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An experiment takes 10 minutes to raise the temperature of water in a container from 00C to 1000C 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 0C, the heat of vapourization according to this experiment will come out to be :

Option: 1

530 cal/g


Option: 2

540 cal/g


Option: 3

550 cal/g


Option: 4

560 cal/g


Answers (1)

best_answer

Assuming heater is supplying heat at a uniform rate of \dot{Q}$ cal/s

So in 10 minutes, total heat transferred =10*60 \times \dot{Q}$=600 \times \dot{Q}
This amount should be equal to heat gained by water$=m c \Delta T$  where mis mass of water and c is the specific heat capacity of water.

So $m c \Delta T=600 \dot{Q}$
m \times 1 \ (c a l g^{-1} C^{o-1}) \times(100-0)=600 \dot{Q} \Rightarrow \dot{Q}=\frac{m}{6}-(\mathrm{i})$

Now as given in the question.

The heater supplies heat at constant temperature to convert water into steam,

and the amount of heat is given by heater in 55 minutes=55$\times 60\dot{Q} =3300\dot{Q}


And this amount should be equal to the heat gained by water to convert into steam at a constant temperature

which is equal to "mL" where L is the latent heat of vaporization

So

3300 \dot{Q}=m L ......(ii)
substituting value of \dot{Q}$ from equation (i) \\ we get \\ $3300 \frac{m}{6}=m L$

So $L=550 \mathrm{cal} / \mathrm{g}$

Posted by

Ritika Kankaria

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