Question

A refrigerator takes 30 minutes to transform 100 g of water at 20°C to ice at −10°C. Determine the average heat extraction rate in watts. Specific heat capacity of ice = 2.1 J g⁻¹ °C⁻¹, Specific heat capacity of water = 4.2 J g⁻¹, Specific latent heat of fusion of ice = 336 J g⁻¹.

Answer 1

Given:

Mass of water, m = 100 g

Initial temperature = 20°C

Final temperature of ice = −10°C

Time = 30 min = 1800 s

Specific heat capacity of water, c_w ​= 4.2 J g⁻¹ °C⁻¹

Specific heat capacity of ice, c_i​ = 2.1 J g⁻¹ °C⁻¹

Latent heat of fusion, L = 336 J g⁻¹

1. Cool water from 20°C to 0°C:

Q₁ ​= mc_w​ΔT

= 100 × 4.2 × 20

= 8400 J

2. Freeze water at 0°C:

Q₂ ​= mL

= 100 × 336

= 33600 J

3. Cool ice from 0°C to −10°C:

Q₃ ​= mc_i​ΔT

= 100 × 2.1 × 10

= 2100 J

4. Total heat extracted:

Q = Q₁ ​+ Q₂ ​+ Q₃​

= 8400 + 33600 + 2100

= 44100 J

5. Average heat extraction rate:

P = `Q/t`

= `44100/1800`

= 24.5 W

Average heat extraction rate = 24.5 W​