Chater 0: Exercises:, 7,,, 8, 4 EOC: 40, 4, 46, 8 E: A gasoline engine takes in.80 0 4 and delivers 800 of work er cycle. The heat is obtained by burning gasoline with a heat of combustion of 4.60 0 4. What is the thermal efficiency? Part B ow much heat is discarded in each cycle? Part C What mass of fuel is burned in each cycle? Part D If the engine goes through 0.0 cycles er second, what is its ower outut in kilowatts? Part E If the engine goes through 0.0 cycles er second, what is its ower outut in horseower?
E7: What comression ratio must an Otto cycle have to achieve an ideal efficiency of 66. if.4? E: A window air-conditioner unit absorbs 9.80 0 4 same time eriod deosits.0 0 What is the ower consumtion of the unit? of heat er minute from the room being cooled and in the of heat into the outside air. Part B What is the energy efficiency rating of the unit? E: Descrition: A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In five minutes of oeration of the engine, the heat rejected by the engine melts 0.0400 kg of... A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In five minutes of oeration of the engine, the heat rejected by the engine melts 0.0400 of ice. During this time, how much work is erformed by the engine?
E8: Descrition: You make tea with 0.0 kg of T water and let it cool to room temerature (0.0 degree(s) C) before drinking it. (a) Calculate the entroy change of the water while it cools. (b) The cooling rocess is essentially isothermal for the air in your... You make tea with 0.0 of 86.0 water and let it cool to room temerature (0.0 ) before drinking it. Calculate the entroy change of the water while it cools. Part B The cooling rocess is essentially isothermal for the air in your kitchen. Calculate the change in entroy of the air while the tea cools, assuming that all the heat lost by the water goes into the air. Part C What is the total entroy change of the system tea + air? Exress your answer using two significant figures.
E4: Descrition: A box is searated by a artition into two arts of equal volume. The left side of the box contains 00 molecules of nitrogen gas; the right side contains 00 molecules of oxygen gas. The two gases are at the same temerature. The artition is... A box is searated by a artition into two arts of equal volume. The left side of the box contains 00 molecules of nitrogen gas; the right side contains 00 molecules of oxygen gas. The two gases are at the same temerature. The artition is unctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free exansion and not change temerature. On average, how many molecules of oxygen will there be in either half of the box? Exress your answer using two significant figures. molecules Part B On average, how many molecules of nitrogen will there be in either half of the box? Exress your answer using two significant figures. molecules Part C What is the change in entroy of the system when the artition is unctured? Part D What is the robability that the molecules will be found in the same distribution as they were before the artition was unctured, that is, 00 nitrogen molecules in the left half and 00 oxygen molecules in the right half? Give the order of magnitude. Exress your answer as the order of magnitude. If your answer is, so you would enter -7, then the order of magnitude is
0.40. IDENTIFY: Use the ideal gas law to calculate and for each state. Use the first law and secific exressions for Q, W, and U for each rocess. Use Eq.(0.4) to calculate e. Q is the net heat flow into the gas. SET UP: γ.40 C R /( γ ) 0.79 J/mol K; C C + R 9.0 J/mol K. The cycle is sketched in Figure 0.40. Figure 0.40 EXECUTE: (a) oint.00 atm.0 0 Pa (given); nrt; nrt (0.0 mol)(8.4 J/mol K)(00 K) 8.6 0 m.0 0 Pa oint rocess at constant volume so 8.6 0 m nrt and n, R, constant imlies / T / T T T ( / ) (.00 atm)(600 K/00 K).00 atm.0 0 Pa oint Consider the rocess, since it is simler than. Process is at constant ressure so nrt and n, R, constant imlies / T / T.00 atm.0 0 Pa ( T / T ) (8.6 0 m )(49 K/00 K) 4. 0 m (b) rocess constant volume ( 0) Q nc T (0.0 mol)(0.79 J/mol K)(600 K 00 K) 80 J 0 and W 0. Then U Q W 80 J rocess Adiabatic means Q 0. U nc T (any rocess), so U (0.0 mol)(0.79 J/mol K)(49 K 600 K) 780 J T 00 K T 600 K T 49 K Then U Q W gives W Q U + 780 J. (It is correct for W to be ositive since is ositive.) rocess For constant ressure W (.0 0 Pa)(8.6 0 m 4. 0 m ) 60 J or W nr T (0.0 mol)(8.4 J/mol K)(00 K 49 K) 60 J, which checks. (It is correct for W to be negative, since is negative for this rocess.) Q nc T (0.0 mol)(9.0 J/mol K)(00 K 49 K) 960 J U Q W 960 J ( 60 K) 400 J or U nc T (0.0 mol)(0.79 J/mol K)(00 K 49 K) 400 J, which checks (c) Wnet W + W + W 0 + 780 J 60 J + 0 J (d) Qnet Q + Q + Q 80 J + 0 960 J + 0 J work outut W 0 J (e) e 0.0 0.%. heat energy inut Q 80 J e(carnot) T / T 00 K/600 K 0.00. C EALUATE: For a cycle U 0, so by U Q W it must be that Qnet Wnet for a cycle. We can also check that U net 0: U net U + U + U 80 J 00 J 0 J 0 e < e(carnot), as it must.
0.4. IDENTIFY: Tb Tc and is equal to the maximum temerature. Use the ideal gas law to calculate T a. Aly the aroriate W exression to calculate Q for each rocess. e. U 0 for a comlete cycle and for an isothermal rocess of an ideal gas. Q SET UP: For helium, C R / and C R /. The maximum efficiency is for a Carnot cycle, and ecarnot TC / T. EXECUTE: (a) Qin Qab + Qbc. Qout Q ca. Tmax Tb Tc 7 C 600 K. aa bb a nrt Ta Tb (600 K) 00 K. b ( moles)(8. J/mol K)(600 K) bb nrtb b 0.0 m. T T.0 0 Pa a b b b b c c b c b (0.0 m ) 0.0997 m a. Tb Tc c Qab nc Tab ( ) c c nrtb c Qbc Wbc d d nrtb ln nrtb ln. b b bc ( mol) 8. J/mol K (400 K) 9.97 0 J ( ) Q 4 (.00 mol) 8. J/mol K (600 K)ln.0 0 J. b Q Q Q Qout Qca nc Tca ( ) b 4 in ab + bc.0 0 J. 4 (.00 mol) 8. J/mol K (400 K).66 0 J. 4 4 (b) Q U + W 0 + W W Q Q.0 0 J.66 0 J 4.4 0 J. in out 4.4 0 J e W Qin 0. %. 4.0 0 J TC (c) e 00 K max ecarnot 0.67 67% T 600 K EALUATE: The thermal efficiency of this cycle is about one-third of the efficiency of a Carnot cycle that oerates between the same two temeratures.
0.46. IDENTIFY: Use Eq.(0.4) to calculate e. SET UP: The cycle is sketched in Figure 0.46. C R / for an ideal gas C C + R 7 R / SET UP: rocess Figure 0.46 Calculate Q and W for each rocess. 0 imlies W 0 & 0 imlies Q nc T nc ( T T ) But nrt and constant says nrt and nrt. ( ) nr( T T ); nr T (true when is constant). Thus Q nc T nc ( / nr) ( C / R) ( C / R) ( ) ( C / R). Q > 0; heat is absorbed by the gas.) Then 0 0 0 0 0 rocess 0 so W ( ) 0( 0 0 ) 00 (W is ositive since increases.) 0 imlies Q nc ( ) T nc T T But nrt and constant says nrt and nrt. ( ) nr( T T ); nr T (true when is constant). Thus Q nc T nc ( / nr) ( C / R) ( C / R) ( ) ( C / R). ( Q > 0; heat is absorbed by the gas.) Then 0 0 0 0 0 rocess 4 0 imlies W 0 0 so Q nc T nc ( / nr) ( C / R)( )( ) ( C / R) 0 0 0 0 0 ( Q < 0 so heat is rejected by the gas.) rocess 4 0 so W ( 4 ) 0( 0 0 ) 00 (W is negative since decreases) 0 so Q nc T nc ( / nr) ( C / R) ( C / R) 0( 0 0 ) ( C / R) 00 ( Q < 0 so heat is rejected by the gas.) total work erformed by the gas during the cycle: W W + W + W + W 0 + + 0 tot 4 4 0 0 0 0 0 0 (Note that W equals the area enclosed by the cycle in the -diagram.) tot total heat absorbed by the gas during the cycle C C C + C Q Q + Q + R R R 0 0 0 0 0 0 ( Q ): eat is absorbed in rocesses and. C + ( C + R) C + R But C C + R so Q 00 00. R R total heat rejected by the gas during the cycle ( Q ): eat is rejected in rocesses 4 and 4. C C C + C Q Q + Q R R R C 4 4 0 0 0 0 0 0 C + ( C + R) C + R But C C + R so QC 00 00. R R W 00 R R efficiency e Q [C + R]/ R ( ) C + R ( R/) + R 9 ( ) 0 0 C. e 0.0 0.% C + R C + R EALUATE: As a check on the calculations note that QC + Q 00 + 00 00 W, as it should. R R
0.8. IDENTIFY and SET UP: First use the methods of Chater 7 to calculate the final temerature T of the system. EXECUTE: 0.600 kg of water (cools from 4.0 C to T) Q mc T (0.600 kg)(490 J/kg K)( T 4.0 C) (4 J/K) T. 0 J 0.000 kg of ice (warms to 0 C, melts, and water warms from 0 C to T) Q mc (0 C (.0 C)) + ml + mc ( T 0 C) ice f water Q + + 4 4 Q 7 J +.67 0 J + (09. J/K) T.88 0 J + (09. J/K) T 0.000 kg (00 J/kg K)(.0 C) 4 0 J/kg (490 J/kg K)( T 0 C) Q system 0 gives 4 (4 J/K) T. 0 J +.88 0 J + (09. J/K) T 0 4 (.74 0 J/K) T 9.48 0 J 4 T (9.48 0 J)/(.74 0 J/K) 4.8 C 08 K EALUATE: The final temerature must lie between.0 C and 4.0 C. A final temerature of 4.8 C is consistent with only liquid water being resent at equilibrium. IDENTIFY and SET UP: Now we can calculate the entroy changes. Use S Q / T for hase changes and the method of Examle 0.6 to calculate S for temerature changes. EXECUTE: ice: The rocess takes ice at C and roduces water at 4.8 C. Calculate S for a reversible rocess between these two states, in which heat is added very slowly. S is ath indeendent, so S for a reversible rocess is the same as S for the actual (irreversible) rocess as long as the initial and final states are the same. S dq / T, where T must be in kelvins For a temerature change dq mc dt so For a hase change, since it occurs at constant T, S dq / T Q / T ± ml / T. T S ( mc / T ) dt mcln( T / T ). Therefore ice ice f water T S mc ln(7 K/8 K) + ml / 7 K + mc ln(08 K/7 K) S ice (0.000 kg)[(00 J/kg K) ln(7 K/8 K) + (4 0 J/kg)/7 K + (490 J/kg K) ln(08 K/7 K)] S ice.9 J/K + 6.7 J/K +.7 J/K 9.4 J/K water: Swater mc ln( T / T ) (0.600 kg)(490 J/kg K)ln(08 K/8 K) 80. J/K For the system, S Sice + Swater 9.4 J/K 80. J/K + J/K EALUATE: Our calculation gives S > 0, as it must for an irreversible rocess of an isolated system.