Question

A refrigerator converts 100 g of water at 20°C to ice at -10°C in 35 minutes. Calculate the average rate of heat extraction in terms of watts.

Given: Specific heat capacity of ice = 2.1 J g^-1°C^-1

Specific heat capacity of water = 4.2 J g^-1°C^-1

Specific latent heat of fusion of ice = 336 J g^-1

Answer 1

Given that

Mass of water converted to ice = m = 100 g

Temperature of water t_w = 20°C

Temperature of ice t_i = –10°C

Total time t = 35 min = 2100 s

Specific heat capacity of ice = 2.1 J g^-1°C^-1

Specific heat capacity of water = 4.2 J g^-1°C^-1

Specific latent heat of fusion of ice = 336 J g^-1

Amount of heat released when 100 g water cools from 20C to 0°C is

Q₁ = mcΔT

= 100 x 4.2 x 20

= 8400 J

Amount of heat released when 100 g water converts to ice at 0°C is

Q₂ = mL

= 100 x 336

= 33600 J

Amount of heat released when 100 g ice cools from 0°C to −10°C is

Q₃ = mcΔT

= 100 x 2.1 x 10

= 2100 J

Hence, the total heat released is 

Q = Q₁ + Q₂ + Q₃

Q =  8400 + 33600 + 2100

Q = 44100 J

Therefore, the average rate of heat extraction is

P = `Q/t = 44100/2100` = 21 W