A refrigerator works between $${4^ \circ }C$$  and $${30^ \circ }C.$$  It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is
(Take, $$1\,cal = 4.2\,Joules$$   )

Solution :
Given, temperature of source,
$$T = {30^ \circ }C = 30 + 273 \Rightarrow {T_1} = 303\,K$$
Temperature of sink, $${T_2} = {4^ \circ }C = 4 + 273$$
$${T_2} = 277\,K$$
As, we know that $$\frac{{{Q_1}}}{{{Q_2}}} = \frac{{{T_1}}}{{{T_2}}} \Rightarrow \frac{{{Q_2} + W}}{{{Q_2}}} = \frac{{{T_1}}}{{{T_2}}}\,\,\left\{ {\because W = {Q_1} - {Q_2}} \right\}$$
where $${{Q_2}}$$ is the amount of heat drawn from the sink (at $${{T_2}}$$ ), $$W$$ is workdone on working substance, $${{Q_1}}$$ is amount of heat rejected to source (at room temperature $${{T_1}}$$ ).
$$\eqalign{ & \Rightarrow W{T_2} + {T_2}{Q_2} = {T_1}{Q_2} \cr & \Rightarrow W{T_2} = {T_1}{Q_2} - {T_2}{Q_2} \cr & \Rightarrow W{T_2} = {Q_2}\left( {{T_1} - {T_2}} \right) \cr & \Rightarrow W = {Q_2}\left( {\frac{{{T_1}}}{{{T_2}}} - 1} \right) \cr & \Rightarrow W = 600 \times 4.2 \times \left( {\frac{{303}}{{277}} - 1} \right) \cr & W = 600 \times 4.2 \times \left( {\frac{{26}}{{277}}} \right) \cr & W = 236.5\,Joules \cr & {\text{Power}} = \frac{{{\text{Work done}}}}{{{\text{Time}}}} = \frac{W}{t} = \frac{{236.5}}{1} = 236.5\,W \cr} $$