Answer to Question #111593 in General Chemistry for Akash

Question

2. Calculate the gross and net calorific value of a coal sample from the following data obtained from a bomb calorimeter. Weight of coal 0.73 gm, weight of water in the calorimeter 1500 gm, water equivalent of calorimeter 470 gm, initial temp 25 C, and final temp, 28 C, percentage of hydrogen in coal 2.5%, latent heat of steam 587 cal/gm.

Answer 1

Energy released from coal = energy gain by water

Let gross calorific value be x joule per gram

Then,

Energy released by coal = $470\times 4.2\times(28-25)=5922 \ J$

Now energy released = gross calorific value multiplied by mass

So,

$$5922= x\times0.72\\or\\x=8225\ \frac{J}{gm}$$

Net calorific value = gross calorific value - heat of vaporisation.

So,

Net calorific value = $8225-(587\times4.2)=5759.6\ \frac{J}{gm}$