Solved 2018 Question Paper ICSE Class 10 Physics — Question 15

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

$3.213 \times 10^4 = \text{m}' \times 3.57 \times 10^5 \\[0.5em] \text{m}' = \dfrac{3.213 \times 10^4}{3.57 \times 10^5} \\[0.5em] \text{m}' = 0.09 \text{ kg} = 90 \text{ g}$

Hence, mass of ice added = 90 g