A system absorbs 20J energy at constant volume .The change in internal energy is?

When a system absorbs energy at constant volume, the change in internal energy \( \Delta U \) is equal to the heat absorbed by the system, as there is no work done at constant volume.

Given:
- Heat absorbed, \( q = 20 \, \text{J} \)
- Process occurs at constant volume

The first law of thermodynamics states:

\[
\Delta U = q + W
\]

Since the volume is constant, \( W = 0 \) (no work done), so:

\[
\Delta U = q = 20 \, \text{J}
\]

Thus, the change in internal energy \( \Delta U \) is **20 J**.

Would you like a more detailed explanation, or have any further questions?

Here are five related questions you might find helpful:
1. How does the first law of thermodynamics apply in processes at constant pressure?
2. What would be the change in internal energy if the system instead lost 20 J of heat at constant volume?
3. How is work calculated when a process occurs at constant pressure?
4. Can internal energy change in an isothermal process, and why?
5. What factors affect the internal energy of an ideal gas?

**Tip:** For processes at constant volume, remember that the change in internal energy equals the heat absorbed or released by the system.

A system absorbs 20J energy at constant volume. The change in internal energy is?

Explore how to calculate the change in internal energy when a system absorbs 20J of heat at constant volume. Learn key thermodynamics concepts, including the First Law of Thermodynamics, and understand how heat and work affect internal energy. Ideal for high school students in Grades 11-12.

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