Question

Water at ${50}^{\circ }C$ is filled in a cubical container of side $1m$. The thickness of the walls of the container is $1mm.$ The container is surrounded by large amount of ice at $0^{\circ }C$. The temperature of the water becomes ${25}^{\circ }C$ in $10ln2$ seconds. Choose the correct options.
[Given, specific heat of water $=1cal/\left(g{m}^{\circ }C\right)$; latent heat of fusion of ice $=80cal/gm$; density of water $=1gm/{cm}^3$; heat capacity of the container =1 $J/K$]

• A. Thermal conductivity of the material is $70J/m^{\circ }C$
• B. Thermal conductivity of the material is $60J/m^{\circ }C$
• C. Mass of the ice melted is $312.5kg$
• D. Mass of the ice melted is $252kg$

Answer 1

Answer: (A), (C)

Mass of the ice melted is $312.5kg$

Let temperature of water at any instant be $T$, then heat current
$$i=\frac{kA}{x}.(T-0)-\left(1\right)$$
where $A=6a^2=6m^2;x=$ thickness $=1mm={10}^{-3}m$
Rate of heat lost from water, $\frac{dQ}{dt}=+ms\frac{dT}{dt}-\left(2\right)$
Ratw of heat lost from water = Rate of heat transfer through the wall
So, we get from $\left(1\right)\&\left(2\right),-ms\frac{dT}{dt}=\frac{kAT}{x}\Rightarrow -{\int }_{{50}^{\circ }}^{{25}^{\circ }}\frac{dT}{T}=\frac{kA}{msx}{\int }_0^{10\ell n2}dt\Rightarrow \ln \left(2\right)=\frac{kA}{msx}\cdot 10\left(\ln 2\right)$ So, $\frac{kA\left(10\right)}{msx}=1$

On Putting values we get $k=\frac{msx}{10A}=\frac{\left({10}^{3}kg\right)(4.2\times {10}^{3}J/{kg}^{\circ }C){10}^{-3}m}{10\times \left(6m^3\right)}=k=70J/m^{\circ }C$
As we know, total heat transferred through the container $=$ total heat absorbed by ice.
$$\begin{array}{c}Q=\int dQ=\int \frac{kA}{x}Tdt\text{Lost by water.}\\ Q=mS\Delta T={10}^3\times 4200\times 25J=\left({10}^{\circ }gm\right)\left(1\text{cal)}\right(25)=m_{ice}L\end{array}$$
$⟹m_{ice}=\left(\frac{{10}^6\times 25}{80}\right)gm=\frac{25000}{80}kg=312.5kg$
mass of ice meited $=312.5kg$