Problem 38 - Pearson Physics

Question

Rank The reservoir temperatures for heat engines A through D are given below. Rank the engines in order of increasing efficiency. Indicate ties where appropriate. \begin{tabular}{|c|c|c|c|c|} \hline Engine & \(\mathbf{A}\) & \(\mathbf{B}\) & \(\mathbf{C}\) & D \\ \hline \(\boldsymbol{T}_{\mathbf{h}}(\mathbf{K})\) & 400 & 440 & 800 & 1240 \\ \hline \(\boldsymbol{T}_{\mathbf{C}} \mathbf{( K )}\) & 200 & 420 & 600 & 1020 \\\ \hline \end{tabular}

Step-by-Step Solution

Answer

B, D, C, A (ascending efficiency)

Step 1: Understand Efficiency Formula

The efficiency of a heat engine is given by the formula: \( \eta = 1 - \frac{T_c}{T_h} \), where \( T_h \) is the hot reservoir temperature and \( T_c \) is the cold reservoir temperature. The efficiency depends on the difference between the hot and cold temperatures.

Step 2: Calculate Efficiency for Engine A

Plug in the values for Engine A: \( T_h = 400 \) K and \( T_c = 200 \) K. Substitute into the efficiency formula: \( \eta_A = 1 - \frac{200}{400} = 0.5 \).

Step 3: Calculate Efficiency for Engine B

Plug in the values for Engine B: \( T_h = 440 \) K and \( T_c = 420 \) K. Substitute into the efficiency formula: \( \eta_B = 1 - \frac{420}{440} \approx 0.0455 \).

Step 4: Calculate Efficiency for Engine C

Plug in the values for Engine C: \( T_h = 800 \) K and \( T_c = 600 \) K. Substitute into the efficiency formula: \( \eta_C = 1 - \frac{600}{800} = 0.25 \).

Step 5: Calculate Efficiency for Engine D

Plug in the values for Engine D: \( T_h = 1240 \) K and \( T_c = 1020 \) K. Substitute into the efficiency formula: \( \eta_D = 1 - \frac{1020}{1240} \approx 0.1774 \).

Step 6: Order the Engines by Efficiency

Compare the efficiency values: \( \eta_B \approx 0.0455 \), \( \eta_D \approx 0.1774 \), \( \eta_C = 0.25 \), \( \eta_A = 0.5 \). The order from least to greatest efficiency is B, D, C, A.

Key Concepts

Reservoir Temperatures

In the realm of heat engines, reservoir temperatures are crucial. They refer to the hot and cold temperatures around which an engine operates. Simply put, the hot reservoir (\(T_h\) is the source from which the engine draws heat, while the cold reservoir (\(T_c\)) is where the engine releases leftover heat.

Understanding these temperatures is key because they directly impact the engine’s ability to convert heat into work. For a heat engine to function efficiently, there must be a significant temperature difference between these two reservoirs. The greater the difference, the more potential the engine has to do work efficiently.

In our example exercise, Engine A has a hot reservoir at 400 K and a cold reservoir at 200 K. This gives us a better perspective regarding the engine’s efficiency potential.

Carnot Efficiency Formula

The Carnot efficiency formula is a staple in evaluating how well a heat engine can perform. It’s given by the equation: \[\eta = 1 - \frac{T_c}{T_h}\]
This simple formula expresses the theoretical maximum efficiency a heat engine can achieve, which is solely dependent on the reservoir temperatures.

To better understand:

• \(\eta\): Represents efficiency, expressed as a decimal or percentage.
• \(T_h\): Temperature of the hot reservoir in Kelvin.
• \(T_c\): Temperature of the cold reservoir in Kelvin.

The closer \(T_c\) is to \(T_h\), the lower the efficiency. Thus, minimizing \(T_c\) and maximizing \(T_h\) is an ideal approach for boosting the engine’s efficiency.

Thermodynamics

Thermodynamics is the science of energy transformations involving heat and work. It lays the groundwork for understanding the behavior of heat engines and their efficiencies.

When we talk about heat engines, we’re dealing with the conversion of thermal energy into mechanical work. This process is governed by the laws of thermodynamics:

• The First Law deals with energy conservation, ensuring that total energy remains constant within a system.
• The Second Law introduces entropy, indicating that spontaneous processes increase the universe’s entropy, affecting the efficiency of energy conversions.

Heat engines exploit the principles of thermodynamics to convert heat. By referencing these laws, we grasp why not all thermal energy converts to work, emphasizing the maximum potential efficiency is achieved when these conversions are ideal, as per the Carnot cycle.

Temperature Difference in Heat Engines

A heat engine’s efficiency highly depends on the temperature difference between the hot and cold reservoirs. The greater this difference, often referred to as the temperature gradient, the more effectively a heat engine can convert heat into work.

Why does this matter? Well, the temperature difference influences the net energy transfer within the engine:

• Large temperature differences increase the engine’s working potential, thus improving efficiency.
• If the temperature difference is minimal, less energy is available for conversion to work, and efficiency drops.

In practical terms, when we look at engines like those in the exercise, an engine operating between 400 K and 200 K would potentially be more efficient than one between 440 K and 420 K, as a greater temperature difference exists in the former. Thus, ensuring a significant temperature difference is crucial for optimizing a heat engine’s performance.