A liquid at 30oC is poured very slowly into a calorimeter that is at a 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 the surrounding)

Correct option is A. 270.00, 120.00
CASE - I: If calorimeter is open and after evaporation liquid escapes
5×S×50+5L=W×30 (1)
80×S×20=W×30. (2)
80×S×20=5×S×50+5L
5L=1350S
L/S=270
CASE - II: If calorimeter is closed (vapour not allowed to escape)
Heat gain=Heat loss
5S(80−30)+5L=W(110−80)
S=Specific heat of liquid
L=Latent heat of liquid
W=Water equivalent of calorimeter
250S+5L=W×30 (1)
Now 80gm liquid is poured
Heat gain=Heat loss
Here final temperature=50oC
80×S×20=5L+5S×30+W×30 (2)
From (1) (2)
L/S=120.