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Question Question 3(i)
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' = = 261.639 g
Hence, mass of ice required = 261.64 g