Question

A vessel containing $100\text{g}$ water at $0^{\circ }C$ is left in the middle of a room. In $15$ minutes, the temperature of the water rises by $2^{\circ }C$. When an equal amount of ice is placed in the vessel, it melts in $10$ hours. The specific heat of fusion of ice is
(Take specific heat capacity of water as $1\text{cal/g.deg}$)

• A. $50\text{cal/g}$
• B. $80\text{cal/g}$
• C. $70\text{cal/g}$
• D. $40\text{cal/g}$

Answer 1

The correct option is B $80\text{cal/g}$

Given that,
Mass of water = Mass of ice, $m=100\text{g}$
Specific heat capacity of water, $s=1\text{cal/g.deg}$
Change in temperature of water in $15\text{min},\Delta T=2^{\circ }C$

The amount of heat required to rise the temperature of $100\text{g}$ of water from $0^{\circ }C$ to $2^{\circ }C$

$=ms\Delta T=100\times 1\times 2=200\text{cal}$

$200\text{cal}$ heat is supplied in $15$ minutes.
The rate of heat supply per second = $\frac{200}{15\times 60}=\frac{2}{9}\text{cal/s}$

So,total heat supplied in $10$ hours,
$Q=10\times 60\times 60\times \frac{2}{9}=8000\text{cal}$

$8000\text{cal}$ melts same amount of ice $(=100\text{g})$.
The heat required to melt $1\text{g}$ of ice = $\frac{8000\text{cal}}{100\text{g}}=80\text{cal/g}$.
So, latent heat of fusion of ice is $80\text{cal/g}$

Hence, option (d) is correct.