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QJEEAdv19Phd3b9: JEE Advanced 2019 Physics - Class 11 JEEAdv19Phd3b9 Solution: A liquid at 30 C is poured very slowly into a Calorimeter that…

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]

Step-by-Step Verified Solution

Case - I 5C ×\times 50 + 5L = C2 ×\times 30 ....(i)

Case - II 80C[50 −- 30] = C2 [80 −- 50] ....(ii)

By Eq. (i) and (ii)

1600C = 250 + 5L

∴\therefore LC=13505=270∘C{L \over C} = {{1350} \over 5} = 270^\circ C

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