`0.562g` of graphite kept in a bomb calorimeter in excess of oxygen at `298K` and 1 atmospheric pressure was burnt according to the equation,
`C_(Graph ite)+O_(2(g))rarr CO_(2(g))`
durgin the reaction, temperature rises from `298K` o `298.89K`. If the heat capacity of the calorimeter and its contents is `20.7 kJ//K`, what is the enthalpy change for the above reaction at `298K` and `1 atm`?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the heat produced (Q) during the reaction.
The heat produced during the reaction can be calculated using the formula:
\[ Q = -C_v \Delta T \]
Where:
- \( C_v \) is the heat capacity of the calorimeter and its contents (given as 20.7 kJ/K).
- \( \Delta T \) is the change in temperature, which is \( 298.89 K - 298 K = 0.89 K \).
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