30 g of ice at 0°C is used to bring down the temperature of a certain mass of water at 70°C to 20°C. Find the mass of water. [Specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹ and specific latent heat of ice = 336 J g⁻¹]
Topic: Principle of mixtures with ice
Answer
Given: ice 30 g at 0 °C; water cooled from 70 °C to 20 °C
c_w = 4.2 J g⁻¹ °C⁻¹, L = 336 J g⁻¹ Let the mass of water be m grams.
HEAT GAINED by the ice: melting at 0 °C = 30 × 336 = 10 080 J
warming 0 °C → 20 °C = 30 × 4.2 × 20 = 2 520 J
total = 10 080 + 2 520 = 12 600 J
HEAT LOST by the water, cooling 70 °C → 20 °C (Δθ = 50 °C):
= m × 4.2 × 50 = 210m
Heat lost = heat gained: 210m = 12 600 m = 12 600 ÷ 210 m = 60 g
Mass of water = 60 g
This one is the usual calorimetry question turned round: the ice mass is given and the WATER mass is the unknown. The method does not change — everything that warms or melts on one side, everything that cools on the other.
The melted ice warms from 0 °C to 20 °C, a rise of 20 °C, while the hot water cools from 70 °C to 20 °C, a fall of 50 °C. Two different values of Δθ, as always.
The melting term, 10 080 J, is by far the largest quantity here — leave it out and you get 12 g instead of 60 g.