Problem 79 - Pearson Physics

Question

Think \& Calculate A heat engine does \(2700 \mathrm{~J}\) of work with an efficiency of \(0.18\). What is (a) the heat taken in from the hot reservoir and (b) the heat given off to the cold reservoir? (c) If the efficiency of the engine is increased, does your answer to part (a) increase, decrease, or stay the same? Explain.

Step-by-Step Solution

Answer

a) 15000 J, b) 12300 J, c) Decreases.

Step 1: Understand Efficiency Formula

The efficiency \( \eta \) of a heat engine is given by the formula: \( \eta = \frac{W}{Q_{in}} \), where \( W \) is the work done by the engine and \( Q_{in} \) is the heat absorbed from the hot reservoir.

Step 2: Calculate Heat Absorbed \( Q_{in} \)

Rearrange the efficiency formula to solve for the heat absorbed \( Q_{in} \): \( Q_{in} = \frac{W}{\eta} = \frac{2700 \mathrm{~J}}{0.18} \). Calculate the value to find \( Q_{in} \).

Step 3: Perform Calculation for \( Q_{in} \)

Compute \( Q_{in} = \frac{2700 \mathrm{~J}}{0.18} = 15000 \mathrm{~J} \). Thus, the heat taken in from the hot reservoir is 15000 J.

Step 4: Calculate Heat Rejected \( Q_{out} \)

Use the formula for heat rejected: \( Q_{out} = Q_{in} - W \). With \( Q_{in} = 15000 \mathrm{~J} \) and \( W = 2700 \mathrm{~J} \), find \( Q_{out} \).

Step 5: Perform Calculation for \( Q_{out} \)

Calculate \( Q_{out} = 15000 \mathrm{~J} - 2700 \mathrm{~J} = 12300 \mathrm{~J} \). Thus, the heat given off to the cold reservoir is 12300 J.

Step 6: Analyze Effect of Increased Efficiency

As the efficiency of the engine increases, the heat absorbed \( Q_{in} \) from the hot reservoir decreases for a given amount of work done.

Key Concepts

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat, work, and energy. It provides a quantitative description of how these energies are transformed and governed by them. An essential aspect is the understanding of different systems and processes:

• Closed Systems: Exchanges energy but not matter with its surroundings.
• Open Systems: Exchanges both energy and matter.
• Isolated Systems: Exchanges neither energy nor matter.

The laws of thermodynamics are fundamental to comprehending heat engines:

• First Law: Energy cannot be created or destroyed, only transformed.
• Second Law: Total entropy, or degree of disorder, always increases in a closed system.

In our example, a heat engine is a device converting heat energy into mechanical work, highlighting energy transformations, a key concept in thermodynamics.

Work-Energy Principle

The work-energy principle is a cornerstone in physics, stating that the work done by forces on an object equals the change in its kinetic energy. In the context of heat engines, it helps explain how energy input from heat becomes useful work output. When we refer to a heat engine performing work, we mean that the system is transferring energy from the hot reservoir and converting part of that energy to perform tasks or produce motion:

• Work Done (W): The energy output from the engine; in our example, it’s 2700 J.
• Heat Input (Q_in): The total heat energy absorbed from the hot reservoir.

The principle helps us quantify and discern inefficiencies, understanding that not all absorbed energy can be converted, indicated by the engine’s efficiency.

Heat Transfer

Heat transfer is crucial in understanding how heat engines work. It describes the movement of thermal energy from one place to another and outlines the methods: conduction, convection, and radiation. In heat engines, the main focus is usually on conduction and convection:

• Conduction: Transfer of heat through a material, essential in parts of the engine where heat moves to perform work.
• Convection: Transfer of heat by the physical movement of fluid; relevant for cooling mechanisms in engines.
• Radiation: Transfer of heat in the form of electromagnetic waves, less relevant in this engineering context.

Understanding how heat moves between the hot and cold reservoirs helps in analyzing efficiency and performance, as demonstrated in calculating the absorbed and rejected heat energies.

Efficiency Formula

The efficiency formula in thermodynamics measures how well a heat engine converts the heat energy it receives into useful work. It is represented by the symbol \( \eta \) and defined by the equation:\[\eta = \frac{W}{Q_{in}} \]Where:

• \( W \) is the work done (2700 J in this case).
• \( Q_{in} \) is the heat energy absorbed from the heat source.

The result is a dimensionless number that ranges from 0 to 1 (or often expressed as a percentage). In practice, due to inherent inefficiencies, such as friction and heat loss, no engine reaches 100% efficiency.In the original exercise, the efficiency is 0.18, showing that only 18% of absorbed heat is converted to work. Increasing efficiency means less heat is required to perform the same amount of work, improving the engine's practical performance and energy usage.