problem17_p100

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

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17.100: See Problem 17.97. Denoting C by , bT a C + = a and b independent of temperature, integration gives. . ) ( 2 ) a( 2 1 2 2 1 2 - + - = T T b T T n Q In this form, the temperatures for the linear part may be expressed in terms of Celsius temperatures, but the quadratic must be converted to Kelvin temperatures,
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Unformatted text preview: K. 500 and K 300 2 1 = = T T Insertion of the given values yields K) 300-K K)(500 mol J mol)(29.5 00 . 3 ( = Q )) K) 300 ( K) 500 )(( K mol J 10 10 . 4 ( 2 2 2 3- +-J. 10 97 . 1 4 =...
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This document was uploaded on 02/05/2008.

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