$M \mathrm{~kg}$ of water at $t^{\circ} \mathrm{C}$ is divided into two parts so that one part of mass $m…
$M \mathrm{~kg}$ of water at $t^{\circ} \mathrm{C}$ is divided into two parts so that one part of mass $m \mathrm{~kg}$ when converted into ice at $0^{\circ} \mathrm{C}$ would release enough heat to vapourise the other part, then $\frac{m}{M}$ is equal to (Specific heat of water $=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$, Latent heat of fusion of ice $=80 \mathrm{cal} \mathrm{g}^{-1}$, Latent heat of steam $=540 \mathrm{cal} \mathrm{g}^{-1}$ )
$640 - t$
$\frac{720-t}{640}$
$\frac{640+t}{720}$
$\frac{640-t}{720}$
Solution
Specific heat of water $=1 \mathrm{cal} \mathrm{g}^{-1 \circ} \mathrm{C}^{-1}$
Latent heat of fusion of ice $=80 \mathrm{cal} \mathrm{g}^{-1}$
Latent heat of steam $=540 \mathrm{cal} \mathrm{g}^{-1}$
$\begin{aligned}
& m \times 80+m \times 1 \times t=(\mathrm{M}-m) \times 1 \times(100-t)+540(\mathrm{M}-m) \\
& \Rightarrow m \times 80+m t=(\mathrm{M}-m) \times 100-(\mathrm{M}-m) t+540 \mathrm{M}-\mathrm{m} \times 540 \\
& \Rightarrow 80 m+m t=\mathrm{M} \times 100-m \times 100-\mathrm{M} t+m t+540 \mathrm{M}-540 \mathrm{~m} \\
& \Rightarrow 80 m+100 m+540 m=640 \mathrm{M}-\mathrm{M} t \\
& \text { or, } 720 m=\mathrm{M}(640-t) \\
& \therefore \frac{m}{M}=\frac{640-t}{720}
\end{aligned}$