Question

On burning 0.83 gm of a fuel in bomb calorimeter, the
temperature of 3,500 gm of water increased from 26.5 to 29.2 0
C.
Water equivalent of calorimeter and latent heat of steam are 385 gm
and 587 cal/gm respectively. If the fuel contains 0.7% hydrogen,
calculate its gross and net calorific value.

Answer 1

Answer:

solution: Here wt. of fuel (x) = 0.83 g of ; wt of water (W) = 3,500 g; water equivalent of calorimeter (w) = 385 g ; (t2- t) = (29.2oC -26.5oC) = 2.7oC ; percentage of hydrogen (H) = 0.7% ; latent heat of steam = 587 cal/gGross calorific value = (W + w) (t1- t2) cal/g x= (3,500 +385) ×2.7 = 12,638 cal/g0.83 NCV= [GCV – 0.09 H ×587] = (12,63 8– 0.09 ×0.7 ×587) cal/g= (12,638 – 37) cal/g = 12,601 cal/g

Explanation:

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Answer 2

Answer:

Gross Calorific Value = 10489 cal/g

Net Calorific value = 10452.02 cal/g

Explanation:

• Bomb Calorimeter is an apparatus used to find the heat of reaction.
• It is generally used to find the calorific value of materials.
• The heat absorbed or released by the reaction is reflected in the change in temperature of water in bomb calorimeter.
• Gross Calorific Value of a fuel is the total amount of heat released on complete combustion. In this, assumption is that all the water vapor is fully condensed.
• Net Calorific Value of a fuel is obtained by subtracting the heat of vaporization of water vapor from Gross Calorific Value.

The calorific value of fuel = $mass_{water} C_{p} (T_{final}- T_{initial})+mass_{cal} C_{cal} (T_{final}- T_{initial})$

$T_{final}$ = 29.2°C

$T_{initial}$ = 26.5°C

$C_{p,water}$ = 1 cal/g

Gross calorific value of fuel = (3500+385)*1*(29.2 - 26.5)

Gross calorific value of fuel = 10489 cal/g

Gross Heat released by fuel = 10489 * 0.83 = 8705.87 cal

Net Calorific value of fuel = Gross calorific value - 587*(0.09*%Hydrogen)

Net Calorific value of fuel = 10489 - 587*(0.09*0.7) =  10452.02 cal/g

Net Heat released by fuel = 10452.02 * 0.83 = 8675.17 cal

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