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When 0.45 kg of ice of 0^(@)C mixed w...

When 0.45 kg of ice of `0^(@)C` mixed with 0.9 kg of water at `55^(@)C` in a container the resulting temperature is `10^(@)C` . Calculate the heat of fusion of ice .

Text Solution

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`c_("water ") = 4186 J kg^(-1) k^(-1)`
`m_(ice) = 0.45 kg, Delta T = 55 - 10 = 45^(@)C`
heat lost by 0.9 kg water when its temperature falls from `55^(@)C` to `10^(@)C`
`= mc_("water ") Delta T = 0.9 xx 4186 xx 45^(@)`
`=169533 J`
Heat required to melt 0.45 kg ice into water at `0^(@)C`
`= mL_(f ) = 0.45 xx L_(f ) J`
Heat required to raise temperature of 0.45 kg water from` 0^(@)C` to `10^(@)C`
`= mc_("water ") Delta T`
`= 0.45 xx 4186 xx (10-0) = 4186 xx 10 xx 0.45`
`=18837 J`
By the principle of calorimetry
Heat gained = Heat lost
`0.45 xx L_(f ) = 169533 - 18837`
`0.45 xx L_(f ) = 169533 - 18837 `
`L_( f) =33 4880 J kg^(-1)`
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