# A chemical engineer studying the properties of fuels placed 1.820 g of a hydrocarbon in the bomb of a calorimeter and filled it with O2 gas. What was the heat of the reaction for the combustion per gram of the fuel?

Jul 11, 2017

$21.597 \frac{\text{kJ}}{g}$

#### Explanation:

Don't let the detail of the data frighten you! We simply want to know how much heat was generated by the reaction. The rest is just how we FIND that value.

The temperature change was 23.55 – 20.00 = 3.55 °C – that applies to both the calorimeter and the water, so we just need to calculate the two different absorbed heats and combine them for the total reaction heat. Note that a change of 3.55 °C is the same as 3.55 °K for later in the problem.

Water: 2550g * 4.184 J/(g-°C) * 3.55°C = 37875.7 J (or 37.875 kJ)
Calorimeter: 403J/(°K) * 3.55°K = 1430.7 J

Combined (Heat generated by the reaction): $37875.7 J + 1430.7 J = 39306.4 J$

$\frac{39306.4 J}{1.820 g f u e l} = 21596.9 \frac{J}{g}$ or $21.597 \frac{\text{kJ}}{g}$