Question

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

• A. $0.495kg$
• B. $0.595kg$
• C. $0.695kg$
• D. $0.795kg$

Answer 1

The correct option is A $0.495kg$

Total heat energy gained by water $m{L}_f+ms\Delta T$

$\left(0^{o}C\right)$ ice $\Rightarrow$ water $\left(0^{o}C\right)$ $0^o$ water $\Rightarrow {27}^{o}C$ water

$=\left(0.1\right)(8\times {10}^4)+\left(0.1\right)\left({10}^3\right)\left(27\right)$

$=107\times {10}^2$ units

Total energy last by container $={\int }_{T_f}^{T_i}{m}_{con}(A+BT)dT$

$={\int }_{300}^{500}{m}_{con}(100+(2\times {10}^{-2})TdT=m_c{\int }_{300}^{500}100dT+m_c{\int }_{300}^{500}(2\times {10}^{-2})TdT$

$=m_c\left(100\right)[T{]}_{300}^{500}+2\times {10}^{-2}{m}_c{\left[\frac{T^2}{2}\right]}_{300}^{500}$

$=m_c\left(100\right)[500-300]+2\times {10}^{-2}{m}_c[\frac{(500{)}^2}{2}-\frac{(300{)}^2}{2}]$

$=\left(m_c\right)[2.16\times {10}^4]$ units

Now, Total heat last = Total heat gained

$m_c(2.16\times {10}^4)=107\times {10}^3$

$m_c=0.495kg$ $\left(A\right)$