Calculate the mass of ice required to lower the temperature of 300 g of water at 40°C to water at 0°C. (Specific latent heat of ice = 336 J g-1, Specific heat capacity of water = 4.2 J g-1 °C-1)
Topic: Heat and calorimetry
Answer
Given: mass of water m = 300 g, cooling from 40 °C to 0 °C
specific heat capacity of water c = 4.2 J g⁻¹ °C⁻¹ specific latent heat of ice L = 336 J g⁻¹ Let the mass of ice be m′.
Heat lost by the water = m c Δθ = 300 × 4.2 × (40 − 0) = 300 × 4.2 × 40 = 50 400 J
Heat gained by the ice in melting = m′ L = m′ × 336
By the principle of calorimetry, heat lost = heat gained: m′ × 336 = 50 400 m′ = 50 400 ÷ 336 m′ = 150 g
Mass of ice required = 150 g
The final temperature is 0 °C, which is also the melting point of ice, so the ice only MELTS — the melted water is never warmed above 0 °C. That is why the right-hand side is m′L alone, with no m′cΔθ term after it.
Compare this with the questions where the mixture ends at 5 °C or 10 °C. There you need BOTH terms: m′L to melt the ice, then m′cΔθ to warm the melted water up to the final temperature. Adding a warming term here, or leaving it out there, is the commonest error in the whole chapter. Read the final temperature first and let it decide how many terms you write.