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(B) The heat extracted from the water to freeze it is $Q_2 = m L = 1 \ g 	imes 80 \ cal/g = 80 \ cal$. Converting this to Joules: $Q_2 = 80 	imes 4.

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$33.6$

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(B) The heat extracted from the water to freeze it is $Q_2 = m L = 1 \ g 	imes 80 \ cal/g = 80 \ cal$. Converting this to Joules: $Q_2 = 80 	imes 4.2 \ J = 336 \ J$. The coefficient of performance $(K)$ of a refrigerator is given by $K = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}$. Here,$T_2 = 0^{\circ} C = 273 \ K$ and $T_1 = 27.3^{\circ} C = 273 + 27.3 = 300.3 \ K$. Substituting the values: $\frac{336}{W} = \frac{273}{300.3 - 273} = \frac{273}{27.3} = 10$. Therefore,$W = \frac{336}{10} = 33.6 \ J$.

The freezing compartment of a refrigerator is at $0^{\circ} C$ and the room temperature is $27.3^{\circ} C$. Calculate the work done by the refrigerator to freeze $1 \ g$ of water at $0^{\circ} C$. (Given: $L_{\text{ice}} = 80 \ cal \ g^{-1}$) (in $J$)

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