problem20_47

University Physics with Modern Physics with Mastering Physics (11th Edition)

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20.47: a) × × = = × = - = Pa) 10 363 ( J, 10 6.52 kJ 1005 kJ 1657 3 5 V p W U J, 10 39 . 8 ) m 2202 . 0 m 4513 . 0 ( 4 3 3 × = - and so J. 10 36 . 7 5 × = + = W U Q b) Similarly, V p U Q - = H ) m 0682 . 0 m 6 Pa)(0.0094 10 (2305 kJ) 1969 kJ 1171 ( 3 3 3 - × + - = J. 10 33 . 9 5 × - = c) The work done during the adiabatic processes must be found indirectly (the coolant is not ideal, and is not always a gas). For the entire cycle,
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Unformatted text preview: , = U and so the net work done by the coolant is the sum of the results of parts (a) and (b), J. 10 97 . 1 5 -The work done by the motor is the negative of this, . 74 . 3 d) J. 10 97 . 1 J 10 1.97 J 10 36 . 7 5 5 5 = = = W Q c K...
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This document was uploaded on 02/06/2008.

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