Question

Calculate the mass of ice required to lower the temperature of 300 g of water 40°C to water at 0°C.

(Specific latent heat of ice = 336 J/g, the Specific heat capacity of water = 4.2J/g°C)

Answer 1

Let m be the mass of the ice to be added.

Heat energy required to melt to lower the temperature is = mL = m x 336

Heat energy imparted by the water in fall of its temperature from 40°C to

0°C = mass of the water x specific heat capacity x fall in temperature

= 300 x 4.2 x 40°C

If there is no loss of heat,

m x 336 J/g = 300 g x 4.2 J/g°C x 40°C

`:. m = (300 xx 4.2 xx 40)/336`

∴ m = 150g