Question

A liquid at ${30}^o$C is poured very slowly into a Calorimeter that is at temperature of ${110}^o$C. The boiling temperature of the liquid is ${80}^o$C. It is found that the first $5$gm of the liquid completely evaporates. After pouring another $80$ gm of the liquid the equilibrium temperature is found to be ${50}^o$C. The ratio of the Latent heat of the liquid to its specific heat will be${}^{\circ }C$. (Neglect the heat exchange with surrounding)

Answer 1

Let $C$ be the specific heat of the liquid, L be the latent heat of vapourisation for the liquid $C_2$ be the heat capacity for calorimeter.
In first case total heat absorbed by the liquid will be equal to the heat given by the calorimeter.

$\begin{array}{c}\text{Case-l}:\text{Total heat supplied by calorimeter will be taken by the liquid to raise its temperature from}{30}^o\text{C to}{80}^o\text{C and rest for evaporation}\\ 5C\times 50+5L=C_2\times 30.....\left(i\right)\\ \text{Case-ll}:\text{Heat supplied will be taken by the liquid to reach the equilibrium temperature of}{80}^o\text{C}\\ 80C[50-30]=C_2[80-50].....\left(ii\right)\end{array}$

By equation $\left(i\right)$ & $\left(ii\right)$
$1600C=250C+5L$

$\therefore \frac{L}{C}=\frac{1350}{5}={270}^{\circ }C$