A colorimeter contains `400 g` of water at a temperature of `5^(@)C`. Then, `200 g` of water at a temperature of `+10^(@)C` and `400 g` of ice at a temperature of `-60^(@)C` are added. What is the final temperature of the contents of calorimeter?
Specific heat capacity of water `-1000` cal`//kg//K`
Specific latent heat of fusion of ice `=80xx1000` cal`//kg`
Relative specific heat of ice `=0.5`

Explanation

Let t be the common temperature above zero celsius.
Heat lost by calorimeter and water added
`=400xx10^(-3)xx1000(5-t)+200xx10^(-3)xx1000(10-t)`
Heat gained by ice `=400xx10^(-3)xx5000(60+t)+400xx10^(-3)xx80000+400xx10^(-3)xx1000t`
Since Heat lost = heat gained `therefore400(5-t)+200(10-t)=200(60+t)+400xx80+400t`
or `t=-24.33^(@)C`
This is contradictory of the assumption that `t` is above zero.
Hence, the final temperature is either `0^(@)C` or less than `0^(@)C. Let the temperature be `t^(*@)C` (less than `0^(@)C).
Then `400xx10^(-3)xx1000xx5+200xx10^(-3)xx100xx10+600xx10^(-3)80000+600xx10^(-3)xx500xxt=400xx500(60-t)`
or `400xx5+200xx10+600xx80+300t=200(60-t)`
or `20+20+480+3t=120-2t`
or `t=-80^(@)C`
This is absurd because the lowest temperature is `-60^(@)C` and we dicard this second asumption also.
So the final temperature is `0^(@)C`.
Let x grams of water be converted into ice.
Heat lost `=400xx10^(-3)xx100xx5+200xx10^(-3)xx1000xx10+x xx10^(-3)xx8000-4000+80x`
Heat gained `=400xx10^(-3)xx500xx60`
or `4000+80x=12000`
or `x=100g`
Hence, the final result is `500g` of ice and `500g` of water at `0^(@)C`.