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Question
50 g of ice at 0°C is added to 300 g of a liquid at 30°C. What will be the final temperature of the mixture when all the ice has melted? The specific heat capacity of the liquid is 2.65 Jg-1 °C-1 while that of water is 4.2 J g-1 °C-1. Specific latent heat of fusion of ice = 336 Jg-1.
Solution
m1 = 50 g, t1 = 0 °C, m2 = 300 g, t2 = 30 °C
Heat energy taken by ice to melt = 50 × 336 J = 16800 J
By law of conservation of energy,
Heat energy given by liquid = Heat taken by water
300 × 2.65 × (30 - t) = 16800 + 50 × t × 4.2
⇒ 23850 - 795 t = 16800 + 210 t
⇒ - 795 t - 210 t = 16800 - 23850
⇒ - 1005 t = - 7050
⇒ t = `7050/1005 = 7.014^circ "C"`
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