20 THE SECOND LAW OF THERMODYNAMICS 20 1 IDENTIFY For a heat engine W Q H Q C H 0 Q Q C 0 We Q H W SET UP EXECUTE a H b 4300 J 2200 J Q W Q C 0 34 34 6500 J Q C e 2200 J 6500 J EVALUATE Since the engine operates on a cycle the net Q equal the net W But to calculate the efficiency we use the heat energy input HQ IDENTIFY For a heat engine Q 0 Q H 0 Q C 20 2 We Q H C W Q H 2600 J Q C 6400 J SET UP H Q 9000 J W EXECUTE a 2600 J 9000 J We Q H b 9000 J 6400 J 0 29 29 20 3 EVALUATE Since the engine operates on a cycle the net Q equal the net W But to calculate the efficiency we use the heat energy input HQ W IDENTIFY and SET UP The problem deals with a heat engine 3700 W Q Power Eq 20 4 to calculate the efficiency e and Eq 20 2 to calculate C EXECUTE a Q and H work output 0 23 23 W t e 16 100 J Use cL is the heat of combustion W Q H 3700 J 16 100 J heat energy input Q W Q Q b C H Q Q W Heat discarded is 16 100 J 3700 J 12 400 J C c HQ is supplied by burning fuel H Q Qm H L c where 16 100 J 4 mL c 0 350 g H the engine goes through 60 0 cycles 4 60 10 J g d W 3700 J per cycle t In 1 00 s P W t P 5 Q EVALUATE C IDENTIFY W Q H 60 0 3700 J 1 00 s 2 22 10 W 1 hp 746 W 12 400 J Q C SET UP For 1 00 s W 222 kW 298 hp In one cycle Q Q Q tot We H 0 Q Q C 0 Q H 3 180 10 J C 20 4 H 3700 J This equals totW for one cycle Copyright 2012 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 20 1 20 2 Chapter 20 20 5 EXECUTE a 6 43 10 J 5 180 10 J 3 0 280 5 WQ H e 6 43 10 J 1 80 10 J 6 43 10 J 5 5 5 H Q C Q W 4 63 10 J b EVALUATE Of the converted to mechanical work and the remaining reservoir IDENTIFY This cycle involves adiabatic ab isobaric bc and isochoric ca processes SET UP ca is at constant volume ab has of heat energy supplied to the engine each second is is discarded into the low temperature and bc is at constant pressure For a constant pressure 4 63 10 J 1 80 10 J Q 0 5 5 p V R so Q pC R Q nC V process W p V and Q nC T pV nRT gives n T p p V If 1 40 the For a constant volume process 0W and T pV nRT gives For a diatomic ideal gas VC R 1 atm 1 013 10 Pa 5 gas is diatomic and n T V p R so Q R 7 2 VC R pC 3 9 0 10 m V p 3 bp bV EXECUTE a aV 3 9 0 10 m 2 0 10 m b Heat enters the gas in process ca since T increases 1 5 atm p V b b p V a a 1 5 atm V b V a and 3 3 p a p b 3 5 2 3 3 2 0 10 m 1 4 12 3 atm For an adiabatic process Q VC R V p 5 2 0 10 m 12 3 atm 1 5 atm 1 013 10 Pa atm 5470 J 2 3 5 3 Q H 5470 J 3723 J Q C 3723 J c Heat leaves the gas in process bc since T increases Q d e pC R W Q H We Q H p V 5 7 1 5 atm 1 013 10 Pa atm 7 0 10 m 3 2 Q 5470 J C 1747 J 5470 J 3723 J 1747 J 0 319 31 9 3 EVALUATE We did not use the number of moles of the gas IDENTIFY Apply 1e 1 SET UP In part b H 10 000 J EXECUTE a 1 Q e The heat discarded is C Q e Q C Q H 1 r 1 1 0 40 9 50 10 000 J 1 0 594 4060 J W Q H 0 594 59 4 e 1 Q C Q H r 1 1e 10 000 J b C Q EVALUATE The work output of the engine is IDENTIFY SET UP r is the compression ratio e EXECUTE a 0 40 b 1 9 6 2 EVALUATE An increase in r gives an increase in e IDENTIFY Convert coefficient of performance K to energy efficiency rating EER SET UP which rounds to 58 1 8 8 2 0 595 an increase of 1 4 4060 J 0 581 and EER 0 40 e HK watts P watts H Btu h P watts 5940 J 20 6 20 7 20 8 Copyright 2012 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 20 9 20 10 20 11 The Second Law of Thermodynamics 20 3 H Btu h P watts EXECUTE 1 Btu h 0 293 W so watts H H Btu h 0 293 K 0 293 0 293 EER and K 3 0 For EER 3 41 3 0 10 2 EER 3 41 K EVALUATE The EER is larger than K but this does not mean that the air conditioner is suddenly better at cooling Use Eq 20 9 to calculate IDENTIFY and SET UP For the refrigerator W and then Eq 20 2 to calculate HQ a EXECUTE Performance coefficient W Q K b SET UP The operation of the device is illustrated in Figure 20 9 3 40 10 J 2 10 1 62 10 J K Q W 4 Q C Eq 20 9 3 4 10 J K and 2 10 C C 4 4 C EXECUTE W Q Q H Q W Q C H Q 5 02 10 J H negative because heat goes out of the system 1 62 10 J 3 40 10 J 4 4 4 Q delivered to the high temperature reservoir is greater than Figure 20 9 Q EVALUATE H the heat taken in from the low temperature reservoir IDENTIFY W Q The …