Question

Complete combustion of a 0.350 g sample of a compound in a bomb calorimeter releases 14.0 kJ of heat. The bomb calorimeter has a mass of 1.20 kg and a specific heat of 3.55 J/(gi°C).

If the initial temperature of the calorimeter is 22.5°C, what is its final temperature?
Use q equals m C subscript p Delta T..
19.2°C
25.8°C
34.2°C
72.3°C

Answer 1

To determine the final temperature of the bomb calorimeter after the combustion of the compound, we can use the formula for heat transfer:

\[
q = m C_p \Delta T
\]

Where:
- \( q \) is the heat absorbed by the calorimeter (in Joules),
- \( m \) is the mass of the calorimeter (in grams),
- \( C_p \) is the specific heat capacity of the calorimeter (in J/g°C),
- \( \Delta T \) is the change in temperature (in °C).

Given data:
- The heat released by combustion: \( q = 14.0 \, \text{kJ} = 14,000 \, \text{J} \)
- Mass of the calorimeter: \( m = 1.20 \, \text{kg} = 1200 \, \text{g} \)
- Specific heat of the calorimeter: \( C_p = 3.55 \, \text{J/g°C} \)
- Initial temperature: \( T_i = 22.5 \, °C \)

Since this heat is absorbed by the calorimeter, we will set \( q \) to this value in the above equation. Rearranging the equation for \( \Delta T \):

\[
\Delta T = \frac{q}{m C_p}
\]

Now substituting the known values:

\[
\Delta T = \frac{14,000 \, \text{J}}{1200 \, \text{g} \times 3.55 \, \text{J/g°C}}
\]

Calculating the denominator:

\[
1200 \, \text{g} \times 3.55 \, \text{J/g°C} = 4260 \, \text{J/°C}
\]

Now calculating \( \Delta T \):

\[
\Delta T = \frac{14,000 \, \text{J}}{4260 \, \text{J/°C}} \approx 3.29 \, °C
\]

Finally, to find the final temperature \( T_f \):

\[
T_f = T_i + \Delta T
\]

\[
T_f = 22.5 \, °C + 3.29 \, °C \approx 25.79 \, °C
\]

Rounding to one decimal place gives us:

\[
T_f \approx 25.8 \, °C
\]

Therefore, the final temperature of the bomb calorimeter is **25.8°C**.