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 ⁰C-1 and Specific latent heat of ice = 336000 J kg-1
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Given,
mass of water m = 170 g = 0.17 kg
initial temperature = 50°C
final temperature = 5°C
fall in temperature = (50 - 5) = 45°C = 45 K
Heat lost by water = m x c x Δt
= 0.17 x 4200 x 45
= 3.213 x 104
If m' kg ice is added, heat gained by it to melt to 0°C = m'L
= m' x 3.36 x 105
Heat gained by it to raise temperature by 5°C = m' x C x Δt
= m' x 4200 x 5
= m' x 2.1 x 104 J
Total heat gained by ice = (m' x 3.36 x 105) + (m' x 2.1 x 104)
= m' x 3.57 x 105 J
By the principle of method of mixtures:
heat lost by water = heat gained by ice
$$\begin{aligned}3.213 \times 10^4 = \text{m}' \times 3.57 \times 10^5 \\[0.5em] \text{m}' = \frac{3.213 \times 10^4}{3.57 \times 10^5} \\[0.5em] \text{m}' = 0.09 \text{ kg} = 90 \text{ g}\end{aligned}$$
Hence, mass of ice added = 90 g