problem17_94

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

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17.94: Equating the heat lost be the soda and mug to the heat gained by the ice and solving for the final temperature = T ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 K kg J 4190 kg 120 . 0 K kg J 910 kg 257 . 0 K kg J 4190 kg 00 . 2 kg J 10 334 C 0 . 15 K kg J 2100 kg 120 . 0 C 0 . 20 K kg J 910 kg 257 . 0 K kg J 4190 kg 00 . 2 3 + + × + ° - ° + C. 1 . 14 ° = Note that the mass of the ice
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Unformatted text preview: ( 29 kg 120 . appears in the denominator of this expression multiplied by the heat capacity of water; after the ice melts, the mass of the melted ice must be raised further to . T...
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This document was uploaded on 02/05/2008.

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