To determine how much heat is absorbed by the reaction in the bomb calorimeter, we will use the equation [tex]\( q = m \cdot C_p \cdot \Delta T \)[/tex], where:
- [tex]\( q \)[/tex] is the heat absorbed,
- [tex]\( m \)[/tex] is the mass of the calorimeter,
- [tex]\( C_p \)[/tex] is the specific heat capacity of the calorimeter material,
- [tex]\( \Delta T \)[/tex] is the change in temperature.
Here are the details provided:
- Initial temperature ([tex]\( T_1 \)[/tex]): [tex]\( 28.50^\circ C \)[/tex]
- Final temperature ([tex]\( T_2 \)[/tex]): [tex]\( 27.45^\circ C \)[/tex]
- Mass of the calorimeter ([tex]\( m \)[/tex]): 1.400 kg
- Specific heat capacity ([tex]\( C_p \)[/tex]): [tex]\( 3.52 \, J/(g \cdot {}^\circ C) \)[/tex]
### Step-by-Step Solution:
1. Convert the mass to grams:
The mass of the calorimeter is given in kilograms, but the specific heat capacity is in [tex]\( J/(g \cdot {}^\circ C) \)[/tex]. Hence, we need to convert the mass from kilograms to grams.
[tex]\[
m = 1.400 \, \text{kg} \times 1000 \, (\text{g/kg}) = 1400 \, \text{g}
\][/tex]
2. Calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
The change in temperature is the final temperature minus the initial temperature.
[tex]\[
\Delta T = T_2 - T_1 = 27.45^\circ C - 28.50^\circ C = -1.05^\circ C
\][/tex]
Note: The negative sign indicates a decrease in temperature.
3. Calculate the heat absorbed:
Using the formula [tex]\( q = m \cdot C_p \cdot \Delta T \)[/tex]:
[tex]\[
q = 1400 \, \text{g} \times 3.52 \, \left( \frac{J}{g \cdot {}^\circ C} \right) \times (-1.05^\circ C)
\][/tex]
[tex]\[
q = 1400 \times 3.52 \times (-1.05) \, J
\][/tex]
[tex]\[
q = -5174.4 \, J
\][/tex]
4. Interpret the result:
The negative sign indicates that the calorimeter is losing heat to the reaction system, meaning that the reaction is absorbing heat. Therefore, we take the absolute value to express the heat absorbed by the reaction:
[tex]\[
\text{Absolute heat absorbed} = 5174.4 \, J
\][/tex]
Given the options:
- 140 J
- 418 J
- 1470 J
- 5170 J
The closest option to our calculated value of 5174.4 J is:
[tex]\[
\boxed{5170 \, J}
\][/tex]
Therefore, the heat absorbed by the reaction is 5170 J.
