Question

$M\, kg$ of water at $t{ }^{\circ} C$ is divided into two parts so that one part of mass $m kg$ when converted into ice at $0^{\circ} C$ would release enough heat to vapourise the other part, then $\frac{m}{M}$ is equal to
(Specific heat of water $=1$ cal $g ^{-1}{ }^{\circ} C ^{-1}$,
Latent heat of fusion of ice $=80$ cal $g ^{-1}$,
Latent heat of steam $=540$ cal $g ^{-1}$ )

• A. $640-t$
• B. $\frac{720-t}{640}$
• C. $\frac{640+t}{720}$
• D. $\frac{640-t}{720}$

Answer 1

Answer: (D)

Given, Specific heat of water $=1$ cal $g^{-1}{ }^{\circ} C ^{-1}$
Latent heat of fusion of ice $=80$ cal $g ^{-1}$
Latent heat of steam $=540$ cal $g^{-1}$
According to the question,
$m \times 80+m \times 1 \times t=(M-m) \times 1 \times (100-t)+540(M-m)$
$\Rightarrow m \times 80+m t=(M-m) \times 100-(M-m) t+540 M-m \times 540$
$\Rightarrow 80 m+m t=M \times 100-m \times 100-M t+m t+540\, M-540 m$
$\Rightarrow 80\, m+100 \, m+540\, m=640\, M-M t$
$\Rightarrow 720\, m=M(640-t)$
$ \Rightarrow \frac{m}{M}=\frac{640-t}{720}$