Question

A refrigerator converts 100 g of water at 20°C to ice at −10°C in 73.5 minutes. If the specific heat capacity of ice and water are 2 J g⁻¹°C⁻¹ and 4.2 J g⁻¹°C⁻¹ and specific latent heat of fusion of ice is 336 J g⁻¹ calculate:

1. heat extracted to bring the water to 0°С.
2. heat extracted to change the water to ice at 0°C.
3. Heat extracted to lower the temperature of ice to −10°C.
4. Average rate of extraction of heat.

Answer 1

Given: m = 100 g, c_w = 4.2 J g⁻¹°C⁻¹, c_ice = 2 J g⁻¹°C⁻¹, L = 336 J g⁻¹

(i) Heat extracted to cool water from 20°C to 0°C:

Q₁ = mc_w (20 − 0)

= 100 × 4.2 × 20

= 8400 J

(ii) Heat extracted to freeze water at 0°C:

Q₂ = mL

= 100 × 336

= 33600 J

(iii) Heat extracted to cool ice from 0°C to -10°C:

Q₃ = mc_ice (0 − (−10))

= 100 × 2 × 10

= 2000 J

(iv) Total heat extracted:

Q = Q₁ + Q₂ + Q₃ 

= 8400 + 33600 + 2000

= 44000 J

Time = 73.5 min = 73.5 × 60 = 4410 s

Average rate of extraction of heat:

$$= \frac{Q}{t}$$

$$= \frac{44000}{4410}$$

$$= 9.98\ \text{J/s} \approx 10\ \text{J/s} = 10\ \text{W}$$

Also, $$= \frac{44000}{73.5} = 598.6\ \text{J/min}$$