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`).

Options

A. `0.495 kg`
B. `0.224 kg`
C. `0.336 kg`
D. `0.621 kg`

Answer

A

Explanation

Heat recived by ice is
`Q_(1)=mL+mCDelta T = 10700cal`
Heat lost be the container is
`Q_(2) = underset(300)overset(500)(int) m_(C) (A+BT)dT = m_(C)[AT +(BT^2)/(2)]_(300)^(500)`
`=21600m_(C)`
By principle of calorimetry, `Q_(1)=Q_(2)`
`rArr m_(c) = 0.495 kg`.