Question

An experiment takes $10$ minutes to raise the temperature of water in a container from $0^{o}C$ to ${100}^{o}C$ 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 $1cal/g^{o}C$, 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

Answer: (B)

550 cal/g

Assuming heater is supplying heat at uniform rate of $\dot{Q}cal/s$
in 10 minutes total heat transferred $=600\times \dot{Q}$
This amount should be equal to heat gained by water $=mc\Delta T$ where m is mass of water and c is specific heat capacity of water.
$mc\Delta T=600\dot{Q}$
$m\times 1cal{g}^{-1}{C^{\circ }}^{-1}\times (100-0)=600\dot{Q}\Rightarrow \dot{Q}=\frac{m}{6}$...(i)
The heater supplies heat at constant temperature to convert water into steam, the amount of heat is given by
$55minutes\times 60\times sec\times \dot{Q}$..(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
$3300\dot{Q}=mL$...(ii)
substituting value of $\dot{Q}$ from equation (i)
$3300\frac{m}{6}=mL$
$L=550cal/g$
Hence correct answer is option B