When 1.5 kg of ice at 0 mixed with 2 kg of water at 70 in a container, the resulting temperature is 5.Then the heat of fusion of ice is (swater=4186Jkg−1K−1)
A
1.42×105Jkg−1
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B
2.42×105Jkg−1
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C
3.42×105Jkg−1
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D
4.42×105Jkg−1
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Solution
The correct option is B3.42×105Jkg−1
Heat lost by water =mwsw(Ti−Tf)
=2×4186×(70−5)=544180J
Heat required to melt ice =mfLf=1.5×Lf
The heat required to raise the temperature of the ice