Question

A burner supplies heat energy at a rate of 434 Js⁻¹ for 60 seconds when 40 g of ice at 0°C changes to water at 75°C. Calculate the latent heat of ice.

Answer 1

Heat supplied by the burner:

Q = 434 × 60

= 26040 J

Heat needed to warm 40 g of water from 0°C to 75°C:

Q₁ = 40 × 4.2 × 75

= 12600 J

Let the specific latent heat of fusion of ice be L J/g.

Heat needed to melt 40 g of ice at 0°C:

Q₂ = 40 L J

Total heat used:

40L + 12600 = 26040

40L = 13440

$$L = \frac{13440}{40}$$

= 336 J/g

Therefore, the specific latent heat of fusion of ice is 336 J/g.