Question

$2\text{kg}$ ice at $-{20}^{\circ }\text{C}$ is mixed with $5\text{kg}$ water at ${20}^{\circ }\text{C}$. Then final amount of water in the mixture would be
Given specific heat of ice $=0.5{\text{cal/g}}^{\circ }\text{C}$,
Specific heat of water $=1{\text{cal/g}}^{\circ }\text{C}$,
Latent heat of fusion for ice $=80\text{cal/g}$.

• A. $6\text{kg}$
• B. $5\text{kg}$
• C. $4\text{kg}$
• D. $2\text{kg}$

Answer 1

The correct option is A $6\text{kg}$

$2\text{kg}$ ice at $-{20}^{\circ }\text{C}+5\text{kg}$ water at ${20}^{\circ }\text{C}$
Let $M\text{kg}$ of ice melts into water
From principle of calorimetry
$Q_{\text{gain}}=Q_{\text{lost}}$
$2\times \frac{1}{2}\times 20+M\times 80=5\times 1\times 20$
$M=1\text{kg}$

Final amount of water is $=5+1=6\text{kg}$