Question

The water of mass 75 g at `100^(@)C` is added to ice of mass `20 g` at `-15^(@)C` . What is the resulting temperature. Latent heat of ice = `80 cal//g` and specific heat of ice = `0.5`.

Answer 1

Let the resulting temperture be `T_(0) .^(@)C`
sp. Heat of water , `s_(1) = 1 cal//g// .^(@)C`
Heat lost by water = `m_(1) s_(1) Delta T_(1)`
` = 75 xx 1 xx (100 - T_(0)) cal`.
Heat gained by ice
(i) From `-15^(@)C to 0^(@)C = m_(2) s_(2) Delta T_(2)`
`= 20 xx 0.5 xx (0+15) = 150 cal`
(ii) In converting into water at `0^(@)C = m_(2)L`
`= 20 xx 80 = 1600 cal`
(iii) in raising the temperature of water formed form `0^(@)C` to `T_(0) .^(@)C`
` = m_(2) s_(1) (T_(0)-0) = 20 xx 1 xx T_(0) = 20 T_(0)` cal
Ac cording to principle of calorimetry,
heat lost = heat gained
`75(100-T_(0)) = 150 + 1600 + 20 T_(0)`
or ` 7500 = 75 T_(0) = 1750 + 20 T_(0)`
or ` 95 T_(0) = 5750 ot T_(0) = (5750)/(95) = 60.5^(@)C`.