Question

In a cold storage, ice melts at the rate of 2 kg per hour when the external temperature is 20°C. The minimum power output of the motor used to drive the refrigerator which just prevents the ice from melting is (latent heat of fusion of ice = 80 cal g$^{-1}$)

• A. $28.5 \text{ W}$
• B. $13.6 \text{ W}$
• C. $9.75 \text{ W}$
• D. $16.4 \text{ W}$

Hint:

For problems involving refrigeration, always use the latent heat formula $Q = mL$ to calculate heat energy and apply the coefficient of performance (COP) formula to find the minimum power required.

Answer 1

The Correct Option is B

Step 1: Understanding the given data - Mass of ice melting per hour, $m = 2$ kg = $2000$ g - Latent heat of fusion of ice, $L = 80$ cal g$^{-1}$ - 1 cal = 4.2 J Step 2: Calculating the total heat required per hour The heat required to melt the ice per hour is given by: $$Q = m L$$ $$Q = (2000 \text{ g}) \times (80 \text{ cal g}^{-1})$$ $$Q = 160000 \text{ cal}$$ Now, converting this to Joules: $$Q = 160000 \times 4.2 \text{ J}$$ $$Q = 672000 \text{ J}$$ Step 3: Calculating the minimum power required Power is given by: $$P = \frac{Q}{t}$$ Since $t = 1$ hour = 3600 seconds, $$P = \frac{672000}{3600}$$ $$P = 186.67 \text{ W}$$ For a refrigerator, the coefficient of performance (COP) for an ideal Carnot refrigerator is: $$COP = \frac{T_C}{T_H - T_C}$$ Here, - $T_C = 0^\circ C = 273$ K - $T_H = 20^\circ C = 293$ K $$COP = \frac{273}{293 - 273} = \frac{273}{20} = 13.65$$ Step 4: Determining the work done The work input is given by: $$W = \frac{Q}{COP}$$ $$W = \frac{186.67}{13.65}$$ $$W \approx 13.6 \text{ W}$$ Thus, the minimum power required is 13.6 W.