# Question 8899d

Jul 12, 2017

$- 0.143 k J$

#### Explanation:

A bomb calorimeter is one in which there is constant volume. Moreover, it is one in which namely combustion reactions occur; hence, the hydrocarbon in question.

In order to measure heat evolved by a reaction, the bomb calorimeter is "calibrated" with known reactions to arrive at the heat capacity of the calorimeter. This was provided to us in the question.

We relate the temperature change and heat capacity of the calorimeter to arrive at the heat evolved by the reaction in question, like so:

${q}_{r x n} = - {C}_{c a l} \Delta T$

Thus,

q = (-69.6J)/(°C)*2.06°C#
$q = - 143.376 J \left(\frac{k J}{{10}^{3} J}\right) \approx - 0.143 k J$