Question

An experiment takes 10 minutes to raise the temperature of water in a container from 0°C to 100°C and another 55 minutes to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking specific heat of water to be 1 cal/g°C , the heat of vapourization according to this experiment will come out to be

(A) 530 cal/g 

(B) 550 cal/g 

(C) 540 cal/g 

(D) 560 cal/g

Answer 1

The correct option is (B) 550 cal/g.

Assuming heater is supplying heat at uniform rate of Q cal/ s in 10 minutes total heat transferred = 600 x Q 

This amount should be equal to heat gained by water = mcΔT where m is mass of water and c is specific heat capacity of water. 

mcΔT = 600Q 

m x 1calg^-1 C^0-1 × (100 − 0) = 600Q 

⇒ Q = m/6 ...(i) 

The heater supplies heat at constant temperature to convert water into steam, the amount of heat is given by 

55 minutes x 60 x sec x ˙ Q ..(i) 

this amount should be equal to heat gained by water to convert into steam at constant temperature = mL where L is latent heat of vaporization 

3300Q = mL ...(ii) 

substituting value of Q from equation (i) 

3300 m/6 = mL 

L = 550 cal/g