A liquid at $30 ° C$ is poured very slowly into a calorimeter that is at temperature of $110 ° C$ . The boiling temperature of the liquid is $80 ° C$ . It is found that the first 5 g of the liquid completely evaporates. After pouring another 80 g of the liquid the equilibrium temperature is found to be $50 ° C$ . The ratio of the latent heat of the liquid to its specific heat will be $x \times 90 ° C$ , then $x =$ ________ [Neglect the heat exchange with surrounding]

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answer is 3.

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Detailed Solution

Let $m_{c a l} =$ mass of calorimeter
$S_{c a l} =$ specific heat of calorimeter
$S_{l i q} =$ specific heat of liquid
L=latent heat of liquid
Step-1:
Heat lost by calorimeter $= m_{c a l} s_{c a l} \times 30$
Heat gained by liquid $= 5 \times s_{l i q} .50 + m L$
$m_{l i q} s_{c a l} 30 = s_{l i q} 50 + 5 L_{f}$
Step-2:
Heat lost by calorimeter $m_{c a l} s_{c a l} \times 30$
Heat lost by liquid $= 80 \times s_{l i q} \times 20$
$m_{c a l} s_{c a l} \times 30 = 80 s_{l i q} 20$