The temperature inside a refrigerator is $${t_2}^ \circ C$$  and the room temperature is $${t_1}^ \circ C.$$  The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

Solution :
For a refrigerator, we know that $$\frac{{{Q_1}}}{W} = \frac{{{Q_1}}}{{{Q_1} - {Q_2}}} = \frac{{{T_1}}}{{{T_1} - {T_2}}}$$
where,
$${{Q_1}} =$$  amount of heat delivered to the room
$$W =$$  electrical energy consumed
$${{T_1}} =$$  room temperature $$= {{t_1} + 273}$$
$${{T_2}} =$$  temperature of sink $$= {{t_2} + 273}$$
$$\therefore \frac{{{Q_1}}}{1} = \frac{{{t_1} + 273}}{{{t_1} + 273 - \left( {{t_2} + 273} \right)}} \Rightarrow {Q_1} = \frac{{{t_1} + 273}}{{{t_1} - {t_2}}}$$