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Question

40g of ice at 0oC is used to bring down the temperature of certain mass of water at 60oCto10oC. Find the mass of water used.
[Specific heat capacity of water =4200 JKg1C1
[Specific latent heat of fusion of ice =336×103Jkg1]

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Solution

Let, mass of water used =m
Since, Heat gained = Heat lost
m1cΔt1=mL+m2cΔt2
m×4.2×(6010)=(40×336)+(40×4.2×(100))
m×4.2×50=(40×336)+1680
Therefore, the mass of water used, m=13440+168042×5=15120210=72g

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