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An engine operating between $100^\circ C$ and $0^\circ C$ takes 453.6 Kcal of heat. How much useful work can be done by it?

(a) 508.8 kJ
(b) 453.6 kcal
(c) 508.8 kcal
(d) None.
Can you answer this question?
 
 

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Answer: 508.8 kJ
$T_2$ = 100 + 273 = 373 K and
$T_1$ = 0 + 273 = 273 K
Q = 453.6 kcal = 453.6 $\times$ 4.184 = 1897.86 kJ
Now, $ \frac{W}{Q} = \frac{T_2 - T_1}{T_2} = 1897.86 (\frac{373-273}{373})$ = 508. 8 kJ
answered Feb 24, 2014 by mosymeow_1
 

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