Question

A refrigerator converts 100 g of water at 20°C to ice
10°C in 73.5 min. Calculate the average rate of
heat extraction in watt. The specific heat capacity of
water is 4.2 J g-1 K-1, specific latent heat of ice is
336 J g-1 and the specific heat capacity of ice is
2.1 J g- K-1​

Answer 1

Answer:Answer:Amount of heat released when 100g of water cools from 20o to 0oC = 100 × 20 × 4.2 = 8400J.  Amount of heat released when 100g of water converts into ice at 0oC = 100 × 336 = 33600J. Amount of heat released when 100g of ice cools from 0oC to -10oC = 100 × 10 × 2.1 = 2100J. Total amount of heat = 8400 + 33600 + 2100 = 44100J. Time taken = 73.5min = 4410s. Average rate of heat extraction (power). P = E/t = 44100/4410 = 10W

Explanation: