Question

A reversible power cycle whose thermal efficiency is \(50 \%\) operates between a reservoir at \(1800 \mathrm{~K}\) and a reservoir at a lower temperature \(T\). Determine \(T\), in \(\mathrm{K}\).

Short answer

The temperature of the lower reservoir is 900 K.

Step-by-step solution

- Understand the efficiency of a reversible cycle

The efficiency of a reversible (Carnot) cycle operating between two reservoirs is given by \[ \text{efficiency} = 1 - \frac{T_{cold}}{T_{hot}} \]where \( T_{hot} \) is the temperature of the hot reservoir and \( T_{cold} \) is the temperature of the cold reservoir.

- Set up the given values

From the problem, the thermal efficiency is given as 50%, or 0.50 in decimal form. Also, \( T_{hot} = 1800 \, \text{K} \).

- Use the efficiency formula to find the unknown temperature

Substitute the given efficiency and the temperature of the hot reservoir into the efficiency formula: \[ 0.50 = 1 - \frac{T}{1800} \]

- Solve for the cold reservoir temperature

Rearrange the equation to isolate \( T \): \[ 0.50 = 1 - \frac{T}{1800} \] \[ \frac{T}{1800} = 0.50 \] \[ T = 0.50 \times 1800 \] \[ T = 900 \, \text{K} \]

Key concepts

Reversible Cycles

In thermodynamics, a reversible cycle is an idealized process that occurs infinitely slowly, ensuring that the system remains in thermodynamic equilibrium throughout. The Carnot cycle, named after French engineer Sadi Carnot, is the most well-known example of a reversible cycle. This cycle consists of two isothermal processes (where temperature remains constant) and two adiabatic processes (where no heat is exchanged with the surroundings). Reversible cycles are significant because they represent the maximum possible efficiency that any heat engine can achieve between two given temperature reservoirs. Understanding these cycles is crucial for studying real-world engines and refrigerators, as it sets a benchmark for their performance.

Thermal Efficiency

Thermal efficiency is a measure of how well an engine or cycle converts heat into work. It is defined as the ratio of the work output to the heat input. For a Carnot cycle operating between two temperature reservoirs, the thermal efficiency is given by the formula:
\[\text{Efficiency} = 1 - \frac{T_{cold}}{T_{hot}}\].
Here, \(T_{ hot }\) and \( T_{ cold } \) represent the temperatures of the hot and cold reservoirs, respectively. The efficiency tells us what fraction of the heat supplied to the system is converted into useful work. In the given exercise, the Carnot cycle's thermal efficiency is 50%, meaning that half of the heat input is converted into work while the rest is expelled to the cold reservoir.

Temperature Reservoirs

A temperature reservoir is an idealized body with a large thermal capacity that can absorb or release heat without undergoing any change in temperature. In the context of the Carnot cycle, two reservoirs are involved:

• The hot reservoir provides heat at a constant high temperature (denoted as \(T_{hot}\) ).
• The cold reservoir absorbs the waste heat at a constant lower temperature (denoted as \(T_{cold}\) ).

Understanding the role of these reservoirs is crucial because the efficiency of the cycle depends on the temperature difference between them. In the given problem, the hot reservoir is at 1800 K, and the task is to find the temperature of the cold reservoir, given the efficiency of the cycle. Using the efficiency formula, we found that the cold reservoir temperature is 900 K. This temperature difference is what drives the cycle and enables it to perform work.