JEE Advance - Physics (2019 - Paper 1 Offline - No. 16)

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]
Answer
270

Explanation

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

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

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