Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![**Problem Statement:**
If a system has 375 kcal of work done to it, and releases \(5.00 \times 10^2\) kJ of heat into its surroundings, what is the change in internal energy (\(\Delta E\) or \(\Delta U\)) of the system?
**Solution:**
To find the change in internal energy (\(\Delta E\)), we use the first law of thermodynamics, which states:
\[
\Delta E = q + w
\]
Where:
- \(q\) is the heat exchanged by the system.
- \(w\) is the work done on the system.
**Given Values:**
- Work done on the system: \(w = 375 \text{ kcal}\)
- Heat released into surroundings: \(q = -500 \text{ kJ}\)
**Conversion:**
1 kcal = 4.184 kJ
Thus, converting work from kcal to kJ:
\[
375 \text{ kcal} \times 4.184 \frac{\text{kJ}}{\text{kcal}} = 1569 \text{ kJ}
\]
Now, substitute the values into the formula:
\[
\Delta E = -500 \text{ kJ} + 1569 \text{ kJ} = 1069 \text{ kJ}
\]
**Conclusion:**
The change in internal energy (\(\Delta E\)) of the system is 1069 kJ.](https://content.bartleby.com/qna-images/question/4cbada35-3122-4a6c-a25c-246bca0ae236/b2fbc305-87cb-4f40-a3e0-e011c72b5846/4w0mtz_processed.png)
Transcribed Image Text:**Problem Statement:**
If a system has 375 kcal of work done to it, and releases \(5.00 \times 10^2\) kJ of heat into its surroundings, what is the change in internal energy (\(\Delta E\) or \(\Delta U\)) of the system?
**Solution:**
To find the change in internal energy (\(\Delta E\)), we use the first law of thermodynamics, which states:
\[
\Delta E = q + w
\]
Where:
- \(q\) is the heat exchanged by the system.
- \(w\) is the work done on the system.
**Given Values:**
- Work done on the system: \(w = 375 \text{ kcal}\)
- Heat released into surroundings: \(q = -500 \text{ kJ}\)
**Conversion:**
1 kcal = 4.184 kJ
Thus, converting work from kcal to kJ:
\[
375 \text{ kcal} \times 4.184 \frac{\text{kJ}}{\text{kcal}} = 1569 \text{ kJ}
\]
Now, substitute the values into the formula:
\[
\Delta E = -500 \text{ kJ} + 1569 \text{ kJ} = 1069 \text{ kJ}
\]
**Conclusion:**
The change in internal energy (\(\Delta E\)) of the system is 1069 kJ.
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