Suppose 19.0 J of heat is added to a system and the net amount of work done by the system is 7.0 J. How does the internal energy of the system change? O It increases by 12.0 J O It decreases by 26.0 J O It increases by 19.0 J O It decreases by 9.0 J O There is no change

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Chapter1: Units, Trigonometry. And Vectors
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**Question:**

Suppose 19.0 J of heat is added to a system and the net amount of work done by the system is 7.0 J. How does the internal energy of the system change?

**Options:**

- ○ It increases by 12.0 J
- ○ It decreases by 26.0 J
- ○ It increases by 19.0 J
- ○ It decreases by 9.0 J
- ○ There is no change

**Explanation:**
This question is based on the First Law of Thermodynamics, which states that the change in internal energy of a system (\(\Delta U\)) is equal to the heat added to the system (\(Q\)) minus the work done by the system (\(W\)):

\[
\Delta U = Q - W
\]

**Calculation:**

Given:
- Heat added, \(Q = 19.0 \, \text{J}\)
- Work done, \(W = 7.0 \, \text{J}\)

\[
\Delta U = 19.0 \, \text{J} - 7.0 \, \text{J} = 12.0 \, \text{J}
\]

Thus, the internal energy of the system increases by 12.0 J.
Transcribed Image Text:**Question:** Suppose 19.0 J of heat is added to a system and the net amount of work done by the system is 7.0 J. How does the internal energy of the system change? **Options:** - ○ It increases by 12.0 J - ○ It decreases by 26.0 J - ○ It increases by 19.0 J - ○ It decreases by 9.0 J - ○ There is no change **Explanation:** This question is based on the First Law of Thermodynamics, which states that the change in internal energy of a system (\(\Delta U\)) is equal to the heat added to the system (\(Q\)) minus the work done by the system (\(W\)): \[ \Delta U = Q - W \] **Calculation:** Given: - Heat added, \(Q = 19.0 \, \text{J}\) - Work done, \(W = 7.0 \, \text{J}\) \[ \Delta U = 19.0 \, \text{J} - 7.0 \, \text{J} = 12.0 \, \text{J} \] Thus, the internal energy of the system increases by 12.0 J.
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