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Question
The temperature of 170 g of water at 50°C is lowered to 5°C by adding a certain amount of ice to it. Find the mass of ice added.
Given: Specific heat capacity of water = 4200 J kg-1 °C-1 and specific latent heat of ice = 336000 J kg-1.
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Solution
Given mass of water = 170 g = 1.17 kg,
Initial temperature = 50°C
Fall in temperature = Δt = (50 - 5) = 45°C = 45K
Heat lost by water = mc Δt
= 0.17 × 4200 × 45
= 3.213 × 104 J
If m' kg ice is added, heat gained by it to melt to 0°C = m'L
= m'C Δt
= m' × 4200 × 5
= m' × 2.1 × 104 J
Total heat gained by ice
= 3.36 × 105 m' + 2.1 × 104 m'
= 3.57 × 105 m' J
By the principle of method of mixtures heat lost by water = heat gained by ice
⇒ 3.213 × 104 = 3.57 × 105 m'
⇒ m' = `(3.213 xx 10^4)/(3.57 xx 10^5)`
⇒ m' = 0.09 kg (90 g)
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