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Question
20 g of ice at 0°C is added to 200g of water at 20°C. Calculate the drop in temperature ignoring the heat capacity of the container. (Specific latent heat of ice = 80 cal/g)
Solution
Let x be the drop in temperature of water
Heat lost by water = 200 × x
Heat gained by ice in melting = 20 × 80 = 1600 calories
Heat gained by the water from ice in rising from 0°C to (20 - x)°C = 20 × 1 × (20 - x)
∵ Heat gained = Heat lost
∴ 200 × x = 20 × 80 + 20 (20 - x)
or 200x + 20x = 400 + 1600
x = `2000/220 = 9.1^circ "C"`
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