Question

Heat energy is supplied at a constant rate to $100g$ of ice at $0^{\circ }C$. The ice is converted into water at $0^{\circ }C$ in $2$ minutes. How much time will be required to raised the temperature of water from $0^{\circ }C$ to ${20}^{\circ }C$? [Given: specific heat capacity of water $=4.2J{g}^{-1}{}^{\circ }{C}^{-1}$, specific latent heat of ice $=336J{g}^{-1}]$

Answer 1

Given: $m=100g,t=2minutes=2\times 60sec$.
Heat energy taken by ice at $0^{\circ }C$ to convert to water at $0^{\circ }C$.
$Q=mL=100\times 336=33600J$
$\therefore P=\frac{Q}{t}=\frac{33600}{2\times 60}=280J/S$

The heat energy required to convert water at $0^{\circ }C$ to ${20}^{\circ }C$.
$Q=mc△t=100\times 4.2\times 20=8400J$
$Q=P\times t$
$8400=280\times t$
$t=30\sec =0.5min$