Question

$250g$ of ice at $0^{\circ }C$ converts into water at $0^{\circ }C$ in a minute when a constant heat is supplied. In how much time (sec) the water thus obtained will change to ${20}^{\circ }C$? The specific latent heat of ice is $336J{g}^{-1}$. The specific heat capacity of water is $4.2J{g}^{-1}{}^{\circ }{C}^{-1}$.

• A. 15

Answer 1

The correct option is A 15
Given:
Mass of the ice, $m_i=250g$
Specific latent heat (of melting) of ice, $L_i=336J{g}^{-1}$
Specific heat capacity of water, $c_w=4.2J{g}^{-1}{}^{\circ }{C}^{-1}$
Time taken by the ice to completely melt into water, $t=1min$
Required rise in temperature of the water, $\Delta t=20-0={20}^{\circ }C$
Heat energy absorbed by the ice to melt, $Q_i=m_i\times L_i=250\times 336=84000J$
This heat is supplied in one minute. The rate of heat supply, $R=84000Jmi{n}^{-1}$
Heat energy required by the water to have the required rise in temperature, $Q_w=m_w\times c_w\times \Delta t$
$⟹Q_w=250\times 4.2\times 20=21000J$
So, the time needed for water to raise its temperature from $0^{\circ }C$ to ${20}^{\circ }C$, $t=\frac{21000}{84000}$
$⟹t=\frac{1}{4}=0.25min=15sec$