Calculate the mass of ice required to lower the temperature of 300 g of water at 40°C to water at 0°C. (Specific latent heat of ice = 336 J g^-1, Specific heat capacity of water = 4.2 J g^-1 °C^-1)

Given,

mass of water m = 300 g

initial temperature = 40°C

final temperature = 0°C

Therefore, fall in temperature = 40 - 0 = 40 °C

Heat lost by water = m x c x Δt

= 300 x 4.2 x 40 = 5.04 x 10⁴ J

If m' g ice is added, heat gained by it to melt to 0 °C = m'L = m' x 336 J

By the principle of method of mixture,

heat lost by water = heat gained by ice

Therefore, 5.04 x 10⁴ = m' x 336

m' = $\dfrac{5.04 \times 10^{4}}{336}$ = 150 g