6(c)
A copper vessel of mass 100 g contains 150 g of water at 50 °C. How much ice is needed to cool it to 5 °C?
Given : Specific heat capacity of copper = 0.4 J g-1 ⁰C-1
Specific heat capacity of water = 4.2 J g-1 ⁰C-1
Specific latent heat of fusion of ice = 336 J g-1
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Heat energy imparted by vessel = 100 x 0.4 x (50 - 5) = 1800 J
Heat energy imparted by water = 150 x 4.2 x (50 - 5) = 28,350 J
Let m g of ice be used, then
Heat energy taken by ice to melt = m x 336 J
Heat energy taken by melted ice to raise it's temperature from 0 °C to 5 °C = m x 4.2 x (5 - 0) = 21m J
By law of conservation of energy
heat energy imparted by vessel and water = heat energy taken by ice and melted ice
1800 + 28,350 = 336m + 21m
357 m = 30,150
m = $\frac{30150}{357}$ = 84.45 g