**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 kJThus, 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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Answered: If a system has 375 kcal of work done…

If a system has 375 kcal of work done to it, and releases 5.00 x 10² kJ of heat into its surroundings, what is the change in internal energy (AE or AU) of the system? AE = kJ

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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Question

**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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