Phys 22: Homework 10 Solutions W A = 5W B Q IN QIN B QOUT A = 2Q OUT 2 QOUT B QIN B A = 3Q IN = QIN B QOUT. e A = W A e B W B A Q IN.

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Grace Morton
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1 HRK 26.7 Summarizing the information given in the question One way of doing this is as follows. W A = 5W Q IN A = Q IN Q OU A = 2Q OU Use e A = W A Q IN = QIN A QOU Q IN A A A and e = W Q IN = QIN QOU Q IN Find the ratio e A /e and then, in turn, e A and e. e A = W A e QIN Q IN A W QIN W e A = 5 W e Q IN e A = 5 e Finally 5( Q IN Q OU = e A e = QIN A = QIN ) = Q IN 2 Q IN = Q OU QOU A QIN Q IN A Q IN QOU 2 QOU QIN Q IN Q IN QOU e = 1 QOU Q IN e = 1 2 Q OU e A = 5 e HRK (a) An engine has efficiency and for a Carnot engine this is equal to e A = 5 9 e = W Q IN = Q IN Q OU Q IN e = H L H HW10 - Page 1 of 12

2 because Q H H = Q L L for a Carnot engine regardless of the working substance. he Carnot efficiency is the best possible efficiency for any engine using a hot reservoir at temperature H and a cool reservoir at L. Since we know the temperatures we immediately have e = e = Since Q IN Q OU = W we can easily find the work done using our knowledge of the efficiency and the heat entering the engine from the hot reservoir 26.17(b) A refrigerator has coefficient of performance e = W Q IN W = (0.199)(568 J) W = J K = Q IN W = Q IN Q OU Q IN where heat is entering the system from the cold reservoir and being expelled at the hot reservoir. For a Carnot refrigerator the K value is as high as possible and equals L K = H L K = K = 4.0 he work supplied to a refrigerator is W = Q OU Q IN. Again, we know the coefficient of performance and Q IN so we have K = Q IN W W = Q IN K W = W = 05.1 J HW10 - Page 2 of 12

3 HRK (a) With reference to the pv diagram of this cycle we see that the intermediate step is superfluous. Heat is expelled as we traverse an isotherm in the direction of decreasing volume. Conversely, heat is absorbed as we traverse an isotherm in the direction of increasing volume. he net effect of expelling and then absorbing the same amount of heat in an isothermal process is to do nothing! (Obviously in reality it would be worse than doing nothing as this would introduce losses in a real (not Carnot) engine). We know that the efficiency of a Carnot engine is related to the temperature of the hot and cool reservoirs 26.22(b) e = H L H e = 1 1 his is just an example of the engine described in (a). he maximum efficiency, which it certainly will not reach, is e = ( ) ( ) ( ) e = 0.58 HW10 - Page of 12

4 HRK (a) Section 1 b a 1 H b a Q H H also = 0 (isothermal) (isothermal) Section 2 0 c b c b 0 (adiabatic) also = ( H L ) (isotherm H isotherm L) Section d c d 1 (isothermal) L c Q L = Q H L H also = 0 (isothermal) HW10 - Page 4 of 12

5 Section 4 0 a d a d 0 (adiabatic) also = ( H L ) (isotherm L isotherm H) 26.1(b) Using Q H = H S we get Q H = (400 K)(0.5 J/K) Q H = 200 J 26.1(c) Using Q L = L S Q L = (250 K)( 0.5 J/K) = 125 J and W = Q H Q L we get W = = 75 J and since the system does useful work on the environment. the work done on the system is W = 75 J As a check the work done by a Carnot engine (see 26.17(a)) is ( ) H L W = Q IN H ( ) W = W = 75 J(on environment, so negative this on system) HRK 26. Use = nc v d HW10 - Page 5 of 12

6 in the normal expression for change in entropy for a reversible process HRK (a) Using the ideal gas law f i =10 =5 =10 =5 =10 =5 nc v d na d na 2 d ] 10 na 5 (4.8 mol) ( J/(mol K 4 ) ).0441 J/K (10 K 5 K ) p 1 V 1 = nr 1 p 2 (V 1 ) = nr 2 ( 2 = 1 since isothermal) p 2 = p 1 Find p using the adiabatic process connecting point 1 and. p 1 V γ 1 = p V γ p = p 1 ( V1 p = p 1 ( 1 (where γ = 7 5 ) 7 5 V ) 7 5 for diatomic gas) p = (0.215)p 1 We can do something similar to relate to (b) 1 V = V 2 5 ( V1 = 1 V = (0.644) 1 Isothermal 1 2 (using pv = nr and p 1 V γ 1 = p V γ to get 1V γ 1 1 = V γ 1 ) 2 5 ( ) V1 W 12 = nr 1 ln = nr 1 ln() he sign indicates the gas did work on the piston. Since E (12) int = W 12 + Q 12 (where E (12) int = 0 since = 0) Q 12 = nr 1 ln () V 1 ) HW10 - Page 6 of 12

7 which is heat that flowed into the gas. Also Const. Vol 2 2 S 12 = 1 S 12 = S 12 = Q 12 1 S 12 = nr 1 ln () 1 S 12 = nr ln () S 12 = 1.1R 1 W 2 = 0 (constant volume process) E (2) int = Q 2 E (2) int = 5 2 nr( 1 ) C v = 5 R for an ideal diatomic gas 2 Also E (2) int = 5 2 nr( 0.56) 1 E (2) int = 0.89R 1 S 2 = S 2 = 2 2 nc v d d S 2 = nc v 2 S 2 = 5 ( ) 2 nr ln 2 S 2 = 5 ( ) 2 nr ln (0.644)1 S 2 = 1.1R 1 Adiabatic 1 W 1 = 1 γ 1 (p f V f p i V i ) W 1 = 5 2 (p 1V 1 p V ) W 1 = 5 2 (p 1V 1 (0.215)p 1 (V 1 )) W 1 = 0.89p 1 V 1 =0.89R 1 E (1) int = W 1 + Q 1 but Q 1 = 0 adiabatic process E (1) int = 0.89p 1 V 1 =0.89R 1 HW10 - Page 7 of 12

8 Also S 1 = S 1 = S 1 = adiabatic process Of course E and S over a complete cycle should be zero and this checks out. HRK o minimize confusion we recall that the sign convention is such that heat flowing into the system is positive as is work done on the system (a) Path segment I(i): E int = 0 Q = W ( ) 2V0 Q = p 0 V 0 ln V 0 isothermal process Path segment I(ii): E int = Q const. vol. process Q = 2 nr(4 0 0 ) C v 2 R Q = 9 2 nr 0 = 9 2 p 0V 0 HW10 - Page 8 of 12

9 Path segment II(i): E int = 0 Q = W Q = p 0 V 0 ln(2) isothermal process 26.41(b) Path segment II(ii): Q = nc p Q = 5 2 R(4 0 0 ) C p = 5 R for a monatomic ideal gas 2 Q = 15 2 R 0 = 15 2 p 0V 0 ( ) 2V0 Path segment I(i): W = p 0 V 0 ln = p 0 V 0 ln (2) isothermal process Path segment I(ii): W = 0 const. vol. process V 0 ( ) V0 /2 Path segment II(i): W = p 0 V 0 ln = p 0 V 0 ln(2) isothermal process Path segment II(ii): W = p V W = (2p 0 )(2V V 0) V 0 W = (2p 0 ) 2 V 0 =p 0 V (c) Path segment I(i): E int = 0 isothermal process Path segment I(ii): E int = nc v E int = 2 R(4 0 0 )= 9 2 R 0 = 9 2 p 0V 0 Path segment II(i): E int = 0 isothermal process Path segment II(ii): E int = 2 R(4 0 0 )= 9 2 R 0 = 9 2 p 0V 0 HW10 - Page 9 of 12

10 26.41(d) Path segment I(i): Q I(i) isothermal process 0 p 0V 0 ln(2) = R ln(2) 0 ncv d Path segment I(ii): ( ) 4V0 nc v ln So total S along path I is 4R ln(2). V 0 nc v ln(4) = R(2 ln(2)) = R ln(2) 2 Path segment II(i): Q II(i) isothermal process 0 p 0V 0 ln(2) = R ln(2) 0 ncp d Path segment II(ii): ( ) 4V0 nc p ln V 0 nc p ln(4) = 5 R(2 ln(2)) = 5R ln(2) 2 So total S along path II is also 4R ln(2) as it must be since path I and path II have the same start and end points. HRK his is an example of irreversible heat transfer. capacity c we have For an object with mass m and specific heat mcd d mc ( ) f mc ln i 26.42(a) he amount of heat lost by the copper block, Q c, must equal the amount gained by the lead block Q l m c c c (400 eqm ) = m l c l ( eqm 200) (.05 kg)(87 J/kg)(400 eqm ) = (.1 kg)(129 J/kg)( eqm 200) eqm = 20 where we have used the heat capacities c c = (87 J/kg) and c l = (129 J/kg) for copper and lead respectively. HW10 - Page 10 of 12

11 26.42(b) We are told that the two block system is insulated so that no heat enters or leaves, therefore Q = 0. We also know that no work is done by (or on) the system therefore W = 0. Using the first Law of thermodynamics E int = Q + W we have 26.42(c) For the lead block E int = 0 and similarly for the copper block S l = S l = f i,l f i,c f = mc l d i,l f = mc c d i,c HRK S l + S c m l c l ln ( ) + m c c c ln (.1 kg)(129 J/(kg K)) ln J/K ( ) ( ) +(.05 kg)(87 J/(kg K)) ln ( ) Note that 107 C = 80 K and 18.6 C = K, and also c Al = 900 J/(kg K) and c w = 4190 J/(kg K) 26.44(a) he heat lost by the aluminum must equal the heat gained by the water (b) (0.196)(900)( eqm 107) = (0.052)(4190)(18.6 eqm ) eqm = 58 C = 1 K f f S Al = i,al = mc Ald i,al ( ) eqm S Al = m Al c Al ln 107 ( ) 1 S Al = (0.196)(900) ln 80 S Al = 24.4 J/K HW10 - Page 11 of 12

12 26.44(c) ( ) eqm S H20 = m H20 c H20 ln 107 ( ) 1 S H20 = (0.052)(4190) ln S H20 = 27.8 J/K 26.44(d) S Al + S H20.4 J/K he total change in entropy is positive as expected for any system undergoing an irreversible process. HRK Since the lake is so much larger than the cube the lake s temperature will not change over the course of this process. he terms in the heat and entropy equations below correspond to (i) warming the ice to ice at 0 C (ii) Melting the ice to water at 0 C and (iii) heating the water to 15 C, respectively. Note that c ice = 2220 J/(kg K) and L f =, 000 J/(kg K) Q into ice cube = Q m i c i (0 i )+m i L + m i c w (15 0) = Q (0.0126)(2220)(0 + 10) + (0.0126)( 10 ) + (0.0126)(4190)(15 0) = Q 5267 J = Q out of Lake out of Lake out of Lake out of Lake S ice = m i c i ln m il f 27 + m ic w ln S ice = (0.0126)(2220) ln (0.0126)( 10 ) 27 S ice = S ice = J/K + (0.0126)(4190) ln Finally S lake = S lake = Qout of Lake S lake = S lake = J/K isothermal process S total = S ice + S lake S total = 0.95 J/K HW10 - Page 12 of 12

Handout 12: Thermodynamics. Zeroth law of thermodynamics

Handout 12: Thermodynamics. Zeroth law of thermodynamics 1 Handout 12: Thermodynamics Zeroth law of thermodynamics When two objects with different temperature are brought into contact, heat flows from the hotter body to a cooler one Heat flows until the temperatures