A Carnot freezer takes heat from water at $$0°C$$ inside it and rejects it to the room at a temperature of $$27°C$$. The latent heat of ice is $$336 \times 10^3$$ J kg$$^{-1}$$. If 5 kg of water at $$0°C$$ is converted into ice at $$0°C$$ by the freezer, then the energy consumed by the freezer is close to:

We start by noting that the freezer first removes heat from the water at $$0^{\circ}\text{C}$$ to turn it into ice at the same temperature. This amount of heat is the latent heat of fusion of water.

The latent heat of ice is given as $$L = 336 \times 10^{3}\ \text{J kg}^{-1}.$$

Mass of water to be frozen is $$m = 5\ \text{kg}.$$

Hence the heat that must be taken out of the water (the heat absorbed from the cold reservoir) is

$$Q_L = mL = 5 \times 336 \times 10^{3} \ \text{J}.$$

Simplifying, we get

$$Q_L = 1680 \times 10^{3}\ \text{J} = 1.68 \times 10^{6}\ \text{J}.$$

Now we recall the formula for the coefficient of performance (COP) of a Carnot refrigerator or freezer. For a Carnot machine operating between a cold reservoir at absolute temperature $$T_L$$ and a hot reservoir at absolute temperature $$T_H$$, the COP is

$$\text{COP} = \frac{Q_L}{W} = \frac{T_L}{T_H - T_L}.$$

The cold reservoir (inside the freezer) is at $$0^{\circ}\text{C},$$ which corresponds to

$$T_L = 0 + 273 = 273\ \text{K}.$$

The hot reservoir (the room) is at $$27^{\circ}\text{C}$$ or

$$T_H = 27 + 273 = 300\ \text{K}.$$

Therefore,

$$T_H - T_L = 300 - 273 = 27\ \text{K}.$$

Substituting these values into the COP expression gives

$$\text{COP} = \frac{273}{27} = 10.111\ldots$$

We take $$\text{COP} \approx 10.11$$ for numerical work.

From the definition of COP, the electrical work (energy) required is

$$W = \frac{Q_L}{\text{COP}}.$$

Substituting $$Q_L = 1.68 \times 10^{6}\ \text{J}$$ and $$\text{COP} = 10.11,$$ we obtain

$$W = \frac{1.68 \times 10^{6}}{10.11}\ \text{J}.$$

Carrying out the division,

$$W \approx 1.662 \times 10^{5}\ \text{J}.$$

Rounding to three significant figures gives

$$W \approx 1.67 \times 10^{5}\ \text{J}.$$

Thus the energy consumed by the Carnot freezer is closest to $$1.67 \times 10^{5}\ \text{J}.$$

Hence, the correct answer is Option D.