Question

The two opposite faces of a cubical piece of iron (thermal conductivity $=0.2\text{CGS units}$) are at ${100}^{\circ }\text{C}$ and $0^{\circ }\text{C}$. The face at $0^{\circ }\text{C}$ is in contact with the ice at $0^{\circ }\text{C}$. If the area of a surface is $4{\text{cm}}^2$ and latent heat of fusion of ice is $80\text{J/g}$, then the mass of ice melted in $10$ minutes will be:

• A. $30\text{gm}$
• B. $300\text{gm}$
• C. $5\text{gm}$
• D. $50\text{gm}$

Answer 1

The correct option is B $300\text{gm}$

According to the question:
Equating latent heat of fusion of ice with heat transfer by conduction through iron (cubical in shape)
$Q=mL=KA\frac{({\theta }_1-{\theta }_2)}{l}t$
$\Rightarrow m=\frac{1}{L}\times KA\frac{({\theta }_1-{\theta }_2)}{l}\times t$
$\Rightarrow m=\frac{1}{80}\times 0.2\times 4\times \frac{(100-0)}{\sqrt{4}}\times 10\times 60$ [$\because l^2=4$ for cubical block]
$\Rightarrow m=\frac{0.2\times 4\times 100\times 600}{80\times 2}=300\text{gm}$