Calculate the mass of ice required to lower the temperature of 600 g of water at 50°C to water at 12°C. (Take, specific latent heat of ice = 366 Jg^-1 and specific heat capacity of water = 4.2 Jg^-1°C^-1)

Given,

Mass of water (m) = 600 g

Initial temperature = 50°C

Final temperature = 12°C

∴ fall in temperature = 50 - 12 = 38 °C

Heat lost by water = m x c x Δt

= 600 x 4.2 x 38 = 95,760 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

∴ 95,760 = m' x 366

m' = $\dfrac{95,760}{366}$ = 261.639 g

Hence, mass of ice required = 261.64 g