Question

A refrigerator works between 2° C and 27° C. To keep the temperature of the refrigerated space constant, 660 calories of heat are to be removed every second. The power required is

• A. 60 W
• B. 55 W
• C. 252 W
• D. 231 W

Answer 1

Answer: (C)

252 W

Answer 2

The temperatures of the cold and hot reservoirs are given in Celsius. We convert them to Kelvin:

Cold reservoir temperature, $T_2 = 2^\circ C = 2 + 273 = 275 K$.

Hot reservoir temperature, $T_1 = 27^\circ C = 27 + 273 = 300 K$.

The rate at which heat is removed from the refrigerated space (cold reservoir) is given as 660 calories per second. We convert this to Joules per second using the conversion factor 1 cal = 4.2 Joules:

Rate of heat removal, $\frac{Q_2}{t} = 660 \text{ cal/s} = 660 \times 4.2 \text{ J/s}$.

$\frac{Q_2}{t} = 2772 \text{ J/s}$.

For an ideal refrigerator (Carnot refrigerator), the coefficient of performance ($\beta$) is given by the ratio of the temperature of the cold reservoir to the temperature difference between the hot and cold reservoirs:

$\beta = \frac{T_2}{T_1 - T_2}$

$\beta = \frac{275 K}{300 K - 275 K} = \frac{275}{25} = 11$.

The coefficient of performance is also defined as the ratio of the heat removed from the cold reservoir ($Q_2$) to the work done by the refrigerator (W):

$\beta = \frac{Q_2}{W}$

Since we are given the rate of heat removal ($Q_2/t$) and we need to find the power required (which is the rate of work done, $P = W/t$), we can write:

$\beta = \frac{Q_2/t}{W/t} = \frac{Q_2/t}{P}$.

Now we can find the power required by rearranging the formula:

$P = \frac{Q_2/t}{\beta}$

Substitute the values of $Q_2/t$ and $\beta$:

$P = \frac{2772 \text{ J/s}}{11}$

$P = 252 \text{ J/s}$

$P = 252 \text{ W}$.

The power required is 252 W.