Question

$22320\, cal$ of heat is supplied to $100\,g$ of ice at $0^{\circ} C$. If the latent heat of fusion of ice is $80 \,cal \,g ^{-1}$ and latent heat of vaporization of water is $540 \,cal \,g ^{-1}$, the final amount of water thus obtained and its temperature respectively are

• A. $8\, g , 100^{\circ} C$
• B. $100 \,g , 90^{\circ} C$
• C. $92\, g , 100^{\circ} C$
• D. $80\, g , 100^{\circ} C$

Answer 1

Answer: (C)

Heat required to convert $100 \,g$ of ice at $0^{\circ} C$ to water at $100^{\circ} C$
$=(100 \,g )\left(80 \,cal \,g ^{-1}\right)+(100 \,g )\left(1 \,cal\, g ^{-1}{ }^{\circ} C ^{-1}\right)\left(100^{\circ} C \right) $
$=8000 \,cal +10000\, cal =18000 \,cal$
But $22320\, cal$ of heat is supplied, so remaining amount of heat
$=22320 \, cal -18000 \,cal =4320 \,cal$
Let the amount of water evaporated by remaining heat be $m$. Then
$m\left(540\, cal \,g ^{-1}\right)=4320\, cal $
or $m=\frac{4320 \,cal }{540 \,cal \,g ^{-1}}=8 \,g$
Thus the final amount of water obtained at $100^{\circ} C =100\, g -8\, g$
$=92\,g $.