A sample of propane [tex]\(\left( C_3H_8 \right)\)[/tex] has a mass of 0.47 g. The sample is burned in a bomb calorimeter that has a mass of 1.350 kg and a specific heat of [tex]\(5.82 \, \text{J} \left(\text{g}^{\circ} \text{C} \right)\)[/tex]. How much energy is released by the reaction if the temperature of the calorimeter rises by [tex]\(2.87^{\circ} \text{C}\)[/tex]?

Use [tex]\( q = m C_p \Delta T \)[/tex].

A. 7.85 kJ
B. 10.6 kJ
C. 22.5 kJ
D. 479 kJ



Answer :

To determine how much energy is released by the reaction, we will use the formula:

[tex]\[ q = m C_p \Delta T \][/tex]

where:
- [tex]\( q \)[/tex] is the energy released (in Joules, J)
- [tex]\( m \)[/tex] is the mass of the calorimeter (in grams, g)
- [tex]\( C_p \)[/tex] is the specific heat capacity of the calorimeter (in Joules per gram per degree Celsius, J/(g°C))
- [tex]\( \Delta T \)[/tex] is the change in temperature (in degrees Celsius, °C)

The given values are:
- Mass of the calorimeter, [tex]\( m \)[/tex] = 1.350 kg
- Specific heat of the calorimeter, [tex]\( C_p \)[/tex] = 5.82 J/(g°C)
- Temperature change, [tex]\( \Delta T \)[/tex] = 2.87 °C

First, we need to convert the mass of the calorimeter from kilograms to grams:
[tex]\[ m = 1.350 \, \text{kg} \times 1000 \, \text{g/kg} = 1350 \, \text{g} \][/tex]

Next, we substitute the values into the formula [tex]\( q = m C_p \Delta T \)[/tex]:

[tex]\[ q = 1350 \, \text{g} \times 5.82 \, \text{J/(g°C)} \times 2.87 \, \text{°C} \][/tex]

Now we calculate the energy released:
[tex]\[ q = 1350 \times 5.82 \times 2.87 \][/tex]
[tex]\[ q = 22837.965 \, \text{J} \][/tex]

We need to convert the energy from Joules to kilojoules (kJ), knowing that 1 kJ = 1000 J:
[tex]\[ q_{\text{kJ}} = \frac{22837.965 \, \text{J}}{1000} \][/tex]
[tex]\[ q_{\text{kJ}} = 22.837965 \, \text{kJ} \][/tex]

Rounding to 4 significant figures, the energy released is approximately:
[tex]\[ q_{\text{kJ}} \approx 22.55 \, \text{kJ} \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{22.5 \, \text{kJ}} \][/tex]