CBSE Class 11-science Answered

A known mass (0.5— 1.0 g) of a fuel is taken in the clean crucible. Then a fine mercury wire, touching the fuel sample, is stretched across the electrodes.

The lid is screwed tightly and filled with oxygen to a pressure of 25 atm. The bomb is lowered into a copper calorimeter containing known mass of water. The water is stirred and the initial temperature is noted. The electrodes are connected to a 6 V battery and the circuit is completed. The sample burns and heat is liberated. Uniform stirring is continued and the maximum temperature attained is recorded.

Calculation

X = mass; in g of fuel sample taken in crucible

W = mass of water in calorimeter

w = water equivalent in g of calorimeter, stirrer, thermometer, bomb

t₁ = initial temperature of water in calorimeter

t₂ = final temperature of water in calorimeter

L  =  higher calorific value of fuel in cal/g

Heat liberated by burning of a fuel = XL

Heat absorbed by water and apparatus = (W + w)(t₂ – t₁)

Heat liberated = heat absorbed

XL = (W + w) (t₂-t₁)

Thus, HCV of the fuel L = (W+w)(t₂–t₁)/x cal/g or kcal/kg

The water equivalent of the calorimeter is determined by burning a fuel of known calorific value and using the above equation. Example, Water equivalent of benzoic acid (HCV = 6325 kcal/7kg) and naphthalene (HCV = 9688 kcal/kg)
If H is the percentage of hydrogen in the fuel

Then 9Hg/100 = Mass of H₂0 from 1 g fuel = 0.09 Hg

Heat taken by water in the form of steam = 0.09 H x 587

LCV = HCV - Latent heat of water formed

       = HCV-0.09 × H × 587 cal/g (or kcal/kg)