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When 0.15 kg of ice at 0 ^(@) 0 is mixe...

When` 0.15 `kg of ice at `0 ^(@) 0` is mixed with `0.30` kg of water at `50 ^(@) C` in a container, the resulting temperature is `6.7 ^(@) C` . Calculate the heat of fusion of ice. `(S_("water") = 4186 J kg^(-1) K^(-1))`

Text Solution

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Heat lost by water ` = m s_(w) (theta_(f) - theta_(1))_(w)`
` = (0.30 kg)(4186 J kg^(-1) K^(-1)) (50.0^(@) C - 6.7 ^(@) C)`
` = 54376.14 J`
Heat required to melt ice ` = m_(2) L_(f) = (0.15 kg) L_(r) `
Heat required to raise temperature of ice water to final temperature ` = m_(1) s_(w) (theta_(f) - theta_(i))_(I)`
` = (0.15 kg) (4186 J kg^(-1) K^(-1)) (6.7 ^(@) C - 0^(@) C)`
` = 4206.93 J`
Heat lost = heat gained
`54376.14 J = (0.15 kg) L_(f) + 4206.93 J`
` L_(f) = 3.34 xx 10^(5) J kg^(-1)`.
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