Question

The temperature inside and outside of the refrigerator is $260\text{K}$ and $315\text{K}$ respectively. Assuming that the refrigerator cycle is reversible, calculate the heat delivered to the surroundings for every joule of work done.

• A. $5\text{J}$
• B. $6.19\text{J}$
• C. $5\text{J}$
• D. $5.73\text{J}$

Answer 1

The correct option is D $5.73\text{J}$

Refrigerator rejects heat to surroundings (high temperature)
Inside temperature; $T_2=260\text{K}$
Outside temperature, $T_1=315\text{K}$
$\Rightarrow$ For the reversible cycle (Carnot's refrigerator),
Coefficient of performance $\left(\beta \right)=\frac{Q_2}{W}=\frac{T_2}{T_1-T_2}$
$(\because W=1\text{Joule}$ )
$\Rightarrow \frac{Q_2}{1}=\frac{260}{315-260}$
$\therefore Q_2=4.73\text{J}$
Hence, for each joule of work done, the heat delivered to the surrounding is,
$Q_1=Q_2+W$
$\Rightarrow Q_1=4.73+1=5.73\text{J}$