An ice cube of mass 0.1 kg at `0^@C` is placed in an isolated container which is at `227^@C`. The specific heat s of the container varies with temperature T according to the empirical relation `s=A+BT`, where `A= 100 cal//kg.K and B = 2xx 10^-2 cal//kg.K^2`. If the final temperature of the container is `27^@C`, determine the mass of the container.
(Latent heat of fusion for water = `8xx 10^4 cal//kg`, specific heat of water `=10^3 cal//kg.K`).

Explanation

Given
Initial temperature of container, `T_(1)`=227+273=500K
Final temperature of container, `T_(2)`=27+273=300K
Specific heat of container, c = A +BT = 100+2xx10^(-2)`T
Let m is mass of container.
`DeltaQ=int_(T_(1))^(T_(2))`mcdT = `int_(500)^(300)(100+2xx10^(-2)T)`dT
`m|100T+2xx10^(-2)T^(2)/2|_(500)^(200)`
`=m[100(300-500)+10^(-2)(300^(2)-500^(2)]`
`=m[-2xx10^(4)+10^(-2)(300+500)(300-500)]`
`=m[-2xx10^(4)-1600]`
=-21600m
Heat gained by ice = mL + `mc_(w)DeltaT`
`=0.1 xx8xx10^(4)+0.1xx10^(3)(27-0)`
=10700 cal
Heat lost by container = Heat gained by ice
21600m =10700
m=0.495kg