If minimum possible work is done by a refrigerator in converting 100 grams of water at 0°C to ice, how much heat (in calories) is released to the surroundings at temperature 27°C (Latent heat of ice = 80 Cal/gram) to the nearest integer?

We have 100 g of water at 0 °C that must be converted to ice at the same temperature. During freezing the refrigerator must remove only the latent heat, so the heat taken out from the water-ice system (the cold reservoir) is

$$Q_C = m\,L = 100\;\text{g}\times 80\;\text{cal g}^{-1}=8000\;\text{cal}.$$

To obtain the minimum possible work input, the refrigerator has to operate reversibly, i.e. as an ideal Carnot refrigerator, between the cold‐reservoir temperature 0 °C and the surrounding (hot reservoir) temperature 27 °C.

First we convert both temperatures to the absolute (Kelvin) scale:

$$T_C = 0\;^\circ\text{C} = 273\;\text{K},\qquad T_H = 27\;^\circ\text{C} = 300\;\text{K}.$$

The Carnot coefficient of performance (COP) for a refrigerator is stated by the formula

$$\text{COP}_{\text{max}} = \frac{T_C}{T_H - T_C}.$$

Substituting the temperatures,

$$\text{COP}_{\text{max}} = \frac{273}{300-273} = \frac{273}{27} = 10.111\;.$$

The work that must be supplied to a refrigerator is related to the heat extracted and the COP by the equation

$$W_{\text{min}} = \frac{Q_C}{\text{COP}_{\text{max}}}.$$

So,

$$W_{\text{min}} = \frac{8000}{10.111} \approx 791.3\;\text{cal}.$$

The heat rejected to the surroundings (the hot reservoir) is the sum of the heat extracted from the cold side and the work put in:

$$Q_H = Q_C + W_{\text{min}} = 8000 + 791.3 = 8791.3\;\text{cal}.$$

Rounding this to the nearest integer gives

$$Q_H \approx 8791\;\text{cal}.$$

Hence, the correct answer is Option A.