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Question
Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32 °C such that the final temperature is 5 °C. Specific heat capacity of calorimeter = 0.4 J g-1 °C-1, Specific heat capacity of water = 4.2 J g-1°C-1, latent heat capacity of ice = 330 J g-1.
Solution
Heat lost by hot water = Heat gained by ice + Heat lost by calorimeter + Heat gained by melted ice
`("MC"Δ_t)_"water" + ("MC"Δ_t)_"calorimeter" = "MLL"_"F" + ("MC"Δ_t)_"Melted ice"`
∴ 150 × 4.2 (32 - 5) + 50 × 0.4 × (32 - 5) = M × 330 + M × 4.2 × (5 - 0)
∴ (150 × 4.2 × 27) + (50 × 0.4 × 27) = 330 M + 21 M
17010 + 540 = 351 M
∴ M = `17550/351`
∴ M = 50
Mass of ice required = 50g.
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