Question

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

• 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, $ms\theta _{1}=MS\theta _{2}+ML$ .
$5\times 1\times \left(20 - 0\right)=2\times 0.5\times 20+x\times 80$
$x=1 \,kg$
$1 \,kg$ of ice melts to form water.
Hence, final amount of water $5 \,kg+1 \,kg=6 \,kg$ .