CHAPTER 9: Heat and the First Law of Thermodynamics Responses to Questions. (a) No. Because the ernal energies of solids and liquids are complicated and include potential energies associated with the bonds between atoms and molecules, two objects may have different ernal energies but the same temperature. Internal energy will also vary with the mass of the object. If two objects that are at different temperatures are placed in contact, there will be a net energy transfer from the hotter object to the colder one, regardless of their ernal energies. (b) Yes. Just as in (a), the transfer of energy depends on the temperature difference between the two objects, which may not be directly related to the difference in ernal energies.. Because the specific heat of water is quite large, water can absorb a large amount of energy with a small increase in temperature. Water can be heated, then easily transported throughout a building, and will give off a large amount of energy as it cools. This makes water particularly useful in radiator systems. 7. When water at 00ºC comes in contact with the skin, energy is transferred to the skin and the water begins to cool. When steam at 00ºC comes in contact with the skin, energy is transferred to the skin and the steam begins to condense to water at 00ºC. Steam burns are often more severe than water burns due to the energy given off by the steam as it condenses, before it begins to cool. 8. Energy is needed to convert water in the liquid state to the gaseous state (latent heat). Some of the energy needed to evaporate molecules on the surface comes from the ernal energy of the water, thus decreasing the water temperature. 9. No. The water temperature cannot go above 00ºC, no matter how vigorously it is boiling. The rate at which potatoes cook depends on the temperature at which they are cooking. 8. Metabolism is a biochemical process by which living organisms get energy from food. If a body is doing work and losing heat, then its ernal energy would drop drastically if there were no other
source of energy. Metabolism (food) supplies this other source of energy. The first law of thermodynamics applies because the energy contributions from metabolism are included in Q. 0. An adiabatic compression is one that takes place with no exchange of heat with the surroundings. During the compression, work is done on the gas. Since no heat leaves the gas, then the work results in an increase in the gas s ernal energy by the same amount, and therefore an increase in its temperature. ΔE = Q W, so if Q = 0 then ΔE = W. Solutions to Problems. Find the mass of warmed water from the volume of water and its density of 000kg m. Then use the fact that kcal of energy raises kg of water by C, and that the water warms by C. m At m At 0 kg m.0 m 0.0 0 m 0. kg kcal bar 0. kg C.8 kcal ;.8 kcal 0.04bars kg C 00 kcal. (a) (b) 4.86 0 J 7 00 Cal.00 J Cal kwh 00 Cal.9 kwh 860 Cal (c) At 0 cents per day, the food energy costs $0.9 per day. It would be practically impossible to feed yourself in the United States on this amount of money. 6. The wattage rating is 0 Joules per second. Note that L of water has a mass of kg. kg kcal 486 J s. 0 L 60C 80s.0 min L kg C kcal 0 J 9. (a) The heat absorbed can be calculated from Eq. 9-. Note that L of water has a mass of kg.
0 m.0 0 kg Q mct L m.0 L 486J kgg C 00 C 0 C 0 J.49. 0 J (b) Power is the rate of energy usage. E Q Q.49 0 J P t 00s 6 min t t P 00W 0. The heat absorbed by all three substances is given by Eq. 9-, Q mc T. Thus the amount of Q mass can be found as m. The heat and temperature change are the same for all three c T substances. Q Q Q m : m : m : : : : : : Cu Al HO c T c T c T c c c 90 900 486 Cu Al HO Cu Al HO 486 486 486 : : 0.7 : 4.6 : : 4.7 : 90 900 486. The heat lost by the horseshoe must be equal to the heat gained by the iron pot and the water. Note that L of water has a mass of kg. m c T T m c T T m c T T shoe Fe shoe eq pot Fe eq pot HO HO eq HO 0.40 kg40j kggc T.0C shoe 0.0 kg40j kggc.0c 0.0C.0 kg486j kggc.0c 0.0C T 0.8 C 0C shoe. The heat lost by the iron must be the heat gained by the aluminum and the glycerin. i i i m c T T m c T T m c T T Fe Fe Fe eq Al Al eq Al gly gly eq gly 0.90 kg 40 J kggc 4C 0.09 kg 900 J kggc 8C 0.0 kg c 8C c gly 0J kggc 00J kggc gly
7. We assume that all of the kinetic energy of the hammer goes o heating the nail. 0 hammer hammer nail Fe m v hammer hammer KE Q m v m c T 0.0 kg 7.m s T.7C 4C m c nail Fe 0.04 kg40j kggc 9. Assume that the heat from the person is only used to evaporate the water. Also, we use the heat of vaporization at room temperature (8 kcal/kg), since the person s temperature is closer to room temperature than 00 o C. Q 80 kcal 0.08 kg 0.kg 0 ml 8kcal kg Q ml m vap L vap 0. Assume that all of the heat lost by the ice cube in cooling to the temperature of the liquid nitrogen is used to boil the nitrogen, and so none is used to raise the temperature of the nitrogen. The boiling po of the nitrogen is o 77 K 96 C. m c ice ice T T m L ice ice nitrogen vap initial final m m c T T g L 00 0 J kg.0 kg00j kg C0C 96C ice ice ice ice initial final nitrogen vap 7. 0 kg. (a) The heater must heat both the boiler and the water at the same time. Q Pt m c m c T t Fe Fe HO HO o o o m c m c T Fe Fe H O H O 80 kg40j kggc 70 kg486j kggc 8C 7 P. 0 J h 4.946 h 4.9 h (b) Assume that after the water starts to boil, all the heat energy goes o boiling the water, and none goes to raising the temperature of the iron or the steam.
70kg.6 0 J kg m L HO vap Q Pt m L t.77 h HO vap 7 P. 0 J h Thus the total time is t t 4.946h.77h 6.67h 7h 4. The heat lost by the aluminum and the water must equal the heat needed to melt the mercury and to warm the mercury to the equilibrium temperature. m c Al Al T T Al eq m c H T T O HO HO eq m c T T m c T T m L c T T Al Al Al eq H O H O H O eq Hg fusion Hg eq melt L c T T m fusion Hg eq melt Hg 0.60 kg900j kggc 0.400 kg486j kggc.80c.06c.00 kg 8J kggc.06c 9.0C 4. 0 J kg. The kinetic energy of the bullet is assumed to warm the bullet and melt it. mv Q mc T T ml Pb melt initial fusion g Pb melt initial fusion v c T T L 0J kg C 7C 0C 0. 0 J kg 60 m s 7. Segment A is the compression at constant pressure. Since the process is at a constant pressure, the path on the diagram is horizontal from.0 L to.0 L. Segment B is the isothermal expansion. Since the temperature is constant, the ideal gas law says that the product P is constant. 0 P atm. 0. B A L 0.0.0.0.0 C
Since the volume is doubled, the pressure must be halved, and so the final po on this segment is at a pressure of 0. atm. The path is a piece of a hyperbola. Segment C is the pressure increase at constant volume. Since the process is at a constant volume, the path on the diagram is vertical from 0. atm to.0 atm.. (a) No work is done during the first step, since the volume is constant. The work in the second step is given by W P. W.00 Pa 0 m P.4 atm 9.L.9 L 480 J atm L (b) Since there is no overall change in temperature, E 0J (c) The heat flow can be found from the first law of thermodynamics. E Q W Q E W 0 480J 480J o the gas. (a) See the diagram. The isobaric expansion (b) 9-9a. is just a horizontal line on the graph. The work done is found from Eq. W P 4 N m 8.00 m.00 m 70J P (N/m ) 00 400 00 00 00 0 A B B 0.0.0 4.0 6.0 8.0 0.0 (m ) The change in ernal energy depends on the temperature change, which can be related to the ideal gas law, P nrt. E nrt nrt nrt P P P W 70J 4.0 0 J
(c) For the isothermal expansion, since the volume expands by a factor of 4, the pressure drops by a factor of 4 to 4 N m. The spreadsheet used for this problem can be found on the Media Manager, with filename PSE4_ISM_CH9.XLS, on tab Problem 9.. (d) The change in ernal energy only depends on the initial and final temperatures. Since those temperatures are the same for process (B) as they are for process (A), the ernal energy change is the same for process (B) as for process (A), 4.0 0 J.. (a) The work done by an ideal gas during an isothermal volume change is given by Eq. 9-8. W 7.00m nrt ln.60 mol 8.4 J molg K 90 K ln 44. J 40 J.0m (b) Since the process is isothermal, there is no ernal energy change. Apply the first law of thermodynamics. E Q W 0 Q W 40J (c) Since the process is isothermal, there is no ernal energy change, and so E 0.. Since the expansion is adiabatic, there is no heat flow o or out of the gas. Use the first law of thermodynamics to calculate the temperature change. E Q W nrt W 0 700 J g W T nr. mol 8.J mol K 40K 4.0 0 K 8. For the path ac, use the first law of thermodynamics to find the change in ernal energy. E Q W 6 J J 8 J ac ac ac Since ernal energy only depends on the initial and final temperatures, this path that starts at a and ends at c. And for any path that starts at c and ends at a, E E 8 J. ca ac E applies to any
(a) Use the first law of thermodynamics to find Q. abc E Q W Q E W 8J 4J 8J abc abc abc abc abc abc (b) Since the work along path bc is 0, W W P P along path da is 0. abc ab b ab b b a. Also note that the work W W P P P W 4J 7J cda cd c cd c d c b a b abc (c) Use the first law of thermodynamics to find Q. abc E Q W Q E W 8J 7 J J cda cda cda cda cda cda (d) As found above, E E E E,a,c ca ac 8J (e) Since E E J, E E J,d,c,d,c and so E E E E E J,a,d,a,c da which then gives E E J 8J J 6J. Use the first law of thermodynamics to da ca find Q. da E Q W Q E W 6J 0 6J da da da da da da. For a diatomic gas with no vibrational modes excited, we find the parameter. C C R R R P 7 C C R For an adiabatic process, we have P P P P P constant. Use this to find the final pressure..4.00atm 0.468atm 0.47 atm.7
Use the ideal gas law to find the final temperature. P P P T T 9K 0.468.7 4 K 9 C T T P. (a) We first find the final pressure from the adiabatic relationship, and then use the ideal gas law to find the temperatures. For a diatomic gas,.4..4 0.0 m P P P P.00 atm 7.77 0 atm 0.70 m P P nrt T T nr P.0 0 Pa0.0 m P T nr (b) nr J.6mol 8.4 mol g K 7.77 0.0 0 Pa 0.70m 94.6 K 9K J.6mol 8.4 mol g K 40.9 K 404 K J E nc T n RT molgk 4.6mol 8.4 94.6 K 40.9 K.9 0 J C (c) Since the process is adiabatic, no heat is transferred. Q 0 (d) Use the first law of thermodynamics to find the work done by the gas. The work done ON the gas is the opposite of the work done BY the gas. E Q W W Q E W on gas 4.9 0 J 4 4 0.9 0 J.9 0 J 89. (a) Since the pressure is constant, the work is found using Eq. 9-9a. W P (b) Use the first law of thermodynamics..0 0 Pa 4.m. m.947 0 J.9 0 J
E Q W 6.0 0 J.9 0 J 4.8 0 J 4.4 0 J (c) See the accompanying graph. P (atm).0 0. 4. (m )