Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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### Thermodynamics Problem

**Problem Statement:**

If a system has \(4.50 \times 10^2\) 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 solve this problem, we can use the First Law of Thermodynamics, which is expressed as:

\[
\Delta E = Q - W
\]

Where:
- \(\Delta E\) is the change in internal energy.
- \(Q\) is the heat absorbed by the system.
- \(W\) is the work done on the system.

**Given:**
- Work done on the system, \(W = 4.50 \times 10^2\) kcal.
- Heat released by the system, \(Q = -5.00 \times 10^2\) kJ.

**Conversion:**
1 kcal = 4.184 kJ

Convert the work done to kJ:
\[ W = 4.50 \times 10^2 \, \text{kcal} \times 4.184 \, \text{kJ/kcal} = 1882.8 \, \text{kJ} \]

**Calculation:**

Substitute the values into the equation:
\[
\Delta E = (-5.00 \times 10^2 \, \text{kJ}) - 1882.8 \, \text{kJ}
\]

Calculate \(\Delta E\):
\[
\Delta E = -500 - 1882.8 = -2382.8 \, \text{kJ}
\]

**Conclusion:**

The change in internal energy of the system is \(-2382.8\) kJ.
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Transcribed Image Text:### Thermodynamics Problem **Problem Statement:** If a system has \(4.50 \times 10^2\) 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 solve this problem, we can use the First Law of Thermodynamics, which is expressed as: \[ \Delta E = Q - W \] Where: - \(\Delta E\) is the change in internal energy. - \(Q\) is the heat absorbed by the system. - \(W\) is the work done on the system. **Given:** - Work done on the system, \(W = 4.50 \times 10^2\) kcal. - Heat released by the system, \(Q = -5.00 \times 10^2\) kJ. **Conversion:** 1 kcal = 4.184 kJ Convert the work done to kJ: \[ W = 4.50 \times 10^2 \, \text{kcal} \times 4.184 \, \text{kJ/kcal} = 1882.8 \, \text{kJ} \] **Calculation:** Substitute the values into the equation: \[ \Delta E = (-5.00 \times 10^2 \, \text{kJ}) - 1882.8 \, \text{kJ} \] Calculate \(\Delta E\): \[ \Delta E = -500 - 1882.8 = -2382.8 \, \text{kJ} \] **Conclusion:** The change in internal energy of the system is \(-2382.8\) kJ.
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