QQuestion
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]
Exam-ready answer • 03 Mark
✓Answer
Heat lost by water = 85 × 4.2 × (30 − 5) = 8925 J.
Let mass of ice be ‘m’ g.
Heat gained by ice = mL + mc△t
= m × 336 + m × 4.2 × 5
= 357m.
By the principle of calorimetry, 8925 = 357m.
m = \frac{8925}{357} = 25 g.
Therefore, the mass of ice required is 25 g.
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