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A liquid at $30^{\circ} C$ is poured very slowly into a Calorimeter that is at temperature of $110^{\circ} C$. The boiling temperature of the liquid is $80^{\circ} C$. 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 $50^{\circ} C$. The ratio of the Latent heat of the liquid to its specific heat will be. . . . .${ }^{\circ} C$. [Neglect the heat exchange with surrounding]
$260$
$250$
$270$
$280$
Solution
Let $m =$ mass of calorimeter,
$x=$ specific heat of calorimeter
$s=$ specifc heat of liquid
$L =$ latent heat of liquid
First $5 g$ of liquid at $30^{\circ}$ is poured to calorimter at $110^{\circ} C$
$\therefore m \times x \times(110-80)=5 \times s \times(80 \times 30)+5 L$
$\Rightarrow mx \times 30=250 s +5 L \ldots \text { (i) }$
Now, $80 g$ of liquid at $30^{\circ}$ is poured into calorimeter at $80^{\circ} C$, the equilibrium temperature reaches to $50^{\circ} C$.
$\therefore m \times x \times(80-30)=80 \times s \times(50-30)$
$\Rightarrow mx \times 30=1600 s \ldots . \text { (ii) }$
From $(i)$ & $(ii)$
$250 s +5 L =1600 s \Rightarrow 5 L =1350 s$
$\Rightarrow \frac{ L }{ s }=270$
