A liquid at 30 ^ o C is poured very slowly into a Calorimeter that...

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
$5S ( 80 - 30 ) + 5L = W ( 110 - 80 ) \to$ (as first $5gm$ liquid is evaporated)
$S =$ Specific heat of liquid
$L =$ Latent heat of liquid
W=Water equivalent of calorimeter
$⇒ 250S + 5L = W \times 30$ ...(i)
Now $80gm$ liquid is poured,
Heat gain = Heat loss
Here final temperature $= 50 ° C$
$∴ 80 \times S \times 20 = 5L + 5S \times 30 + W \times 30$ ...(ii)
From (i) and (ii)
$\dfrac{L}{S} = 120Answer$
Case $2$ : If calorimeter is open and after evaporation liquid escapes
$5 \times S \times 50 + 5L = W \times 30$ ...(i)
(as first $5gm$ liquid is evaporated)
$80 \times S \times 20 = W \times 30$ ...(ii)
(after pouring $80gm$ liquid, the equilibrium temperature is $50 ° C$
$⇒ 80 \times S \times 20 = S \times S \times 50 + 5L$ (using (i) and (ii)
$SL = 1350S$
$\dfrac{L}{S} = 270$