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In thermodynamics, the thermal efficiency of a power cycle is a measure of how well the cycle converts heat energy (from a hot source) into work. A higher thermal efficiency indicates a more effective power cycle. The formula for thermal efficiency, denoted as \( \eta\), is calculated by the following equation:

\[ \eta = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}} \]
Here, \( Q_{\text{in}} \) represents the heat received from the hot reservoir, and \( Q_{\text{out}} \) represents the heat rejected to the cold reservoir. For our given exercise:

• \( Q_{\text{in}} = 100\) kJ
• \( Q_{\text{out}} = 40\) kJ

Substituting these values into the formula, we get:

\[ \eta = 1 - \frac{40}{100} = 0.6 \]
Converting \( 0.6 \) to a percentage, the thermal efficiency of the cycle is \( 60\% \).
Understanding this concept is crucial as it helps in maximizing the efficiency of power cycles in real-world applications.

Heat transfer is the process in which thermal energy is exchanged between physical systems, depending on the temperature and mode of transfer. In thermodynamic cycles, heat transfer plays a pivotal role:

• **Heat received (\( Q_{\text{in}} \)):** The energy absorbed from the hot reservoir.
• **Heat rejected (\( Q_{\text{out}} \)):** The energy released to the cold reservoir.

For the reversible power cycle in our exercise:

• Heat received: \( Q_{\text{in}} = 100\) kJ
• Heat rejected: \( Q_{\text{out}} = 40\) kJ

This means that from a total of \( 100\) kJ of thermal energy, \( 40\) kJ is rejected, and the remaining energy is converted into useful work. Efficient heat transfer mechanisms ensure optimal performance of power cycles, reducing energy loss and improving overall efficiency.

In thermodynamics, temperatures in calculations are usually converted to the Kelvin scale, as it is the absolute temperature scale. The relation between Celsius and Kelvin is straightforward:

\[ T_{\text{K}} = T_{\text{°C}} + 273.15 \]
In our exercise, the temperature of the hot reservoir \( ( T_{H}) \) is initially given in Celsius. Converting it to Kelvin:

\[ T_{\text{H}} = 327^\text{°C} + 273.15 = 600.15 \text{ K} \]
In thermodynamic problems, it is crucial to ensure temperatures are in the correct units to avoid computational errors. At the end of the problem, we often convert back to Celsius for more intuitive understanding. For instance, the temperature of the cold reservoir \( (T_{\text{C}}) \) is:

\[ T_{\text{C}} = 240 \text{K} - 273.15 = -33.15^\text{°C} \]

The efficiency of a reversible cycle can also be expressed in terms of the temperatures of the hot and cold reservoirs. The relationship is given by:

\[ \eta = 1 - \frac{T_{\text{C}}}{ T_{\text{H}}} \]
Here, \( T_{\text{C}}\) is the temperature of the cold reservoir and \( T_{\text{H}}\) is the temperature of the hot reservoir, both in Kelvin. Using the given problem's results, where \( \eta = 0.6\) and \( T_{\text{H}} = 600\) K:

• \( 0.6 = 1 - \frac{T_{\text{C}}}{600} \)
• Solving for \( T_{\text{C}} \):
  \[ \frac{T_{\text{C}}}{600} = 1 - 0.6 = 0.4 \]
  \[ T_{\text{C}} = 0.4 \times 600 = 240 \text{ K}\ \]

This equation demonstrates that the greater the difference in temperature between reservoirs, the higher the efficiency. Conversely, closer temperatures mean lower efficiency, emphasizing the importance of maintaining significant temperature differences for optimal performance in thermodynamic cycles.