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Sagot :
Certainly! Let's go through the detailed, step-by-step solution to determine the final temperature of the calorimeter.
1. Understand the given data:
- Heat released ([tex]\( q \)[/tex]): [tex]\( 24.0 \, \text{kJ} \)[/tex]. Since we usually deal with energy in joules when performing specific heat calculations, we'll convert this:
[tex]\[ q = 24.0 \, \text{kJ} = 24,000 \, \text{J} \][/tex]
- Mass of the calorimeter ([tex]\( m \)[/tex]): [tex]\( 1.30 \, \text{kg} \)[/tex]. We need to convert this into grams for consistency with the specific heat capacity units:
[tex]\[ m = 1.30 \, \text{kg} = 1,300 \, \text{g} \][/tex]
- Specific heat capacity ([tex]\( C_p \)[/tex]): [tex]\( 3.41 \, \text{J/(g°C)} \)[/tex]
- Initial temperature ([tex]\( T_i \)[/tex]): [tex]\( 25.5 \, ^\circ \text{C} \)[/tex]
2. Equation:
We use the equation relating heat ([tex]\( q \)[/tex]), mass ([tex]\( m \)[/tex]), specific heat ([tex]\( C_p \)[/tex]), and change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ q = m \cdot C_p \cdot \Delta T \][/tex]
Rearrange to solve for the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = \frac{q}{m \cdot C_p} \][/tex]
3. Substitute the values into the rearranged equation:
[tex]\[ \Delta T = \frac{24,000 \, \text{J}}{1,300 \, \text{g} \cdot 3.41 \, \text{J/(g°C)}} \][/tex]
Calculate the denominator:
[tex]\[ m \cdot C_p = 1,300 \, \text{g} \times 3.41 \, \text{J/(g°C)} = 4,433 \, \text{J/°C} \][/tex]
Then, calculate [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = \frac{24,000 \, \text{J}}{4,433 \, \text{J/°C}} \approx 5.414 \, ^\circ \text{C} \][/tex]
4. Determine the final temperature ([tex]\( T_f \)[/tex]):
The final temperature is the initial temperature plus the change in temperature:
[tex]\[ T_f = T_i + \Delta T \][/tex]
Substitute the known values:
[tex]\[ T_f = 25.5 \, ^\circ \text{C} + 5.414 \, ^\circ \text{C} \approx 30.914 \, ^\circ \text{C} \][/tex]
5. Conclusion:
Therefore, the final temperature of the calorimeter is approximately [tex]\( 30.9 \, ^\circ \text{C} \)[/tex], which matches closely with the given possible answer:
[tex]\[ \boxed{30.9 \, ^\circ \text{C}} \][/tex]
1. Understand the given data:
- Heat released ([tex]\( q \)[/tex]): [tex]\( 24.0 \, \text{kJ} \)[/tex]. Since we usually deal with energy in joules when performing specific heat calculations, we'll convert this:
[tex]\[ q = 24.0 \, \text{kJ} = 24,000 \, \text{J} \][/tex]
- Mass of the calorimeter ([tex]\( m \)[/tex]): [tex]\( 1.30 \, \text{kg} \)[/tex]. We need to convert this into grams for consistency with the specific heat capacity units:
[tex]\[ m = 1.30 \, \text{kg} = 1,300 \, \text{g} \][/tex]
- Specific heat capacity ([tex]\( C_p \)[/tex]): [tex]\( 3.41 \, \text{J/(g°C)} \)[/tex]
- Initial temperature ([tex]\( T_i \)[/tex]): [tex]\( 25.5 \, ^\circ \text{C} \)[/tex]
2. Equation:
We use the equation relating heat ([tex]\( q \)[/tex]), mass ([tex]\( m \)[/tex]), specific heat ([tex]\( C_p \)[/tex]), and change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ q = m \cdot C_p \cdot \Delta T \][/tex]
Rearrange to solve for the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = \frac{q}{m \cdot C_p} \][/tex]
3. Substitute the values into the rearranged equation:
[tex]\[ \Delta T = \frac{24,000 \, \text{J}}{1,300 \, \text{g} \cdot 3.41 \, \text{J/(g°C)}} \][/tex]
Calculate the denominator:
[tex]\[ m \cdot C_p = 1,300 \, \text{g} \times 3.41 \, \text{J/(g°C)} = 4,433 \, \text{J/°C} \][/tex]
Then, calculate [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = \frac{24,000 \, \text{J}}{4,433 \, \text{J/°C}} \approx 5.414 \, ^\circ \text{C} \][/tex]
4. Determine the final temperature ([tex]\( T_f \)[/tex]):
The final temperature is the initial temperature plus the change in temperature:
[tex]\[ T_f = T_i + \Delta T \][/tex]
Substitute the known values:
[tex]\[ T_f = 25.5 \, ^\circ \text{C} + 5.414 \, ^\circ \text{C} \approx 30.914 \, ^\circ \text{C} \][/tex]
5. Conclusion:
Therefore, the final temperature of the calorimeter is approximately [tex]\( 30.9 \, ^\circ \text{C} \)[/tex], which matches closely with the given possible answer:
[tex]\[ \boxed{30.9 \, ^\circ \text{C}} \][/tex]
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