# Efficiency & the Carnot Cycle: Equations & Examples

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• 0:03 What is the Carnot Cycle?
• 3:27 Calculating Efficiency
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Instructor: David Wood

David has taught Honors Physics, AP Physics, IB Physics and general science courses. He has a Masters in Education, and a Bachelors in Physics.

After watching this video, you will be able to explain the Carnot Cycle, including what it represents and how it works, and calculate the efficiency of a particular Carnot engine. A short quiz will follow.

## What Is the Carnot Cycle?

A heat engine is a device that produces motion from heat and includes gasoline engines and steam engines. These devices vary in efficiency. The Carnot Cycle describes the most efficient possible heat engine, involving two isothermal processes and two adiabatic processes. It is the most efficient heat engine that is possible within the laws of physics.

The second law of thermodynamics states that it is impossible to extract heat from a hot reservoir and use it all to do work; some must be exhausted into a cold reservoir. Or, in other words, no process can be 100% efficient because energy is always lost somewhere. The Carnot Cycle sets the upper limit for what is possible, for what the maximally-efficient engine would look like.

Before we go through the details of the Carnot Cycle, we should define two important terms: isothermic and adiabatic. An isothermic process is a process where the temperature remains constant and the volume and pressure vary relative to each other. An adiabatic process is a process where no heat enters or leaves the system to or from a reservoir and the temperature, pressure, and volume are all free to change, relative to each other.

The Carnot Cycle can be described by this pressure-volume graph:

Stage 1 is an isothermal expansion, where the volume increases and the pressure decreases at a constant temperature. This is where heat enters the system from the hot reservoir (i.e., from your power source) to keep the temperature constant. The gas inside the engine is allowed to naturally expand and push a piston.

Stage 2 is an adiabatic expansion, where the hot reservoir is now taken away. The gas continues to expand, causing the pressure and temperature to decrease. Stages 1 and 2 are where the engine actually does useful work.

Stage 3 is an isothermal compression, where the volume decreases and the pressure increases at a constant temperature. The temperature is kept constant by putting it in contact with a cold reservoir. Here, you actually have to do work by physically pushing the piston to compress the gas.

Lastly, stage 4 is an adiabatic compression, where you remove the cold reservoir. The volume continues to decrease but without the cold reservoir, this leads to both the pressure and temperature increasing. Stages 3 and 4 both involve you doing work on the system; you are expending energy to make this happen by pushing down on the piston.

The useful work that comes out of the Carnot Cycle is the difference between the work done by the engine in stages 1 and 2 and the work done (or the energy wasted) by you in stages 3 and 4. The Carnot Cycle, with its two isothermal processes and two adiabatic processes, is the most favorable case. In other words, the cycle that produces that largest difference between these values allowed by the laws of physics.

Unfortunately, the Carnot Cycle is not practical in real life. If you built a Carnot engine in your car, it would indeed be super-efficient. But the process would operate so ridiculously slowly that you would be overtaken by pedestrians!

## Calculating Efficiency

The efficiency of a Carnot engine is entirely a function of the temperatures of the hot and cold reservoirs you use. It can be calculated using this equation, where TH is the temperature of the hot reservoir in Kelvin, and TC is the temperature of the cold reservoir in Kelvin:

So, all you do is take the difference in temperature between the reservoirs, divide it by the temperature of the hot reservoir, and then multiply it by 100 to turn it into a percentage. It's important to note that the temperatures must be in Kelvin for it to work. So, if you have a temperature in degrees Celsius for example, you need to add 273 to that value to convert into Kelvin.

## Example

Maybe this would be clearer if we went through an example. Let's say you are operating a Carnot engine with a hot reservoir at a temperature of 100 degrees Celsius and a cold reservoir at a temperature of 0 degrees Celsius. What is the efficiency of the Carnot engine?

Well, first of all we should write down what we know. But we can't do that until we convert our temperatures into Kelvin. 100 degrees Celsius in Kelvin is 100 plus 273, which comes out as 373 Kelvin. 0 degrees Celsius in Kelvin is 0 plus 273, which is 273 Kelvin. So TH is 373, and TC is 273.

Plug these numbers into the efficiency equation (373 minus 273, divided by 373, multiplied by 100) gives us an efficiency of approximate 27%. So, with reservoirs at these admittedly similar temperatures, that is the most efficient that an engine could ever be. As you may have noticed from the equation, the larger the difference in temperature between the two reservoirs, the greater the efficiency.

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The text passage provides an explanation of the Carnot Cycle, including its stages and the calculation of the efficiency of a Carnot engine. It includes a specific example of calculating the efficiency of a Carnot engine with given temperatures.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Efficiency & the Carnot Cycle: Equations & Examples

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• 0:03 What is the Carnot Cycle?
• 3:27 Calculating Efficiency
• 4:09 Example
• 5:20 Lesson Summary

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Lesson Transcript
Instructor: David Wood

David has taught Honors Physics, AP Physics, IB Physics and general science courses. He has a Masters in Education, and a Bachelors in Physics.

After watching this video, you will be able to explain the Carnot Cycle, including what it represents and how it works, and calculate the efficiency of a particular Carnot engine. A short quiz will follow.

## What Is the Carnot Cycle?

A heat engine is a device that produces motion from heat and includes gasoline engines and steam engines. These devices vary in efficiency. The Carnot Cycle describes the most efficient possible heat engine, involving two isothermal processes and two adiabatic processes. It is the most efficient heat engine that is possible within the laws of physics.

The second law of thermodynamics states that it is impossible to extract heat from a hot reservoir and use it all to do work; some must be exhausted into a cold reservoir. Or, in other words, no process can be 100% efficient because energy is always lost somewhere. The Carnot Cycle sets the upper limit for what is possible, for what the maximally-efficient engine would look like.

Before we go through the details of the Carnot Cycle, we should define two important terms: isothermic and adiabatic. An isothermic process is a process where the temperature remains constant and the volume and pressure vary relative to each other. An adiabatic process is a process where no heat enters or leaves the system to or from a reservoir and the temperature, pressure, and volume are all free to change, relative to each other.

The Carnot Cycle can be described by this pressure-volume graph:

Stage 1 is an isothermal expansion, where the volume increases and the pressure decreases at a constant temperature. This is where heat enters the system from the hot reservoir (i.e., from your power source) to keep the temperature constant. The gas inside the engine is allowed to naturally expand and push a piston.

Stage 2 is an adiabatic expansion, where the hot reservoir is now taken away. The gas continues to expand, causing the pressure and temperature to decrease. Stages 1 and 2 are where the engine actually does useful work.

Stage 3 is an isothermal compression, where the volume decreases and the pressure increases at a constant temperature. The temperature is kept constant by putting it in contact with a cold reservoir. Here, you actually have to do work by physically pushing the piston to compress the gas.

Lastly, stage 4 is an adiabatic compression, where you remove the cold reservoir. The volume continues to decrease but without the cold reservoir, this leads to both the pressure and temperature increasing. Stages 3 and 4 both involve you doing work on the system; you are expending energy to make this happen by pushing down on the piston.

The useful work that comes out of the Carnot Cycle is the difference between the work done by the engine in stages 1 and 2 and the work done (or the energy wasted) by you in stages 3 and 4. The Carnot Cycle, with its two isothermal processes and two adiabatic processes, is the most favorable case. In other words, the cycle that produces that largest difference between these values allowed by the laws of physics.

Unfortunately, the Carnot Cycle is not practical in real life. If you built a Carnot engine in your car, it would indeed be super-efficient. But the process would operate so ridiculously slowly that you would be overtaken by pedestrians!

## Calculating Efficiency

The efficiency of a Carnot engine is entirely a function of the temperatures of the hot and cold reservoirs you use. It can be calculated using this equation, where TH is the temperature of the hot reservoir in Kelvin, and TC is the temperature of the cold reservoir in Kelvin:

So, all you do is take the difference in temperature between the reservoirs, divide it by the temperature of the hot reservoir, and then multiply it by 100 to turn it into a percentage. It's important to note that the temperatures must be in Kelvin for it to work. So, if you have a temperature in degrees Celsius for example, you need to add 273 to that value to convert into Kelvin.

## Example

Maybe this would be clearer if we went through an example. Let's say you are operating a Carnot engine with a hot reservoir at a temperature of 100 degrees Celsius and a cold reservoir at a temperature of 0 degrees Celsius. What is the efficiency of the Carnot engine?

Well, first of all we should write down what we know. But we can't do that until we convert our temperatures into Kelvin. 100 degrees Celsius in Kelvin is 100 plus 273, which comes out as 373 Kelvin. 0 degrees Celsius in Kelvin is 0 plus 273, which is 273 Kelvin. So TH is 373, and TC is 273.

Plug these numbers into the efficiency equation (373 minus 273, divided by 373, multiplied by 100) gives us an efficiency of approximate 27%. So, with reservoirs at these admittedly similar temperatures, that is the most efficient that an engine could ever be. As you may have noticed from the equation, the larger the difference in temperature between the two reservoirs, the greater the efficiency.

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### Earning College Credit

Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
user
The text passage provides an explanation of the Carnot Cycle, including its stages and the calculation of the efficiency of a Carnot engine. It includes a specific example of calculating the efficiency of a Carnot engine with given temperatures.

TYPE: SINGLE_QUESTION_WITH_ANSWER