Question

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

Answer 1

Amount of heat released when 100g of water cools from 20°C to 0°C = 100 × 20 × 4.2 = 8400J.

Amount of heat released when 100g of water converts into ice at 0°C = 100 × 336 = 33600J.

Amount of heat released when 100g of ice cools from 0°C to -10°C = 100 × 10 × 2.1 = 2100J.

Total amount of heat = 8400 + 33600 + 2100

= 44100J.

Time taken = 73.5 min

= 4410 s.
Average rate of heat extraction (power)

`"p" = "E"/"t"`

= `44100/4410`

= 10 J/s

= 10 W