A liquid at 30 o C is poured very slowly into a Calorimeter that is at temperature of 110 o C . The boiling temperature of the liquid is 80 o C . It is found that the first 5 g m of the liquid completely evaporates. After pouring another 80 g m of the liquid the equilibrium temperature is found to be 50 o C . The ratio of the Latent heat of the liquid to its specific heat will be ______ o C . [Neglect the heat exchan

Step-by-step 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 5 S ( 80 - 30 ) + 5 L = W ( 110 - 80 ) → (as first 5 g m liquid is evaporated) S = Specific heat of liquid L = Latent heat of liquid W=Water equivalent of calorimeter ⇒ 250 S + 5 L = W × 30 ...(i) Now 80 g m liquid is poured, Heat gain = Heat loss Here final temperature = 50 ° C ∴ 80 × S × 20 = 5 L + 5 S × 30 + W × 30 ...