Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at `32^@`C such that the final temperature is `5^@`C. Specific heat capacity of calorimeter = 0.4 J `g^(-1)^@C^(-1)`, specific heat capacity of water = `4.2" J "g^(-1)""^@C^(-1)`, latent heat capacity of ice = 330 J `g^(-1)`.

Explanation

Heat energy imparted by calorimeter and water contained in it in cooling from `32^(@)C" to "5^(@)C` is used in melting ice and then raising the temperature of melted ice from `0^(@)C" to "5^(@)C`.
For cold body : Ice at `0^(@)C` to water at `5^(@)C`.
Heat gained = `mL+mcDeltaT`
= `mxx330+mxx4.2xx(5-0)`
= `330m+mxx4.2xx5`
= 330 m + 21 m
= 351 m joule.
For hot body : (Water + Calorimeter) at `32^(@)C" to "5^(@)C`.
Heat lost = `m_(1)c_(1)DeltaT_(1)+m_(2)c_(2)DeltaT_(2)`
= `150xx4.2xx(32-5)+50xx0.4xx(32-5)`
= `150xx4.2xx27+50xx0.4xx27`
= 17010 + 540 = 17550 joule.
From the principle of calorimetry, if the system is fully insulated then,
Heat gained by cold body = Heat lost by hot body
351m = 17550
`m=17550/351g`
= 50 g.
`therefore` The mass of ice needed = 50 g.