### Thermodynamics and Internal Energy#### Problem Statement:If a system has 20.5 J of heat added to it, and the system does 75.5 J of work on its surroundings, what is the change in the internal energy of the system, in J?#### Explanation:In thermodynamics, the change in the internal energy of a system can be determined using the First Law of Thermodynamics, which is stated as:\[\Delta U = Q - W\]where:- \(\Delta U\) is the change in internal energy- \(Q\) is the heat added to the system- \(W\) is the work done by the system on its surroundingsGiven:- Heat added (\(Q\)) = 20.5 J- Work done by the system (\(W\)) = 75.5 J#### Solution:Substitute the given values into the equation:\[\Delta U = 20.5 \, \text{J} - 75.5 \, \text{J} = -55 \, \text{J}\]#### Conclusion:The change in the internal energy of the system is \(-55 \, \text{J}\). This means that the internal energy of the system decreases by 55 Joules.

Given: Heat consumed by the system is 20.5 J, and the work done by the system on the surrounding is…

Answered: If a system has 20.5 J of heat added to…

If a system has 20.5 J of heat added to it, and the system does 75.5 J of work on its surroundings, what is the change in the internal energy of the system, in J?

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

### Thermodynamics and Internal Energy

#### Problem Statement:
If a system has 20.5 J of heat added to it, and the system does 75.5 J of work on its surroundings, what is the change in the internal energy of the system, in J?

#### Explanation:
In thermodynamics, the change in the internal energy of a system can be determined using the First Law of Thermodynamics, which is stated as:

\[
\Delta U = Q - W
\]

where:
- \(\Delta U\) is the change in internal energy
- \(Q\) is the heat added to the system
- \(W\) is the work done by the system on its surroundings

Given:
- Heat added (\(Q\)) = 20.5 J
- Work done by the system (\(W\)) = 75.5 J

#### Solution:
Substitute the given values into the equation:

\[
\Delta U = 20.5 \, \text{J} - 75.5 \, \text{J} = -55 \, \text{J}
\]

#### Conclusion:
The change in the internal energy of the system is \(-55 \, \text{J}\). This means that the internal energy of the system decreases by 55 Joules.

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