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Q. The inside and outside temperatures of a refrigerator are $273\, K$ and $303\, K$ respectively. Assuming that refrigerator cycle is reversible, for every joule of work done the heat delivered to the surrounding will be

AMUAMU 2006

Solution:

$\beta=\frac{Q_{2}}{W}=\frac{T_{L}}{T_{H}-T_{L}}$
$Q_{2} =\frac{273 \times 1}{303-273}=\frac{273}{30}=9\, J$
Heat delivered to the surrounding
$Q_{1} =Q_{2}+W$
$=9 + 1 = 10\, J$