First slide

Refrigerator

Question

The temperature of inside and outside of a refrigerator 273 k and 303 k respectively. Assuming that the refrigerator cycle is reversible, for every Joule of work done, the heat delivered to the surrounding will be nearly

Moderate

A

10 J
B

20 J
C

30 J
D

50 J

Solution

$\begin{array}{l} Cop = \frac{Heat removed from inside (Q_{2})}{Work done} = \frac{273}{303 - 273} \\ \Rightarrow Q_{2} = \frac{273}{30} \times 1 Joule = 9 .1 Joule \\ ∴ Q_{1} = Q_{2} + Work done = (9 .1 + 1) J = 10 .1 Joule \end{array}$

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