Question

Heat is being supplied at a constant rate to the sphere of ice, which is melting at the rate of $0.1\text{g/s}$. It melts completely in $100\text{s}$. The rate of rise of temperature thereafter will be-
[Take $L=80\text{cal/g}$]

• A. $0.4{}^{\circ }{\text{C s}}^{-1}$
• B. $2.1{}^{\circ }{\text{C s}}^{-1}$
• C. $3.2{}^{\circ }{\text{C s}}^{-1}$
• D. $0.8{}^{\circ }{\text{C s}}^{-1}$

Answer 1

The correct option is D $0.8{}^{\circ }{\text{C s}}^{-1}$

Given, rate of melting of ice is,

$\frac{dm}{dt}=0.1{\text{g s}}^{-1}$

The amount of heat required to melt ice is,

$Q=mL.......\left(1\right)$

On differentiating $\left(1\right)$ w.r.t time, we get,

$\frac{dQ}{dt}=L\frac{dm}{dt}$

$\frac{dQ}{dt}=80\times 0.1=8\text{cal/s}$

Total mass of water,
$m=\frac{dm}{dt}\times t=0.1\times 100=10\text{g}$

Also, we know $Q=msdT$

$\Rightarrow \frac{dQ}{dt}=ms\frac{dT}{dt}$

$\Rightarrow 8=10\times 1\times \frac{dT}{dt}$

$\therefore \frac{dT}{dt}=\frac{8}{10}=0.8{}^{\circ }{\text{C s}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{D}\right)$ is the correct answer.