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Question
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 Jg-1 °C-1
The Specific heat capacity of water = 4.2 Jg-1 °C-1
The Specific latent heat of fusion ice = 336 Jg-1
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Solution
Heat energy lost by the vessel and water contained in it in cooling the water from 50°C to 5°C is used in heating ice to melt it and then to raise its temperature from 0°C to 5°C.
Now, heat energy lost by the copper vessel is
Qc = mcccΔt = 100 x 0.4 x (50 - 5)
Qc = 1800 J
Similarly, heat energy lost by water is
Qw = mwcwΔt = 150 x 4.2 x (50 - 5)
Qw = 28350 J
Hence, the total heat energy lost is
QL = 1800 + 28350 = 30150 J
Let m g of ice be used to cool water. So, heat gained by ice is
QI = mLice + mcwΔt
QI = 336m + m x 4.2 x 5 = 336m + 21m = 357mJ
Therefore, from the principle of calorimetry, the mass of ice is
QL = QI
∴ 357m = 30150
∴ m = `30150/357` = 84.45 g
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