A liquid at 30   o C is poured very slowly into a Calorimeter that is at temperature of 110   o C …

A liquid at 30 oC is poured very slowly into a Calorimeter that is at temperature of 110 oC. The boiling temperature of the liquid is 80 oC. 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 oC. The ratio of the Latent heat of the liquid to its specific heat will be ______  oC.
[Neglect the heat exchange with surrounding]

Solution

Note: The information about condition of calorimeter is not given in question, so following two cases arise -
Case 1 : If calorimeter is closed (vapour not allowed to escape)
Heat gain = Heat loss
5S(80-30)+5L=W(110-80)→ (as first 5 gm liquid is evaporated)
S= Specific heat of liquid
L= Latent heat of liquid
W=Water equivalent of calorimeter
⇒250S+5L=W×30 ...(i)
Now 80 gm liquid is poured,
Heat gain = Heat loss
Here final temperature =50°C
∴80×S×20=5L+5S×30+W×30 ...(ii)
From (i) and (ii)
LS=120 Answer
Case 2 : If calorimeter is open and after evaporation liquid escapes
5×S×50+5L=W×30 ...(i)
(as first 5 gm liquid is evaporated)
80×S×20=W×30 ...(ii)
(after pouring 80 gm  liquid, the equilibrium temperature is 50°C
⇒80×S×20=S×S×50+5L (using (i) and (ii)
SL=1350 S
LS=270 ;

Asked in: JEE Advanced 2019 (Paper 1)

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