85 g of water at 30°C is cooled to 5°C by adding certain mass of ice. Find the mass of ice required. [Specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹, Specific latent heat of fusion = 336 J g⁻¹]
Topic: Principle of calorimetry with ice
Answer
Given: water 85 g at 30 °C, cooled to 5 °C
c_w = 4.2 J g⁻¹ °C⁻¹, L = 336 J g⁻¹ Let the mass of ice be m grams, added at 0 °C.
HEAT LOST by the water, cooling 30 °C → 5 °C (Δθ = 25 °C):
= 85 × 4.2 × 25 = 8 925 J
HEAT GAINED by the ice: melting at 0 °C = m × 336
warming 0 °C → 5 °C = m × 4.2 × 5 = 21m
total = 336m + 21m = 357m
Heat lost = heat gained: 357m = 8 925 m = 8 925 ÷ 357 m = 25 g
Mass of ice required = 25 g
The familiar two-column layout: what cools on the left, what melts and then warms on the right.
The ice has two terms because the final temperature is 5 °C, above the melting point. Leaving out the 21m gives 26.6 g. Contrast this with a question whose final temperature is 0 °C, where the melted ice never warms and the second term genuinely does not exist.
There is no vessel mentioned here, so the lost side has a single term. Read the question for what is present before you start writing.