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MASTERING PHYSICS CHAPTER 19 CONCEPTS

# MASTERING PHYSICS CHAPTER 19 CONCEPTS - 19.1 Model The heat...

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19.1. Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycle consists of three individual processes. Visualize: Please refer to Figure Ex19.1. Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p -versus- V graph. We get W out 3 kPa m 10 J = () × = 1 2 6 200 100 10 The heat energy transferred into the engine is Q H = 120 J. Because WQ Q out H C =− , the heat energy exhausted is QQW C H out 120 J 10 J 110 J = = (b) The thermal efficiency of the engine is η == = W Q out H 10 J 120 J 0 0833 . Assess: Practical engines have thermal efficiencies in the range ≈− 01 04 .. .

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19.2. Model: The heat engine follows a closed cycle, which consists of four individual processes. Visualize: Please refer to Figure Ex19.2. Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p -versus- V graph. We get W out 3 kPa 100 kPa m 30 J =− () × = 400 100 10 6 The heat energy leaving the engine is Q C 90 J 25 J 115 J =+= . The heat input is calculated as follows: WQ QQ Q W out H C H C out 115 J 30 J 145 J =−⇒=+ = + = (b) The thermal efficiency of the engine is η == = W Q out H 30 J 145 J 0 207 . Assess: Practical engines have thermal efficiencies in the range ≈− 01 04 .. .
19.3. Solve: (a) During each cycle, the heat transferred into the engine is Q H 55 kJ = , and the heat exhausted is Q C 40 kJ = . The thermal efficiency of the heat engine is η =− = 1 1 0 273 Q Q C H 40 kJ 55 kJ . (b) The work done by the engine per cycle is WQ Q out H C 55 kJ 40 kJ 15 kJ =−= =

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19.4. Solve: During each cycle, the work done by the engine is W out 20 J = and the engine exhausts Q C 30 J = of heat energy. Because WQ Q out H C =− , QW Q H out C 20 J 30 J 50 J =+ = + = Thus, the efficiency of the engine is η = 11 0 4 0 Q Q C H 30 J 50 J .
19.5. Solve: (a) The engine has a thermal efficiency of η == 40 0 40 %. and a work output of 100 J per cycle. The heat input is calculated as follows: =⇒= = W QQ Q out HH H 100 J 250 J 040 . (b) Because WQ Q out H C =− , the heat exhausted is QQW C H out 250 J 100 J 150 J = =

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19.6. Solve: The coefficient of performance of the refrigerator is K Q W QW W == = = C in H in in 50 J 20 J 20 J 150 .
19.7. Solve: (a) The heat extracted from the cold reservoir is calculated as follows: K Q W Q Q =⇒ = = C in C C 50 J 200 J 40 . (b) The heat exhausted to the hot reservoir is QQW HC in J J 250 J =+= + = 200 50

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19.8. Model: Assume that the car engine follows a closed cycle. Solve: (a) Since 2400 rpm is 40 cycles per second, the work output of the car engine per cycle is W out 500 kJ s 1 s 40 cycles kJ cycle = 12 5 . (b) The heat input per cycle is calculated as follows: η =⇒ = = W Q Q out H H kJ 62.5 kJ 12 5 020 . . The heat exhausted per cycle is QQW CH in kJ kJ kJ =−= = 62 5 12 5 50 ..
19.9. Solve: The amount of heat discharged per second is calculated as follows: η == + ⇒= = () W Q W QW out H out C out C out MW W 1 1 900 1 032 1 1 913 10 9 .

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MASTERING PHYSICS CHAPTER 19 CONCEPTS - 19.1 Model The heat...

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