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The image describes a Carnot cycle heat engine that operates between two thermal reservoirs at temperatures \(T_1 = 900 \, \text{K}\) and \(T_2 = 300 \, \text{K}\), with the work done per cycle \(W = 2000 \, \text{J}\). The question asks how to calculate the heat rejected to the cooler reservoir \(Q_2\).

The formula given in the image is:
\[
Q_2 = W \left( 1 - \frac{1}{\eta_C} \right)
\]
where \(\eta_C\) is the Carnot efficiency of the cycle, defined as:
\[
\eta_C = 1 - \frac{T_2}{T_1}
\]

### Derivation:

1. **Carnot Efficiency:**
   - The Carnot efficiency \(\eta_C\) represents the maximum theoretical efficiency for a heat engine operating between two temperatures \(T_1\) (hot reservoir) and \(T_2\) (cold reservoir). It is given by:
   \[
   \eta_C = 1 - \frac{T_2}{T_1}
   \]
   
2. **Work Done in Carnot Cycle:**
   - The work done \(W\) by the heat engine is the difference between the heat absorbed from the hot reservoir \(Q_1\) and the heat rejected to the cold reservoir \(Q_2\). Therefore:
   \[
   W = Q_1 - Q_2
   \]
   
3. **Relationship Between Work and Heat:**
   - The efficiency \(\eta_C\) is also the ratio of the work done \(W\) to the heat absorbed from the hot reservoir \(Q_1\):
   \[
   \eta_C = \frac{W}{Q_1}
   \]
   This gives:
   \[
   Q_1 = \frac{W}{\eta_C}
   \]

4. **Calculating Heat Rejected:**
   - Using the first law of thermodynamics:
   \[
   Q_2 = Q_1 - W
   \]
   Substituting \(Q_1 = \frac{W}{\eta_C}\):
   \[
   Q_2 = \frac{W}{\eta_C} - W
   \]
   Factoring out \(W\):
   \[
   Q_2 = W \left( \frac{1}{\eta_C} - 1 \right)
   \]
   Simplifying:
   \[
   Q_2 = W \left( 1 - \frac{1}{\eta_C} \right)
   \]

This is the formula given in the image.

### Calculation Example:
If we plug in the values from the problem:

- \(T_1 = 900 \, \text{K}\)
- \(T_2 = 300 \, \text{K}\)
- \(W = 2000 \, \text{J}\)

1. Calculate the Carnot efficiency \(\eta_C\):
   \[
   \eta_C = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3}
   \]

2. Now, calculate \(Q_2\):
   \[
   Q_2 = 2000 \, \text{J} \left( 1 - \frac{1}{\frac{2}{3}} \right) = 2000 \, \text{J} \left( 1 - \frac{3}{2} \right) = 2000 \, \text{J} \left( -\frac{1}{2} \right)
   \]
   So:
   \[
   Q_2 = 1000 \, \text{J}
   \]
Thus, the heat rejected to the cooler reservoir is \(Q_2 = 1000 \, \text{J}\).

---

Do you have any further questions or need clarification on any steps?

### Related Questions:
1. How does the Carnot efficiency change if the temperature difference increases?
2. What is the maximum possible efficiency for a Carnot engine?
3. How is the first law of thermodynamics applied in heat engine cycles?
4. Can real engines achieve Carnot efficiency, and why or why not?
5. What would happen to \(Q_2\) if the work done by the engine increased?

### Tip:
Always remember that the Carnot efficiency depends solely on the temperatures of the hot and cold reservoirs. Real-world engines are less efficient due to irreversibilities like friction and heat losses.

Wie viel Wärmeenergie Q_2 wird bei jedem Umlauf an das kältere Wärmereservoir abgegeben?

Calculate the heat rejected Q2 in a Carnot cycle where the machine operates between temperatures T1 = 900 K and T2 = 300 K, performing 2000 J of work per cycle. Learn how to use the Carnot efficiency formula to find Q2.

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