Question

An ice cube of mass 0.1 kg at $0^{\circ }$ C is placed in an isolated container which is at ${227}^{\circ }$C. The specific heat s of the container varies with temperature T according to the empirical relation s = A + BT, where A = 100 $\frac{cal}{kg-K}$ and B = $2\times {10}^{-2}\frac{cal}{kg-K^2}$. If the final temperature of the container is ${27}^{\circ }$, determine the mass (in Kg) of the container. (Latent heat of fusion for water = $8\times {10}^4\frac{cal}{kg}$, specific heat of water = ${10}^3\frac{cal}{kg-K}$. Answer upto 2 decimal places.

Answer 1

Let m be the mass of the container,
Initial temperature of container,
$T_i$ = (227 + 273) = 500 K
and final temperature of container
$T_f$ = (27 + 273) = 300 K
Now, heat gained by the ice cube = heat lost by the container
$\therefore \left(0.1\right)(8\times {10}^4)+\left(0.1\right)\left({10}^3\right)\left(27\right)=-m{\int }_{500}^{300}(A+BT)dT$
or 10700 = -m ${[AT+\frac{B{T}^2}{2}]}_{500}^{300}$
After substituting the values of A and B and the proper limits, we get
m = 0.495 kg