The temperature of 170g of water at 50°C is lowered to 5°C by adding a certain amount of ice to it. Find the mass of ice added.
Given mass of water = 170 g = 0.17 kg
T1 = 50°C, T2 = 5°C,
specific heat capacity of water = 4200 J kg –1°C–1
Sp. Latent heat of ice = 336000 J kg–1
Let mass of ice added be x then we have heat lost by water = Heat used by ice.
⇒ water at 50°C to 5°C = ice at °C to water at °C + water at 0°C to 5°C
`=> 0.17 xx 4200 xx (50 - 5)`
`= x xx 336000 + x xx 4200 xx 5 = 32130`
`= 357000 x`
or, `x = 32130/357000`
`= 0.09 Kg`
`x = 90 g`
we should add 90 g of ice to water.