific heat of the chemicals and of the spray is the same as that of water, \(4.19 \times 10{^3} J/kg \cdot K\), and that the initial temperature of the chemicals is 20\(^\circ\)C.
4 step solution
Problem 93
You have 1.50 kg of water at 28.0\(^\circ\)C in an insulated container of negligible mass. You add 0.600 kg of ice that is initially at -22.0\(^\circ\)C. Assume that no heat exchanges with the surroundings. (a) After thermal equilibrium has been reached, has all of the ice melted? (b) If all of the ice has melted, what is the final temperature of the water in the container? If some ice remains, what is the final temperature of the water in the container, and how much ice remains?
5 step solution
Problem 94
A thirsty nurse cools a 2.00-L bottle of a soft drink (mostly water) by pouring it into a large aluminum mug of mass 0.257 kg and adding 0.120 kg of ice initially at -15.0\(^\circ\)C. If the soft drink and mug are initially at 20.0\(^\circ\)C, what is the final temperature of the system, assuming that no heat is lost?
6 step solution
Problem 95
A copper calorimeter can with mass 0.446 kg contains 0.0950 kg of ice. The system is initially at 0.0\(^\circ\)C. (a) If 0.0350 kg of steam at 100.0\(^\circ\)C and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?
7 step solution
Problem 96
Styrofoam bucket of negligible mass contains 1.75 kg of water and 0.450 kg of ice. More ice, from a refrigerator at -15.0\(^\circ\)C, is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.884 kg. Assuming no heat exchange with the surroundings, what mass of ice was added?
8 step solution
Problem 97
In a container of negligible mass, 0.0400 kg of steam at 100\(^\circ\)C and atmospheric pressure is added to 0.200 kg of water at 50.0\(^\circ\)C. (a) If no heat is lost to the surroundings, what is the final temperature of the system? (b) At the final temperature, how many kilograms are there of steam and how many of liquid water?
6 step solution
Problem 99
A carpenter builds a solid wood door with dimensions 2.00 m \(\times\) 0.95 m \(\times\) 5.0 cm. Its thermal conductivity is k = 0.120 W/m \(\cdot\) K. The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional 1.8-cm thickness of solid wood. The inside air temperature is 20.0\(^\circ\)C, and the outside air temperature is -8.0\(^\circ\)C. (a) What is the rate of heat flow through the door? (b) By what factor is the heat flow increased if a window 0.500 m on a side is inserted in the door? The glass is 0.450 cm thick, and the glass has a thermal conductivity of 0.80 W/m \(\cdot\) K. The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 cm of glass.
8 step solution
Problem 100
One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 W is required to keep the interior surface of the box 65.0 C\(^\circ\) (about 120 F\(^\circ\)) above the temperature of the outer surface. The total area of the box is 2.18 m\(^2\), and the wall thickness is 3.90 cm. Find the thermal conductivity of the material in SI units.
5 step solution
Problem 101
Compute the ratio of the rate of heat loss through a single-pane window with area 0.15 m\(^2\) to that for a double-pane window with the same area. The glass of a single pane is 4.2 mm thick, and the air space between the two panes of the double-pane window is 7.0 mm thick. The glass has thermal conductivity 0.80 W /m \(\cdot\) K. The air films on the room and outdoor surfaces of either window have a combined thermal resistance of 0.15 m\(^2 \cdot\) K/W.
5 step solution
Problem 102
Rods of copper, brass, and steel-each with crosssectional area of 2.00 cm\(^2\)-are welded together to form a Y-shaped figure. The free end of the copper rod is maintained at 100.0\(^\circ\)C, and the free ends of the brass and steel rods at 0.0\(^\circ\)C. Assume that there is no heat loss from the surfaces of the rods. The lengths of the rods are: copper, 13.0 cm; brass, 18.0 cm; steel, 24.0 cm. What is (a) the temperature of the junction point; (b) the heat current in each of the three rods?
7 step solution
Problem 106
The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about 1.50 kW/m\(^2\). The distance from the earth to the sun is \(1.50 \times 10{^1}{^1} m\), and the radius of the sun is \(6.96 \times 10{^8} m\). (a) What is the rate of radiation of energy per unit area from the sun's surface? (b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?
3 step solution
Problem 108
A metal sphere with radius 3.20 cm is suspended in a large metal box with interior walls that are maintained at 30.0\(^\circ\)C. A small electric heater is embedded in the sphere. Heat energy must be supplied to the sphere at the rate of 0.660 J/s to maintain the sphere at a constant temperature of 41.0\(^\circ\)C. (a) What is the emissivity of the metal sphere? (b) What power input to the sphere is required to maintain it at 82.0\(^\circ\)C? What is the ratio of the power required for 82.0\(^\circ\)C to the power required for 41.0\(^\circ\)C? How does this ratio compare with 2\(^4\)? Explain.
7 step solution
Problem 109
You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass 68 kg and surface area 1.85 m\(^2\) produces energy at a rate of up to 1300 W, 80% of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around 33\(^\circ\)C instead of the usual 30\(^\circ\)C. (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is 40.0\(^\circ\)C (104\(^\circ\)F)? (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10{^6} J/kg\). (e) How many 750-mL bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of 1.0 kg.
5 step solution
Problem 112
At a chemical plant where you are an engineer, a tank contains an unknown liquid. You must determine the liquid's specific heat capacity. You put 0.500 kg of the liquid into an insulated metal cup of mass 0.200 kg. Initially the liquid and cup are at 20.0\(^\circ\)C. You add 0.500 kg of water that has a temperature of 80.0\(^\circ\)C. After thermal equilibrium has been reached, the final temperature of the two liquids and the cup is 58.1\(^\circ\)C. You then empty the cup and repeat the experiment with the same initial temperatures, but this time with 1.00 kg of the unknown liquid. The final temperature is 49.3\(^\circ\)C. Assume that the specific heat capacities are constant over the temperature range of the experiment and that no heat is lost to the surroundings. Calculate the specific heat capacity of the liquid and of the metal from which the cup is made.
7 step solution
Problem 113
During your mechanical engineering internship, you are given two uniform metal bars \(A\) and \(B\), which are made from different metals, to determine their thermal conductivities. Measuring the bars, you determine that both have length 40.0 cm and uniform cross-sectional area 2.50 cm\(^2\). You place one end of bar \(A\) in thermal contact with a very large vat of boiling water at 100.0\(^\circ\)C and the other end in thermal contact with an ice-water mixture at 0.0\(^\circ\)C. To prevent heat loss along the bar's sides, you wrap insulation around the bar. You weigh the amount of ice initially and find it to be 300 g. After 45.0 min has elapsed, you weigh the ice again and find that 191 g of ice remains. The ice-water mixture is in an insulated container, so the only heat entering or leaving it is the heat conducted by the metal bar. You are confident that your data will allow you to calculate the thermal conductivity \(k_A\) of bar \(A\). But this measurement was tedious-you don't want to repeat it for bar \(B\). Instead, you glue the bars together end to end, with adhesive that has very large thermal conductivity, to make a composite bar 80.0 m long. You place the free end of A in thermal contact with the boiling water and the free end of \(B\) in thermal contact with the ice-water mixture. As in the first measurement, the composite bar is thermally insulated. You go to lunch; when you return, you notice that ice remains in the ice-water mixture. Measuring the temperature at the junction of the two bars, you find that it is 62.4\(^\circ\)C. After 10 minutes you repeat that measurement and get the same temperature, with ice remaining in the ice-water mixture. From your data, calculate the thermal conductivities of bar \(A\) and of bar \(B\).
5 step solution
Problem 114
Consider a poor lost soul walking at 5 km/h on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of 280 W, and almost all of this energy is converted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to \(k'A{_s}{_k}{_i}{_n}(T{_a}{_i}{_r} - T{_s}{_k}{_i}{_n})\), where \(k'\) is 54 J/h \(\cdot\) C\(^\circ\) \(\cdot\) m\(^2\), the exposed skin area \(A{_s}{_k}{_i}{_n}\) is 1.5 m\(^2\), the air temperature \(T{_a}{_i}{_r} \)is 47\(^\circ\)C, and the skin temperature \(T{_s}{_k}{_i}{_n}\) is 36\(^\circ\)C; (iii) the skin absorbs radiant energy from the sun at a rate of 1400 W/m\(^2\); (iv) the skin absorbs radiant energy from the environment, which has temperature 47\(^\circ\)C. (a) Calculate the net rate (in watts) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is \(e\) = 1 and that the skin temperature is initially 36\(^\circ\)C. Which mechanism is the most important? (b) At what rate (in L/h) must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at 36\(^\circ\)C is \(2.42 \times 10{^6}\) J/kg.) (c) Suppose instead the person is protected by light-colored clothing \((e \approx 0)\) so that the exposed skin area is only 0.45 m\(^2\). What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.
8 step solution
Problem 117
Careful measurements show that the specific heat of the solid phase depends on temperature (Fig. P17.117). How will the actual time needed for this cryoprotectant to come to equilibrium with the cold plate compare with the time predicted by using the values in the table? Assume that all values other than the specific heat (solid) are correct. The actual time (a) will be shorter; (b) will be longer; (c) will be the same; (d) depends on the density of the cryoprotectant.
4 step solution
Problem 119
To measure the specific heat in the liquid phase of a newly developed cryoprotectant, you place a sample of the new cryoprotectant in contact with a cold plate until the solution's temperature drops from room temperature to its freezing point. Then you measure the heat transferred to the cold plate. If the system isn't sufficiently isolated from its room-temperature surroundings, what will be the effect on the measurement of the specific heat? (a) The measured specific heat will be greater than the actual specific heat; (b) the measured specific heat will be less than the actual specific heat; (c) there will be no effect because the thermal conductivity of the cryoprotectant is so low; (d) there will be no effect on the specific heat, but the temperature of the freezing point will change.
4 step solution