When `1.5 kg` of ice at `0^(@)C` mixed with `2` kg of water at `70^(@)C` in a container, the resulting temperature is `5^(@)C` the heat of fusion of ice is (`s_("water") = 4186 j kg^(-1)K^(-1))`A. `1.42 xx 10^(5) j kg^(-1)`B. `2.42 xx 10^(5) j kg^(-1)`C. `3.42 xx 10^(5) j kg^(-1)`D. `4.42 xx 10^(5) j kg^(-1)`

When `1.5 kg` of ice at `0^(@)C` mixed with `2` kg of water at `70^(@)

1.	When `1.5 kg` of ice at `0^(@)C` mixed with `2` kg of water at `70^(@)C` in a container, the resulting temperature is `5^(@)C` the heat of fusion of ice is (`s_("water") = 4186 j kg^(-1)K^(-1))`A. `1.42 xx 10^(5) j kg^(-1)`B. `2.42 xx 10^(5) j kg^(-1)`C. `3.42 xx 10^(5) j kg^(-1)`D. `4.42 xx 10^(5) j kg^(-1)`
Answer» Correct Answer - C Heat lost by water `= m_(w)s_(w) (T_(i) - T_(f))` `= 2 xx 4186 xx (70-5) = 544180 j` Heat required to rise temperature of melt ice `= m_(i)L_(f) = 1.5 xx L_(f)` Heat required to rise temperature of ice `= m_(i)s_(w) (T_(f) - T_(0)) = 1.5 (4186) xx (5-0^(@)) = 31395 j` By the principle of calorimetry Heat lost = heat gained `544180 = 1.5 L_(f) + 31395` `:. L_(f) = (512785)/(1.5) = 341856.67 = 3.42 xx 10^(5) j kg^(-1)`

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