# Can Heat Engine Efficiency Be Related to Temperature?

• vetgirl1990
In summary: Qh = W + Qc. Also, in order to apply the formula to a problem, you need to know the temperatures. In the problem, you know that Th = 200C and Tc = 80C, so you can find the efficiency of a Carnot engine operating between these two temperatures. In summary, the efficiency of a Carnot Engine can be described by the relationship Tc/Th = Qc/Qh, leading to the formula e(Carnot) = 1 - Tc/Th. This efficiency can be applied to heat engines, with the formula e(heat engine) = W / Qh = 1 - Qc/Qh. The reason for this
vetgirl1990
The efficiency of a Carnot Engine is described by the relationship: Tc/Th = Qc/Qh, so that e(Carnot) = 1 - Tc/Th

For heat engines, can their efficiency also be related to temperature as well?
Or is the description of their efficiency just: e(heat engine) = W / Qh = 1 - Qc/Qh

I am inclined to say that the only reason that a Carnot Engine's efficiency can be related to temperature like that, is because of the cyclic nature of the Carnot Cycle... But I'm not entirely sure.

The reason I am asking this question, is because I am trying to understand how to solve the following problem: "A heat engine operating between 200C and 80C achieves 20% of the maximum possible efficiency. What energy input will enable the engine to perform 10kJ of work?"
Tc/Th = Qc/Qh, W = Qh-Qc
Therefore, Tc/Th = (Qh - W) / Qh --> Tc/Th = 1 - W/Qh
So plugging in the above values, Qh = 16.7kJ

My solution is only valid if I made the correct assumption that Carnot efficiency can be applied to a heat engine efficiency.

vetgirl1990 said:
The efficiency of a Carnot Engine is described by the relationship: Tc/Th = Qc/Qh, so that e(Carnot) = 1 - Tc/Th

I'm not sure how you derive the Carnot efficiency from Tc/Th = Qc/Qh, but the final formula is correct.

vetgirl1990 said:
For heat engines, can their efficiency also be related to temperature as well?
Or is the description of their efficiency just: e(heat engine) = W / Qh = 1 - Qc/Qh

ηC = W / Qh = (Qh - Qc) / Qh = 1 - Qc / Qh = 1 - (Tc ⋅ Δs) / (Th ⋅ Δs) = 1 - Tc / Th

vetgirl1990 said:
I am inclined to say that the only reason that a Carnot Engine's efficiency can be related to temperature like that, is because of the cyclic nature of the Carnot Cycle... But I'm not entirely sure.

The Carnot efficiency is the theoretical maximal efficiency for thermodynamic cycles (and only for thermodynamic cycles) - so yes.

vetgirl1990 said:
The reason I am asking this question, is because I am trying to understand how to solve the following problem: "A heat engine operating between 200C and 80C achieves 20% of the maximum possible efficiency. What energy input will enable the engine to perform 10kJ of work?"
Tc/Th = Qc/Qh, W = Qh-Qc
Therefore, Tc/Th = (Qh - W) / Qh --> Tc/Th = 1 - W/Qh
So plugging in the above values, Qh = 16.7kJ

My solution is only valid if I made the correct assumption that Carnot efficiency can be applied to a heat engine efficiency.

1) Your calculations are not correct. I recommend to take the Carnot efficiency as you stated it at the beginning of your post (ηC = 1 - Tc / Th).
2) The Carnot efficiency is the maximum possible efficiency - what if the cycle only achieves 20 % of it?
3) Don't forget, that the temperatures in the statement are in °C.

The reason why the efficiency of a Carnot process can be expressed in terms of two temperatures is the fact that it is the only process where heat exchange only takes place at two well defined temperatures. In other processes, temperature changes continuously during heat exchange (e.g. along isochores).

DrDu said:
The reason why the efficiency of a Carnot process can be expressed in terms of two temperatures is the fact that it is the only process where heat exchange only takes place at two well defined temperatures. In other processes, temperature changes continuously during heat exchange (e.g. along isochores).
As DrDu says, the Carnot cycle is a reversible cycle between two temperatures, and you can prove that the efficiency of the cycle is as stated by you. I have just a couple of points to clarify.
The Carnot cycle is a heat engine. It is the most efficient possible one.
efficiency e = W/Qh is not just a description, it follows from the definition of efficiency as (useful work output/heat input to engine)

## What is the Carnot engine?

The Carnot engine is a theoretical thermodynamic cycle that is considered to be the most efficient engine possible. It was developed by French physicist Nicolas Carnot in the 1820s and is based on the principles of thermodynamics.

## How does the Carnot engine work?

The Carnot engine operates by converting heat energy into mechanical work. It consists of four main processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. These processes are carried out by a working fluid, such as a gas, inside a closed system.

## What is the efficiency of a Carnot engine?

The efficiency of a Carnot engine is given by the Carnot efficiency formula, which is equal to the temperature difference between the hot and cold reservoirs divided by the temperature of the hot reservoir. This means that the efficiency of a Carnot engine can never be 100% as there will always be some energy loss in the form of heat.

## What factors affect the efficiency of a Carnot engine?

The efficiency of a Carnot engine is primarily affected by the temperature difference between the hot and cold reservoirs. The larger the temperature difference, the higher the efficiency. Other factors that can affect efficiency include the type of working fluid and the design of the engine.

## How is the Carnot engine used in real life?

While the Carnot engine is a theoretical concept and cannot be built in its exact form, its principles are used in modern-day engines and machines. For example, the Carnot cycle is used in power plants and refrigeration systems to improve efficiency and reduce energy waste.

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