A piece of ice of mass 60 g is dropped into 140 g of water at 50°C. Calculate the final temperature of water when all the ice has melted. (Assume no heat is lost to the surrounding) Specific heat capacity of water = 4.2 Jg–1k–1 Specific latent heat of fusion of ice = 336 Jg–1

Given: ice 60 g at 0 °C dropped into water 140 g at 50 °C

c_w = 4.2 J g⁻¹ °C⁻¹, L = 336 J g⁻¹ Let the final temperature be T °C.

= 140 × 4.2 × (50 − T) = 588(50 − T)

HEAT GAINED by the ice, in two stages: melting the ice at 0 °C = 60 × 336 = 20 160 J warming that melted water from 0 °C to T = 60 × 4.2 × T = 252T

Heat lost = heat gained: 588(50 − T) = 20 160 + 252T 29 400 − 588T = 20 160 + 252T 29 400 − 20 160 = 252T + 588T 9 240 = 840T T = 9 240 ÷ 840 T = 11 °C

Final temperature of the mixture = 11 °C

Here the unknown is the temperature itself, so it appears on BOTH sides of the equation — once in the (50 − T) for the cooling water and once in the T for the warming melted ice. Collect the T terms on one side and the numbers on the other.

The ice still does two things: it melts at 0 °C, then the melted water warms to T. Leaving out the melting term gives a badly wrong answer, because 20 160 J is the largest single quantity in the problem.

Before solving, check that all the ice CAN melt: melting it needs 20 160 J, and the water cooling all the way to 0 °C could supply 140 × 4.2 × 50 = 29 400 J. Since 29 400 > 20 160 there is enough heat, so the ice all melts and the final temperature is above 0 °C. If there had not been enough, the answer would simply be 0 °C with some ice left unmelted — always worth testing first.