A copper vessel of mass 100 g contains 150 g of water at 50 °C. How much ice is needed to cool it to 5 °C? Given : Specific heat capacity of copper = 0.4 J g-1 ⁰C-1 Specific heat capacity of water = 4.2 J g-1 ⁰C-1 Specific latent heat of fusion of ice = 336 J g-1
Topic: Nuclear physics / radioactivity
Answer
Given: copper vessel, mass 100 g, c_Cu = 0.4 J g⁻¹ °C⁻¹
water, mass 150 g, c_w = 4.2 J g⁻¹ °C⁻¹ initial temperature 50 °C, final temperature 5 °C specific latent heat of fusion of ice L = 336 J g⁻¹ Let the mass of ice be m grams, added at 0 °C.
HEAT LOST by the vessel and the water, cooling 50 °C → 5 °C (Δθ = 45 °C):
vessel = 100 × 0.4 × 45 = 1 800 J water = 150 × 4.2 × 45 = 28 350 J total = 1 800 + 28 350 = 30 150 J
HEAT GAINED by the ice, in two stages: melting the ice at 0 °C = m × 336
warming the melted water 0 °C → 5 °C = m × 4.2 × 5 = 21m
total = 336m + 21m = 357m
Heat lost = heat gained: 357m = 30 150 m = 30 150 ÷ 357 m = 84.45 g
Mass of ice required ≈ 84.5 g
Four marks, and the structure is what earns them. Write HEAT LOST on one side and HEAT GAINED on the other, list every body on each side, then equate.
Two things are easy to drop. The copper vessel loses heat too — it cools from 50 °C to 5 °C along with the water, and ignoring it gives 84.5 → 80.8 g. And the ice does two things, not one: it melts at 0 °C AND the melted water then warms to 5 °C. Leaving out that second stage gives 30 150 ÷ 336 = 89.7 g.
Note the temperature drop is 45 °C for the bodies cooling, but the melted ice warms through only 5 °C, because it starts at 0 °C. Each body gets its own Δθ.