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A sample of propane [tex]$\left( C_3 H_8 \right)$[/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]$5.82 \text{ J} / \left( \text{g} \cdot { }^{\circ} \text{C} \right)$[/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. 47.9 kJ


Sagot :

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

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

Here,
- [tex]\( q \)[/tex] stands for the heat energy released,
- [tex]\( m \)[/tex] represents the mass of the calorimeter,
- [tex]\( C_p \)[/tex] is the specific heat capacity of the calorimeter,
- [tex]\( \Delta T \)[/tex] is the change in temperature.

Given:
- The mass of the calorimeter, [tex]\( m \)[/tex], is 1.350 kg. First, we need to convert this mass to grams (since the specific heat capacity is in J/(g·°C)):
[tex]\[ 1.350 \, \text{kg} \times 1000 \, \text{g/kg} = 1350 \, \text{g} \][/tex]

- The specific heat capacity, [tex]\( C_p \)[/tex], is 5.82 J/(g·°C).

- The change in temperature, [tex]\( \Delta T \)[/tex], is 2.87°C.

Using the heat transfer formula:
[tex]\[ q = m C_p \Delta T \][/tex]
[tex]\[ q = 1350 \, \text{g} \times 5.82 \, \frac{\text{J}}{\text{g} \cdot \, ^{\circ} \text{C}} \times 2.87 \, ^{\circ} \text{C} \][/tex]

Multiply these values:
[tex]\[ q = 1350 \times 5.82 \times 2.87 \][/tex]
[tex]\[ q = 22549.59 \, \text{J} \][/tex]

To convert Joules to kilojoules (since 1 kJ = 1000 J):
[tex]\[ q = \frac{22549.59 \, \text{J}}{1000} \][/tex]
[tex]\[ q = 22.55 \, \text{kJ} \][/tex]

Therefore, the energy released by the reaction is approximately 22.5 kJ. So, the correct answer is:
[tex]\[ \boxed{22.5 \, \text{kJ}} \][/tex]
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