Question

$Mkg$ of water at $t{}^{\circ }C$ is divided into two parts so that one part of mass $mkg$ 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,
$m\times 80+m\times 1\times t=(M-m)\times 1\times (100-t)+540(M-m)$
$\Rightarrow m\times 80+mt=(M-m)\times 100-(M-m)t+540M-m\times 540$
$\Rightarrow 80m+mt=M\times 100-m\times 100-Mt+mt+540M-540m$
$\Rightarrow 80m+100m+540m=640M-Mt$
$\Rightarrow 720m=M(640-t)$
$\Rightarrow \frac{m}{M}=\frac{640-t}{720}$