Question

1 kg ice at $-{10}^{\circ }C$ is mixed with 1 kg of water at ${50}^{\circ }C$. When the mixture attains equilibrium, find the mass of ice in final mixture.
$[L_f=80kcal/kg;Sice=\frac{1}{2}kcal/k{g}^{\circ }C;S_{water}=1kcal/k{g}^{\circ }C]$

• A. $\frac{45}{80}kg$
• B. $\frac{35}{80}kg$
• C. $\frac{25}{40}kg$
• D. $\frac{35}{40}kg$

Answer 1

The correct option is B $\frac{35}{80}kg$

$M_{ice}\times S_{ice}\times \Delta T+m{L}_f=M_{water}\times \Delta T1\times \frac{1}{2}\times 10+m\times 80=1\times 50\Rightarrow m=\frac{45}{80}or,1-m=\frac{35}{80}$