Physics 41 Chapter 22 HW Serway 7th Edition Conceptual Questions: 1, 3, 8, 12 Problems: 9, 13, 20, 23, 27, 39, 48, 54, 55 Conceptual Questions: 1, 3, 8, 12 Q22.1 First, the efficiency of the automobile engine cannot exceed the Carnot efficiency: it is limited by the temperature of burning fuel and the temperature of the environment into which the exhaust is dumped. Second, the engine block cannot be allowed to go over a certain temperature. Third, any practical engine has friction, incomplete burning of fuel, and limits set by timing and energy transfer by heat. Q22.3 A higher steam temperature means that more energy can be extracted from the steam. For a constant temperature TThc− T c heat sink at Tc , and steam at Th , the efficiency of the power plant goes as = 1− and is maximized for a high TThh Th . Q22.8 (a) When the two sides of the semiconductor are at different temperatures, an electric potential (voltage) is generated across the material, which can drive electric current through an external circuit. The two cups at 50°C contain the same amount of internal energy as the pair of hot and cold cups. But no energy flows by heat through the converter bridging between them and no voltage is generated across the semiconductors. (b) A heat engine must put out exhaust energy by heat. The cold cup provides a sink to absorb output or wasted energy by heat, which has nowhere to go between two cups of equally warm water. Q22.12 (a) For an expanding ideal gas at constant temperature, the internal energy stays constant. The gas must absorb by heat the same amount of energy that it puts out by work. Then its ΔQ ⎛V ⎞ entropy change is ΔS = = nRln ⎜ 2 ⎟ T V ⎝ 1 ⎠ (b) For a reversible adiabatic expansion ΔQ = 0 , and ΔS = 0 . An ideal gas undergoing an irreversible adiabatic expansion can have any positive value for ΔS up to the value given in part (a). Problems: 9, 13, 20, 23, 24, 27, 39, 48, 54, 55 P22.9 Tc = 703 K Th = 2 143 K ΔT 1 440 (a) ec == =67.2% Th 2143 5 (b) Qh =×1.40 10 J, WQeng = 0.420 h 4 Weng 5.88× 10 J P == =58.8 kW Δt 1 s 13. An ideal gas is taken through a Carnot cycle. The isothermal expansion occurs at 250°C, and the isothermal compression takes place at 50.0°C. The gas takes in 1 200 J of energy from the hot reservoir during the isothermal expansion. Find (a) the energy expelled to the cold reservoir in each cycle and (b) the net work done by the gas in each cycle. P22.13 Isothermal expansion at Th = 523 K Isothermal compression at Tc = 323 K Gas absorbs 1 200 J during expansion. ⎛⎞T ⎛⎞323 (a) QQ==c 1200 J74 =1 J ch⎜⎟ ⎝⎠⎜⎟ ⎝⎠Th 523 (b) WQQeng =−=hc()1 200 − 741 J459 = J T 4.00 Q P22.23 ()COP ==c =0.013 8 =c Carnot refrig ΔTW289 ∴=W 72.2 J per 1 J energy removed by heat. Section 22.5 Gasoline and Diesel Engines P22.27 In a cylinder of an automobile engine, just after combustion, the gas is confined to a volume of 50.0 cm3 and has an initial pressure of 3.00 × 106 Pa. The piston moves outward to a final volume of 300 cm3 and the gas expands without energy loss by heat. (a) If γ = 1.40 for the gas, what is the final pressure? (b) How much work is done by the gas in expanding? γ γ (a) PVii= P f V f γ ⎛⎞ 3 1.40 Vi 6 ⎛⎞50.0 cm PPfi==×⎜⎟()3.00 10 Pa ⎜⎟3 =244 kPa ⎜⎟V 300 cm ⎝⎠f ⎝⎠ V i γ ⎛⎞Vi (b) WPdV= PP= i ⎜⎟ ∫ ⎝⎠V Vi Integrating, γ −1 0.400 ⎛⎞15⎡⎤⎛⎞V ⎡⎤⎛⎞0.0 cm3 ⎢⎥i 65− 3 WPV=−=××−⎜⎟ii1 ⎜⎟ ()2.50() 3.00 10 Pa() 5.00 10 m⎢⎥ 1 ⎜⎟3 γ − 13⎢⎥⎜⎟V 00 cm ⎝⎠⎣⎦⎝⎠f ⎣⎦⎢⎥⎝⎠ = 192 J ⎛⎞Vf P22.39 Δ=SnRln⎜⎟ = R ln2 = 5.76 J K ⎝⎠Vi There is no change in temperature for an ideal gas. FIG. P22.39 P22.20 (a) First, consider the adiabatic process DA→ : γ 53 γ γ ⎛⎞VA ⎛⎞10.0 L PVDD= PV AA so PPDA==⎜⎟1 400 kPa⎜⎟=712 kPa ⎝⎠VD ⎝⎠15.0 L ⎛⎞⎛⎞nRTDAγ nRT γ Also ⎜⎟⎜⎟VVDA= ⎝⎠⎝⎠VVDA γ −1 23 ⎛⎞VA ⎛⎞10.0 or TTDA==⎜⎟ 720 K ⎜⎟ =549 K ⎝⎠VD ⎝⎠15.0 Now, consider the isothermal process CD→ : TTCD==549 K ⎡⎤γ γ ⎛⎞VVVPDADA⎛⎞ ⎛⎞ VA PPCD==⎜⎟⎢⎥ PA⎜⎟ ⎜⎟ = VVVVVγ −1 ⎝⎠CDCC⎣⎦⎢⎥⎝⎠ ⎝⎠ D 1 400 kPa() 10.0 L 53 P ==445 kPa C 24.0 L() 15.0 L 23 γ γ Next, consider the adiabatic process BC→ : PVBB= PV CC γ PVAA ⎛⎞VA But, PC = γ −1 from above. Also considering the isothermal process, PPBA= ⎜⎟ VVCD ⎝⎠VB γ ⎛⎞VPVAAγ ⎛⎞Aγ VVAC 10.0 L() 24.0 L Hence, PVA ⎜⎟B= ⎜⎟γ −1 VC which reduces to VB == =16.0 L ⎝⎠VVVBC⎝⎠D VD 15.0 L ⎛⎞VA ⎛⎞10.0 L Finally, PPBA==⎜⎟1 400 kPa ⎜⎟ =875 kPa ⎝⎠VB ⎝⎠16.0 L State P(kPa) V(L) T(K) A 1 400 10.0 720 B 875 16.0 720 C 445 24.0 549 D 712 15.0 549 (b) For the isothermal process A → B : Δ=EnCTint V Δ=0 ⎛⎞VB ⎛⎞16.0 so QWnRT=− = ln⎜⎟= 2.34 mol() 8.314 J mol⋅ K() 720 K ln ⎜⎟= +6.58 kJ ⎝⎠VA ⎝⎠10.0 For the adiabatic process BC→ : Q = 0 ⎡⎤3 Δ=EnCTT() −=2.34 mol() 8.314 J mol⋅ K() 549 − 720 K =− 4.98 kJ int VC B ⎣⎦⎢⎥2 and WQE=− +Δint =0 +() − 4.98 kJ = − 4.98 kJ For the isothermal process CD→ : Δ=EnCTint V Δ=0 ⎛⎞VD ⎛⎞15.0 and QWnRT=− = l n⎜⎟= 2.34 mol() 8.314 J mol⋅ K() 549 K ln ⎜⎟= −5.02 kJ ⎝⎠VC ⎝⎠24.0 Finally, for the adiabatic process DA→ : Q = 0 ⎡⎤3 Δ=EnCTT() − =2.34 mol() 8.314 J mol⋅ K() 720 − 549 K =+ 4.98 kJ int VA D ⎣⎦⎢⎥2 and WQE=− +Δint =0 + 4.98 kJ = + 4.98 kJ Process Q(kJ) W(kJ) ΔEint (kJ) A → B +6.58 –6.58 0 BC→ 0 –4.98 –4.98 CD→ –5.02 +5.02 0 DA→ 0 +4.98 +4.98 ABCDA +1.56 –1.56 0 The work done by the engine is the negative of the work input. The output work Weng is given by the work column in the table with all signs reversed. Weng −W 1.56 kJ (c) e==ABCD = =0.237 or 23.7% QQhAB→ 6.58 kJ Tc 549 ec =−1 =− 1 = 0.237 or 23.7% Th 720 P22.24 COP= 0.100COPCarnot cycl e or QQhh⎛⎞ ⎛ 1 ⎞ ==0.100⎜⎟0.100 ⎜ ⎟ WW Carnot efficiency ⎝⎠Carnot cycle ⎝ ⎠ Qh ⎛⎞Th ⎛⎞293 K FIG. P22.24 ==0.100⎜⎟ 0.100⎜⎟=1.17 WTT⎝⎠hc−−⎝⎠293 K 268 K Thus, 1.17 joul es of energy enter the room by heat for each joul e of w ork done. 25. An ideal refrigerator or ideal heat pump is equivalent to a Carnot engine running in reverse. That is, energy Qc is taken in from a cold reservoir and energy Qh is rejected to a hot reservoir. Th – Tc (a) Show that the work that must be supplied to run the refrigerator or heat pump is W = Q T c c Tc (b) Show that the coefficient of performance of the ideal refrigerator is COP = T – T h c ⎡⎤ ()Q h P22.25 (a) For a complete cycle, Δ=E 0 and WQ= −= Q Q⎢⎥ −1 . int hcc⎢⎥Q ⎣⎦c Q T We have already shown that for a Carnot cycle (and only for a Carnot cycle) h = h . Q c Tc ⎡⎤TThc− Therefore, WQ= c ⎢⎥. ⎣⎦Tc (b) We have the definition of the coefficient of performance for a refrigerator, Q COP = c . W T Using the result from part (a), this becomes COP = c . TThc− −1000 J P22.48 Δ=S hot 600 K +750 J Δ=S col d 350 K (a) Δ=Δ+Δ=SSU hot Scol d 0.476 J K T1 (b) ec =−1 = 0.417 T2 WeQeng ==ch 0.417() 1 000 J = 417 J (c) Wnet =−=417 J 250 J 167 J TS1Δ=U 350 K() 0.476 J K= 167 J *P22.54 (a) For the isothermal process AB, the work on the gas is ⎛⎞VB WPVAB=− A A ln⎜⎟ ⎝⎠VA 53− 3⎛⎞50.0 WAB =−5() 1.013 × 10 Pa() 10.0× 10 m l n ⎜⎟ ⎝⎠10.0 3 WAB =−8.15 × 10 J where we have used 1.00 atm=× 1.013 105 Pa FIG. P22.54 and 1.00 L=× 1.00 10−33 m 53− 33 WPVBC=− B Δ =−()1.013 × 10 Pa⎡⎤⎣⎦() 10.0− 50.0 × 10 m =+ 4.05 × 10 J 3 WCA = 0 and WWWeng =−AB − BC =4.10 × 10 J = 4.10 kJ (b) Since AB is an isothermal process, ΔEint, AB = 0 3 and QWAB=− AB =8.15 × 10 J 3R 5R For an ideal monatomic gas, C = and C = V 2 P 2 53− PV (1.013×× 10)( 50.0 10 ) 5.06× 103 TT==BB = = BAnR R R 53− PV (1.013×× 10)( 10.0 10 ) 1.01× 103 Also, T ==CC = C nR R R ⎛ 3 ⎞⎛ 5.06×103 −1.01×103 ⎞ QCA =nCV ΔT =1.00⎜ R⎟⎜ ⎟=6.08 kJ ⎝2 ⎠⎝ R ⎠ so the total energy absorbed by heat is QQAB+= CA 8.15 kJ + 6.08 kJ = 14.2 kJ 55 (c) QnCTnRTPV=Δ=Δ=Δ() BC P 22B BC 5 534− QBC =×()1.013 10⎡⎤() 10.0 −×=−×=− 50.0 10 1.01 10 J 10.1 kJ 2 ⎣⎦ 3 WWeng eng 4.10× 10 J (d) e== =4 =0.288 or 28.9% QQQhABCA+×1.42 10 J (e) A Carnot engine operating between Thot = TA = 5060/R and Tcold = TC = 1010/R has efficiency 1 − Tc/Th = 1 − 1/5 = 80.0%.