Question

A vessel of negligible thermal capacity contains 4 kg of water at 40°C. Into the vessel is added 3 kg of ice at 0°C. Calculate the (a) heat energy given out by water to attain temperature of 0°C (b) mass of ice which melts (c) mass of water at 0°C.

[Take sp. heat capacity of water = 4200 Jkg⁻¹°C⁻¹ and sp. latent heat of fusion of ice = 340 × 10³ Jkg⁻¹]

Answer 1

(a) Heat energy given out by water at 40°C = mcθ_f

= 4 kg × 4200 Jkg⁻¹°C⁻¹ × 40°C

= 672000 J

(b) Heat energy absorbed by ice to melt = Heat energy given out by water

⇒ mL = 672000 J

⇒ m × 340 × 10³ J kg⁻¹ = 672000 J

\[m =\frac{672000}{340 \times 10^{3}} \mathrm{~kg}\]

∴ m = 1.978 kg

Mass of ice which melts = 1.978 kg

(c) Mass of water at 0°C = mass of water cooled + mass of water formed from ice

= 4 kg + 1.978 kg

= 5.978 kg