Determine the final temperature of a 377 g sample of iron when 2.64 kJ of heat is added. The initial temperature of the sample was 18.8 oC. The specific heat capacity of iron is 0.449 J/g-K.
Determine the final temperature of a 377 g sample of iron when 2.64 kJ of heat is added. The initial temperature of the sample was 18.8 oC. The specific heat capacity of iron is 0.449 J/g-K.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Determining the Final Temperature of a Heated Iron Sample
In this example, we aim to calculate the final temperature of a 377 g sample of iron when 2.64 kJ of heat is added. The initial temperature of the sample was 18.8°C. The specific heat capacity of iron is given as 0.449 J/g·K.
#### Problem Statement:
"Determine the final temperature of a 377 g sample of iron when 2.64 kJ of heat is added. The initial temperature of the sample was 18.8°C. The specific heat capacity of iron is 0.449 J/g·K."
#### Given Data:
- Mass of iron sample (m): 377 g
- Heat added (q): 2.64 kJ (2640 J)
- Initial temperature (T₁): 18.8°C
- Specific heat capacity of iron (c): 0.449 J/g·K
#### Solution:
To determine the final temperature (T₂), we can use the formula for heat transfer:
\[ q = m \cdot c \cdot \Delta T \]
Where:
- \( q \) is the heat added,
- \( m \) is the mass of the sample,
- \( c \) is the specific heat capacity,
- \( \Delta T \) is the change in temperature (T₂ - T₁).
Rearranging to solve for \( \Delta T \):
\[ \Delta T = \frac{q}{m \cdot c} \]
Substitute the given values into the equation:
\[ \Delta T = \frac{2640 \text{ J}}{377 \text{ g} \cdot 0.449 \text{ J/g·K}} \]
\[ \Delta T \approx 15.4 \text{ K} \]
Now, calculate the final temperature:
\[ T₂ = T₁ + \Delta T \]
\[ T₂ = 18.8°C + 15.4 \text{ K} \]
\[ T₂ \approx 34.2°C \]
Therefore, the final temperature of the iron sample after adding 2.64 kJ of heat is approximately 34.2°C.
**Note:** In the context of this calculation, an increase in temperature (ΔT) in degrees Celsius is equivalent to an increase in temperature in Kelvin since the size of a degree](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b0fedb7-adbc-492c-aa14-9f5d12342848%2F87d43bef-19fa-4830-bf39-afe2bc4fb09a%2Fs0ubbkq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Determining the Final Temperature of a Heated Iron Sample
In this example, we aim to calculate the final temperature of a 377 g sample of iron when 2.64 kJ of heat is added. The initial temperature of the sample was 18.8°C. The specific heat capacity of iron is given as 0.449 J/g·K.
#### Problem Statement:
"Determine the final temperature of a 377 g sample of iron when 2.64 kJ of heat is added. The initial temperature of the sample was 18.8°C. The specific heat capacity of iron is 0.449 J/g·K."
#### Given Data:
- Mass of iron sample (m): 377 g
- Heat added (q): 2.64 kJ (2640 J)
- Initial temperature (T₁): 18.8°C
- Specific heat capacity of iron (c): 0.449 J/g·K
#### Solution:
To determine the final temperature (T₂), we can use the formula for heat transfer:
\[ q = m \cdot c \cdot \Delta T \]
Where:
- \( q \) is the heat added,
- \( m \) is the mass of the sample,
- \( c \) is the specific heat capacity,
- \( \Delta T \) is the change in temperature (T₂ - T₁).
Rearranging to solve for \( \Delta T \):
\[ \Delta T = \frac{q}{m \cdot c} \]
Substitute the given values into the equation:
\[ \Delta T = \frac{2640 \text{ J}}{377 \text{ g} \cdot 0.449 \text{ J/g·K}} \]
\[ \Delta T \approx 15.4 \text{ K} \]
Now, calculate the final temperature:
\[ T₂ = T₁ + \Delta T \]
\[ T₂ = 18.8°C + 15.4 \text{ K} \]
\[ T₂ \approx 34.2°C \]
Therefore, the final temperature of the iron sample after adding 2.64 kJ of heat is approximately 34.2°C.
**Note:** In the context of this calculation, an increase in temperature (ΔT) in degrees Celsius is equivalent to an increase in temperature in Kelvin since the size of a degree
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