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Chemistry · THERMODYNAMICS
हिन्दीIn a bomb calorimeter having heat capacity \( 20.7\text{ kJ K}^{-1} \), 2.40 g of graphite is completely burnt in excess oxygen at 298 K according to \( \text{C}(\text{graphite}) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \). Given that the internal energy change for combustion of graphite is \( -2.48 \times 10^2\text{ kJ mol}^{-1} \), the rise in temperature of the calorimeter is:
Correct answer · Option 3
2.40 K
Explanation
Moles of carbon = \( 2.40/12.0 = 0.20\text{ mol} \). Heat released = \( 0.20 \times 248\text{ kJ} = 49.6\text{ kJ} \). Temperature rise \( \Delta T = q / C_v = 49.6 / 20.7 = 2.396\text{ K} \approx 2.40\text{ K} \).
