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`).

Answer

D

Explanation

Heat lost by container
`Q=int_(500)^(300)(A+BT)dt =-21600`m cal.
Heat gained by ice in melting `Q_(1)=0.1xx80000=8000` cal
Heat gained by water when its temp rises from `0^(@)C` to `27^(@)C`
`Q_(2)=0.1xx1000xx27=2700` cal
Total heat gained `=10700` cal
`21600m=10700 :.m=0.49kg`