The coefficient of performance ($\beta$) of a refrigerator is given by $\beta = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}$, where $Q_2$ is the heat extracted from the refrigerator (cold reservoir), $W$ is the work done (electrical energy consumed), $T_1$ is the absolute temperature of the room (hot reservoir), and $T_2$ is the absolute temperature inside the refrigerator. The total heat delivered to the room is $Q_1 = Q_2 + W$. We need to find the amount of heat delivered per joule of work done, which is $\frac{Q_1}{W}$. $\frac{Q_1}{W} = \frac{Q_2 + W}{W} = \frac{Q_2}{W} + 1 = \beta + 1$ Substituting the expression for $\beta$: $\frac{Q_1}{W} = \frac{T_2}{T_1 - T_2} + 1 = \frac{T_2 + T_1 - T_2}{T_1 - T_2} = \frac{T_1}{T_1 - T_2}$ Converting the given Celsius temperatures to Kelvin: $T_1 = t_1 + 273$ $T_2 = t_2 + 273$ Now, substitute the temperatures into the ratio: $\frac{Q_1}{W} = \frac{t_1 + 273}{(t_1 + 273) - (t_2 + 273)} = \frac{t_1 + 273}{t_1 - t_2}$ For each joule of electrical energy consumed ($W = 1\text{ J}$), the heat delivered is $\frac{t_1 + 273}{t_1 - t_2}$.