Question

Combustion of 1 mole graphite releases \(2.48 \times 10^2\) kJ of energy. What will be the temperature of a bomb calorimeter if 1 g of graphite is burnt at 298 K, given heat capacity = 10.35 kJ/K?

• A. 298 K
• B. 296 K
• C. 300 K
• D. 299 K

Hint:

In bomb calorimetry, use \(q = C\Delta T\) and always convert moles before applying enthalpy values.

Answer 1

The Correct Option is C

Step 1: Molar combustion energy.
1 mole of graphite releases 248 kJ.

Step 2: Moles burned.
1 g graphite = 1/12 ≈ 0.0833 mol.

Step 3: Heat liberated.
q = 248 × 0.0833 ≈ 20.67 kJ.

Step 4: Temperature increase.
ΔT = q/C = 20.67/10.35 ≈ 2 K.

Step 5: Final temperature.
T_final = 298 + 2 = 300 K.

Step 6: Conclusion.
The final temperature is 300 K.