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Chemistry · THERMODYNAMICS
हिन्दीCombustion of 1.0 g of graphite in a bomb calorimeter produces a temperature rise from 298 K to 299 K. If the heat capacity of the calorimeter is \( 20.7\text{ kJ K}^{-1} \), what is the internal energy change (\( \Delta U \)) for the combustion of 1 mol of graphite?
Correct answer · Option B
\( -2.48 \times 10^2\text{ kJ mol}^{-1} \)
Explanation
Heat evolved per gram = \( -20.7\text{ kJ K}^{-1} \times 1\text{ K} / 1\text{ g} = -20.7\text{ kJ g}^{-1} \). For 1 mole (12.0 g), \( \Delta U = -20.7 \times 12.0 = -2.48 \times 10^2\text{ kJ mol}^{-1} \).
