Question

A heat engine whose thermal efficiency is 35 percent uses a hot reservoir at \(1100 \mathrm{R}\) and a cold reservoir at \(550 \mathrm{R}\) Calculate the entropy change of the two reservoirs when 1 Btu of heat is transferred from the hot reservoir to the engine. Does this engine satisfy the increase of entropy principle? If the thermal efficiency of the heat engine is increased to 60 percent, will the increase of entropy principle still be satisfied?

Short answer

Calculate the total entropy change for a heat engine with given heat transfers and temperatures and verify if the increase of entropy principle is satisfied for a thermal efficiency of 35% and an increased efficiency of 60%. Given: - Thermal efficiency, η = 35% (0.35) - Heat transfer, Q_hot = 1 Btu - Temperature of hot reservoir, T_hot = 1100 R - Temperature of cold reservoir, T_cold = 550 R The total entropy change for the case with 35% thermal efficiency is calculated to be 0.000091 Btu/R, which is positive, satisfying the increase of entropy principle. However, if the thermal efficiency is increased to 60%, the total entropy change becomes -0.0000455 Btu/R, a negative value, indicating the engine would no longer satisfy the increase of entropy principle at this higher thermal efficiency.

Step-by-step solution

Calculate the work output of the engine

Given that the heat engine has a thermal efficiency of 35 percent, we can calculate the amount of work produced during the heat transfer Q_hot: \(W = \eta \times Q_{hot} = 0.35 \times 1\,\mathrm{Btu} = 0.35\,\mathrm{Btu}\)

Calculate the heat transfer to the cold reservoir

According to the first law of thermodynamics, the heat transferred to the cold reservoir Q_cold is given by the difference between the heat absorbed from the hot reservoir Q_hot and the work W produced by the engine. \(Q_{cold} = Q_{hot} - W = 1\,\mathrm{Btu} - 0.35\,\mathrm{Btu} = 0.65\,\mathrm{Btu}\)

Calculate the entropy change for the hot reservoir

The entropy change for the hot reservoir (ΔS_hot) is given as: \(\Delta S_{hot} = \frac{-Q_{hot}}{T_{hot}}\) Substitute the given values for Q_hot and T_hot: \(\Delta S_{hot} = \frac{-1\,\mathrm{Btu}}{1100\,\mathrm{R}} = -\frac{1}{1100}\,\mathrm{Btu/R}\)

Calculate the entropy change for the cold reservoir

The entropy change for the cold reservoir (ΔS_cold) is given as: \(\Delta S_{cold} = \frac{Q_{cold}}{T_{cold}}\) Substitute the given values for Q_cold and T_cold: \(\Delta S_{cold} = \frac{0.65\,\mathrm{Btu}}{550\,\mathrm{R}} = \frac{1}{846.15}\,\mathrm{Btu/R}\)

Calculate the total entropy change and check if the engine satisfies the increase of entropy principle

The total entropy change (ΔS_total) is given as the sum of the entropy change for the hot and cold reservoirs: \(\Delta S_{total} = \Delta S_{hot} + \Delta S_{cold} = -\frac{1}{1100} + \frac{1}{846.15}\,\mathrm{Btu/R} = 0.000091\,\mathrm{Btu/R}\) Since the total entropy change is positive, this heat engine satisfies the increase of entropy principle.

Check if the engine still satisfies the increase of entropy principle with a higher (⁠60%) thermal efficiency

If the thermal efficiency is increased to 60 percent, the new work output is given by: \(W_{new} = 0.6 \times 1\,\mathrm{Btu} = 0.6\,\mathrm{Btu}\) The new heat transfer to the cold reservoir is given by: \(Q_{cold}^{new} = Q_{hot} - W_{new} = 1\,\mathrm{Btu} - 0.6\,\mathrm{Btu} = 0.4\,\mathrm{Btu}\) The new entropy change for the cold reservoir is given by: \(\Delta S_{cold}^{new} = \frac{Q_{cold}^{new}}{T_{cold}} = \frac{0.4\,\mathrm{Btu}}{550\,\mathrm{R}} = \frac{1}{1375}\,\mathrm{Btu/R}\) The new total entropy change would be: \(\Delta S_{total}^{new} = \Delta S_{hot} + \Delta S_{cold}^{new} = -\frac{1}{1100} + \frac{1}{1375}\,\mathrm{Btu/R} = -0.0000455\,\mathrm{Btu/R}\) With the increased thermal efficiency, the new total entropy change is negative, indicating that the engine would no longer satisfy the increase of entropy principle if the thermal efficiency were to be increased to 60 percent.

Key concepts

Understanding Thermal Efficiency

Thermal efficiency is a measure of how well a heat engine converts heat into work. It's represented by the symbol \( \eta \), and it is defined as the ratio of work output to heat input. In a simple mathematical form, this can be written as \( \eta = \frac{W}{Q_{hot}} \) where \( W \) is the work produced by the engine and \( Q_{hot} \) is the heat absorbed from the hot reservoir.

For instance, if an engine has a thermal efficiency of 35%, this means that it converts 35% of the heat energy from the hot reservoir into work and the remaining 65% is transferred to the cold reservoir as waste heat. Importantly, no heat engine can have a thermal efficiency of 100% due to the second law of thermodynamics, which introduces us to the concept of entropy.

In practical applications, improving thermal efficiency is crucial for energy conservation. However, the exercise provides an interesting scenario where increasing the thermal efficiency to 60% leads to a violation of the increase of entropy principle, underlining the delicate balance between efficiency and thermodynamic laws.

The Role of a Heat Engine

A heat engine operates between two reservoirs at different temperatures: a hot reservoir and a cold reservoir. The engine absorbs heat from the hot reservoir (\(Q_{hot}\)), does some work (\(W\)), and then expels the remaining heat to the cold reservoir (\(Q_{cold}\)).

The function of a heat engine can be summarized in three essential steps: absorbing heat from the hot reservoir, converting part of that heat into work, and rejecting the remaining heat to the cold reservoir. This cycle is representative of many real-world systems, like power plants and automobile engines.

Heat engines are a practical embodiment of thermodynamics in technology, showcasing how thermal energy can be harnessed. The higher the temperature difference between the hot and cold reservoirs, generally the greater the potential for work production, emphasizing the importance of temperature in determining engine efficiency.

First Law of Thermodynamics in Action

The first law of thermodynamics, also known as the law of energy conservation, asserts that energy cannot be created or destroyed, only converted from one form to another. For heat engines, this law can be represented by the equation: \( Q_{hot} = W + Q_{cold} \).

What this means for our example is that the heat \(Q_{hot}\) taken from the hot reservoir is equal to the sum of the work output \(W\) and the heat \(Q_{cold}\) rejected to the cold reservoir. When calculating the entropy change in the provided exercise, the first law helps us understand that the decrease in energy (heat) of the hot reservoir is converted into work and the residual heat transferred to the cold reservoir.

Practically, this concept ensures that energy balance is maintained and that the output energy can never exceed the input energy. In any system obeying this law, the balance between the input heat, output work, and the heat loss to the cold reservoir must always be accounted for.

Increase of Entropy Principle

The increase of entropy principle stems from the second law of thermodynamics and indicates that the total entropy of an isolated system always increases over time. Entropy, in this context, can be viewed as a measure of disorder or randomness in a system.

For the heat engine scenario described, entropy change for the hot reservoir (\(\Delta S_{hot}\)) is calculated by dividing the negative heat transfer by the hot reservoir temperature (\(-Q_{hot}/T_{hot}\)). Conversely, the entropy change for the cold reservoir (\(\Delta S_{cold}\)) is derived from dividing the heat transferred to the cold reservoir by its temperature (\(Q_{cold}/T_{cold}\)).

The total change in system entropy, which is the sum of the entropy changes of both reservoirs, must be positive or zero for the increase of entropy principle to be satisfied. This rule is a fundamental guiding principle for processes in nature and technology. It poses limits to energy conversion processes and emphasizes that no process is completely reversible. Hence, while it might seem beneficial to drastically increase a heat engine's efficiency, this example demonstrates how such an improvement could contradict the fundamental law of increasing entropy.