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(D) The temperature inside the refrigerator is $T_2 = (t_2 + 273) \, K$ and the room temperature is $T_1 = (t_1 + 273) \, K$.For an ideal refrigerator

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$\frac{t_1 + 273}{t_1 - t_2}$

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(D) The temperature inside the refrigerator is $T_2 = (t_2 + 273) \, K$ and the room temperature is $T_1 = (t_1 + 273) \, K$.For an ideal refrigerator (Carnot cycle),the ratio of heat rejected to the room $(Q_1)$ to the heat absorbed from the cold reservoir $(Q_2)$ is given by $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$.From the first law of thermodynamics,the work done $W$ is $W = Q_1 - Q_2$,which implies $Q_2 = Q_1 - W$.Substituting this into the ratio: $\frac{Q_1}{Q_1 - W} = \frac{T_1}{T_2}$.Rearranging for $\frac{Q_1}{W}$: $\frac{Q_1 - W}{Q_1} = \frac{T_2}{T_1} \Rightarrow 1 - \frac{W}{Q_1} = \frac{T_2}{T_1}$.$\frac{W}{Q_1} = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$.Therefore,the heat delivered to the room per unit of work done is $\frac{Q_1}{W} = \frac{T_1}{T_1 - T_2}$.Substituting $T_1 = t_1 + 273$ and $T_2 = t_2 + 273$,we get $\frac{Q_1}{W} = \frac{t_1 + 273}{(t_1 + 273) - (t_2 + 273)} = \frac{t_1 + 273}{t_1 - t_2}$.

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:

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