Question

How many grams of ice at ` -14 .^(@)C` are needed to cool 200 gram of water form `25 .^(@)C` to `10 .^(@)C`? Take specific heat of ice `=0.5 cal g^(-1) .^(@)C^(-1)` and latant heat of ice = `80 cal g^(-1)` .

Answer 1

Here, `m_(ice) = ? M_(w0 = 200 g`,
`s_(ice) = 0.5 cal g^(-1) .^(@)C^(-1), L_(ice) = 80 cal g^(-1)`
Heat lost by water in cooling from `25^(@)C` to `10^(@)C` is
`Q_(1) = m_(s) xx s_(w) xx Delta T_(1) = 200 xx 1 xx (25 -10)`
`= 3000 cal`.
Heat gained by ice at `-14^(@)C` to change into water at `10^(@)C` is
`Q_(2) = m_(ice) s_(ice) Delta T_(2) +m_(ice) L_(ice) +m_(ice) xx s_(w) xx DeltaT_(3)`
`= m xx 0.5 xx [0-(-14)] + m xx 80`
`+m xx 1 xx(10-0)`
`=97 m cal`
As heat lost = heat gained , so, `Q_(1) = Q_(2)`
` 3000 = 97 m or m = 3000/97 = 31 g`.