A liquid at 30ºC is poured very slowly into a Calorimeter that is at temperature of 110ºC. The boiling temperature of the liquid is 80ºC. It is found that the first 5gm of the liquid completely evaporates. After pouring another 80gm of the liquid the equilibrium temperature is found to be 50ºC. The ratio of the Latent heat of the liquid to its specific heat will be ___ºC.
[ Neglect the heat exchange with surrounding and assume calorimeter is open]

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answer is 270.

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Detailed Solution

Step 1:Heat lost by calorimeter $= m_{c a l} S_{c a l} \times (110 - 80)$
Heat gained by liquid $= 5 \times S_{l i q} . (80 - 30) + 5 L_{f}$

$m_{C a l} S_{C a l} 30 = S_{l i q} 250 + 5 L_{f}___(1)$

Step 2: Heat lost by calorimeter = $m_{c a l} S_{c a l} \times (80 - 50)$
Heat gained by liq = $80 \times S_{l i q} \times (50 - 30)$

$m_{C a l} S_{C a l} \times 30 = 80 S_{l i q} 20___(2)$

$\begin{array}{l} 1600 S_{l i q} - 250 S_{l i q} = 5 L_{f} \\ \Rightarrow 1350 S_{l i q} = 5 L_{f} \\ \Rightarrow \frac{L_{f}}{S_{l i q}} = \frac{1350}{5} = 270 \end{array}$

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