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=2 xx 10^(-2) "cal/kg" K^2`. If the final temperature of the container is `27^@ C`, the mass of the container is (Latent heat of fusion of water `= 8 xx 10^(4)` cal/kg, specific heat of water `=10^3` cal/kg K)

Answer

`0.5`kg

Explanation

Let m be the mass of the container . Initial temperature of container, `T_(1) = (27 + 273) = 500K` and final temperature of container,
`T_(t) = (27 + 273) =300K`
Now, heat gained by the ice cube= heat lost by the container i.e.,(mass of ice)(latent heat of fusion of ice)+(mass of ice) (specific heat of water)
`T_(f) = (227 + 273) = 500K`
`(300K -273K) = -munderset(T_(f))overset(T_(t))intS.dT`
Substituting the values , we have
`(0.1)(8xx10^(4)) + (0.1) (10^(3))(27)`
=`-m underset(500)overset(300)int (A + BT)dT "of" 10700 = -m[AT + (BT^(2))/(2)]_(500)^(300)`
After substituting the values of A and B and the proper limits.
we get `m=0.495`kg.