test2s - Chemistry 2740 Spring 2011 Test 2 Solutions 1. Any...

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Unformatted text preview: Chemistry 2740 Spring 2011 Test 2 Solutions 1. Any of the following would be acceptable: • Allows us to define an absolute entropy scale • Implies that it is impossible to cool an object to absolute zero 2. (a) η * = T l T h- T l = 187 . 15 K 293 . 15- 187 . 15 K = 1 . 77 = q removed w . ∴ q removed = η * w = 1 . 77(31 MJ) = 55 MJ . This is the heat removed from the freezer compartment. This heat, plus the work, are equal to the heat that must be dumped into the room: q dumped =- ( q removed + w ) =- (55 + 31 MJ) =- 86 MJ . (b) A real (irreversible) refrigerator must be less efficient than a reversible refrigerator. Therefore, if it consumes 31 MJ of work, it cannot extract as much heat from the freezer. The total heat dumped into the room will therefore be less than the 86 MJ calculated above. If this seems paradoxical, it is because a reversible refrigerator would actually consume less electricity to extract the same amount of heat to the room. To put it another way, by treating the freezer as ideal, we overestimatedroom....
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This note was uploaded on 03/03/2012 for the course CHEM 2740 taught by Professor Roussel during the Fall '07 term at Lethbridge College.

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test2s - Chemistry 2740 Spring 2011 Test 2 Solutions 1. Any...

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