Question

$2kg$ of ice at $-20°C$ is mixed with $5kg$ of water at $20°C$ . The water content of the final mixture is (Latent heat of ice $=80 \, kcalkg^{- 1}$ , the specific heat of water $=1kcalkg^{- 1 }^\circ C^{- 1}$ and specific heat of ice $=0.5 \, kcal \, kg^{- 1}^\circ C^{- 1}$ )

• A. $7 \, kg$
• B. $6 \, kg$
• C. $4 \, kg$
• D. $3 \, kg$

Answer 1

Answer: (B)

Here, $m$ , $s$ and $\theta _{1}$ are the mass, specific heat and temperature difference for water respectively. Similarly, $M$ , $S$ and $\theta _{2}$ are the corresponding quantities for ice. Here, $L$ is the latent heat of fusion.
We have, the heat lost by water $=ms\theta $ , heat gained by ice $=MS\theta +ML$ .
From the principle of calorimetry, we have, the heat lost by water=heat gained by ice.
Hence, $ms\theta _{1}=MS\theta _{2}+ML$ .
$5\times 1\times \left(20 - 0\right)=2\times 0.5\times 20+x\times 80$
$x=1kg$
$1kg$ of ice melts to form water.
Hence, final amount of water $5kg+1kg=6kg$ .