The temperature insides & outside of refrigerator are $260 K$ and $315 K$ respectively. Assuming that the refrigerator cycle is reversible, calculate the heat delivered to the surrounding for every joule of work done.

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Solution

Heat extract from refrigerator, $Q_{c}$
Heat deliver to surrounding, $Q_{h}$
Temperature of refrigerator, $T_{2}$
Temperature of surrounding, $T_{1}$
For refrigerator,
$C . O . P = \frac{T_{2}}{T_{1} - T_{2}} = \frac{Q_{c}}{W}$
$\Rightarrow \frac{Q_{c}}{W} = \frac{260}{315 - 260} = 4.64$
$\Rightarrow Q_{c} = W \times 4.64$
Heat deliver to surrounding
$Q_{h} = Q_{c} + W = W (4.64 + 1)$
$\Rightarrow \frac{Q_{h}}{W} = 5.64$

$5.64 J$ heat deliver to surrounding per unit joule work done.

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The temperature insides & outside of refrigerator are 260 K and 315 K respectively. Assuming that the refrigerator cycle is reversible, calculate the heat delivered to the surrounding for every joule of work done.

The temperature insides & outside of refrigerator are $260 K$ and $315 K$ respectively. Assuming that the refrigerator cycle is reversible, calculate the heat delivered to the surrounding for every joule of work done.