Question

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

• A. $8g,{100}^{\circ }C$
• B. $100g,{90}^{\circ }C$
• C. $92g,{100}^{\circ }C$
• D. $82g,{100}^{\circ }C$

Answer 1

The correct option is C $92g,{100}^{\circ }C$

Let, $L_f=$ Latent Heat of fusion,

$L_v=$ Latent Heat of Vaporization,

$s=$ specific heat, and

$d\theta =$ change in Temperature

To melt $100g$ ice at $273K$, heat required $=m{L}_f=8000cal$

To convert 100g water at 273K to 100g water at 373K:

Heat required $=msd\theta =100\times 1\times 100=10000cal$

To convert 100g water at 373K to steam at 373K, heat required = $m{L}_v=54000cal$

Thus, total heat required to convert 100g ice at 273K to 100g water at 373K is $8000+10000=18000cal$

But to convert the same quantity of ice to steam at 373K, heat required will be $72000cal$

The heat supplied is $22320cal$, therefore, all of ice is not converted to steam. But all of it is necessarily converted to water at 373K first and then some of this water is converted to steam. The additional heat available after all ice has become water at 373K is $22320-18000=4320cal$.

Thus if $M$ grams of water is converted to steam then, $M{L}_v=4320\Rightarrow M=\frac{4320}{540}=8g$.

So, the final mixture is $92g$ water and $8g$ steam in equilibrium at $373K$.