s_chapter_19 - Chapter 19 The Second Law of Thermodynamics...

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1779 Chapter 19 The Second Law of Thermodynamics Conceptual Problems 1 Modern automobile gasoline engines have efficiencies of about 25%. About what percentage of the heat of combustion is not used for work but released as heat? ( a ) 25%, ( b ) 50%, ( c ) 75%, ( d ) 100%, ( e ) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Q h . The percentage of the heat of combustion (heat absorbed from the high- temperature reservoir) is the ratio of Q c to Q h . We can use the relationship between W, Q h , and Q c ( c h Q Q W = ) to find Q c / Q h . Use the definition of efficiency and the relationship between W, Q h , and Q c to obtain: h c h c h h 1 Q Q Q Q Q Q W = = = ε Solving for Q c / Q h yields: = 1 h c Q Q Substitute for to obtain: 75 . 0 25 . 0 1 h c = = Q Q and ( ) c is correct. 2 If a heat engine does 100 kJ of work per cycle while releasing 400 kJ of heat, what is its efficiency? ( a ) 20%, ( b ) 25%, ( c ) 80%, ( d ) 400%, ( e ) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Q h . We can use the relationship between W, Q h , and Q c ( c h Q Q W = ) to express the efficiency of the heat engine in terms of Q c and W . Use the definition of efficiency and the relationship between W, Q h , and Q c to obtain: W Q Q W W Q W c c h 1 1 + = + = =
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Chapter 19 1780 Substitute for Q c and W to obtain: 2 . 0 kJ 100 kJ 400 1 1 = + = ε and ( ) a is correct. 3 If the heat absorbed by a heat engine is 600 kJ per cycle, and it releases 480 kJ of heat each cycle, what is its efficiency? ( a ) 20%, ( b ) 80%, ( c ) 100%, ( d ) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Q h . We can use the relationship between W, Q h , and Q c ( c h Q Q W = ) to express the efficiency of the heat engine in terms of Q c and Q h . Use the definition of efficiency and the relationship between W, Q h , and Q c to obtain: h c h c h h 1 Q Q Q Q Q Q W = = = Substitute for Q c and Q h to obtain: 2 . 0 kJ 600 kJ 480 1 = = and ( ) a is correct. 4 Explain what distinguishes a refrigerator from a heat pump. Determine the Concept The job of a refrigerator is to move heat from its cold interior to the warmer kitchen environment. This process moves heat in a direction that is opposite its natural direction of flow, analogous to the use of a water pump to pump water out of a boat. The term heat pumps is used to describe devices, such as air conditioners, that are used to cool living and working spaces in the summer and warm them in the winter. 5 [SSM] An air conditioner’s COP is mathematically identical to that of a refrigerator, that is, c AC ref COP COP == Q W . However a heat pump’s COP is defined differently, as h hp COP = Q W . Explain clearly why the two COPs are defined differently. Hint : Think of the end use of the three different devices.
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This note was uploaded on 01/29/2012 for the course PHYS 213 taught by Professor Oshea during the Fall '08 term at Kansas State University.

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s_chapter_19 - Chapter 19 The Second Law of Thermodynamics...

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