The efficiency of a heat engine at a heater temperature of 900 K: calculation and physical meaning

Tasks for calculating the efficiency of heat engines are a classic example of the application of the laws of thermodynamics in technology. Most often, the conditions give the temperatures of the heater (T₁) and the refrigerator (T₂), and you need to find the maximum possible efficiency factor (COP) of an ideal heat engine. Typical wording: “The heater temperature of a heat engine is 900 K, the temperature of the refrigerator is 3 times less.”. At first glance, everything is simple, but here are nuances that are important to consider when making a decision.

In this article we will not only analyze the step-by-step calculation of efficiency for this condition, but also explain the physical meaning of the quantities, consider real examples of heat engines (from steam turbines to refrigerators), and also warn you against common mistakes. The material will be useful to students of technical universities, heating engineers and anyone interested in thermodynamics.

⚠️ Attention: The article uses absolute temperatures in Kelvins (K). If in your problem the temperatures are specified in degrees Celsius, they must be converted to Kelvin using the formula T(K) = t(°C) + 273,15 before calculations.

1. Problem condition: parsing the data

Given:

• 🔥 Heater temperature T₁ = 900 K.
• ❄️ The temperature of the refrigerator is 3 times less than that of the heater.

The first step is to correctly determine T₂. Here, many make the mistake of dividing 900 K by 3 and getting 300 K. However, it is important to clarify: “3 times less” can be interpreted as T₂ = T₁ / 3 (absolute ratio) or T₂ = T₁ - 3T₁ (relative decrease). In physics, “3 times less” is almost always meant division, therefore:

T₂ = 900 K / 3 = 300 K

But what if the problem meant a reduction of 300%? Then T₂ = 900 K - 3 × 900 K = -1800 Kwhich is absurd (negative absolute temperatures are impossible in classical thermodynamics). Therefore always clarify the wording, especially if the task is from a non-standard source.

⚠️ Attention: If the temperature of the refrigerator is below 0 K, this is a signal about incorrect interpretation of the condition. In real heat engines T₂ > 0 K, and for an ideal Carnot cycle T₂ ≥ 0 K.

2. Formula for the efficiency of an ideal heat engine

The maximum efficiency of a heat engine is determined Carnot cycle and is calculated by the formula:

η_max = (T₁ - T₂) / T₁ = 1 - (T₂ / T₁)

Where:

• η_max — maximum efficiency (dimensionless value, often expressed in %);
• T₁ — heater temperature (K);
• T₂ — refrigerator temperature (K).

We substitute our values:

η_max = 1 - (300 K / 900 K) = 1 - 0.333... ≈ 0.6667

To convert to a percentage, multiply by 100:

η_max ≈ 66,67%

This is a theoretical limit. Real heat engines (for example steam turbines or internal combustion engines) have significantly lower efficiency due to friction losses, non-ideal heat transfer and other factors.

3. The physical meaning of the result

The obtained value η_max ≈ 66,67% means that even under ideal conditions maximum 2/3 of the heat received from the heater can be converted into useful work. The remaining third is inevitably transferred to the refrigerator (the environment).

Why does this happen? This is a consequence second law of thermodynamics, which reads: “It is impossible to create a heat engine that would have an efficiency of 100%, that is, completely convert heat into work.”. Part of the energy is always dissipated in the form of “waste” heat.

Examples from real life:

• 🚗 Internal combustion engines (ICE) in cars have an efficiency of ~20–40%. The rest of the fuel combustion energy is spent on heating the exhaust gases, radiator, and friction.
• ⚡ Steam turbines at thermal power plants achieve an efficiency of ~30–50%, but taking into account losses in generators and transformers, the final efficiency of the station rarely exceeds 40%.
• ❄️ Refrigerators and air conditioners work “on the contrary”: their effectiveness is assessed by coefficient of performance (COP), which can be greater than 1 (unlike efficiency).

⚠️ Attention: In tasks they are sometimes asked to find real efficiency, and not the maximum. In this case, additional data is required (for example, machine power or amount of heat supplied/removed). Without them, only η_max can be calculated according to Carnot.

4. Practical application: where such calculations are used

Knowledge about the efficiency of heat engines is used in a variety of areas:

In refrigeration technology (for example, in compressors of household refrigerators), the principles of thermodynamics work “in the opposite direction”: energy is spent on transferring heat from a cold body (freezer) to a warm one (surrounding air). Here, instead of efficiency, it is used coefficient of Performance (COP), which shows how much heat is transferred per 1 J of work expended.

Why is the efficiency higher in rocket engines than in an internal combustion engine?

In rocket engines, the temperature of the heater (combustion products) reaches 3000–4000 K, and the refrigerator is space with a temperature of ~3 K. The temperature difference is enormous, so the theoretical efficiency is close to 100%. However, in practice it is limited by losses due to incomplete combustion of fuel, heat transfer through the body and other factors.

5. Typical mistakes when solving efficiency problems

Even in a simple problem with data T₁ = 900 K i T₂ = T₁ / 3 students often make mistakes:

• ❌ Incorrect temperature conversion. For example, if the condition gives °C, but does not indicate that they have already been converted to Kelvin. Always check the units!
• ❌ Confusion with “3 times less.” As discussed earlier, this may mean division or subtraction. In physics, division is usually implied.
• ❌ Use of Carnot efficiency for real machines. Formula η = (T₁ - T₂)/T₁ gives maximum possible Efficiency. Real efficiency is always lower.
• ❌ Ignoring the sign of temperatures. In the Carnot formula, temperatures should be in Kelvin and positive. Negative or zero values will lead to absurd results (η > 100%).

⚠️ Attention: If the task specifies heater power (Q₁) and refrigerator power (Q₂), the efficiency is calculated differently:

η = (Q₁ - Q₂) / Q₁

This is real efficiencywhich may differ greatly from η_max according to Carnot.

6. Connection with refrigerators: reverse Carnot cycle

It is interesting that refrigerators and air conditioners work according to reverse Carnot cycle. Their efficiency is assessed not by efficiency, but by coefficient of performance (ε):

ε = Q₂ / A = T₂ / (T₁ - T₂)

Where:

• Q₂ — heat removed from the cooled volume;
• A — work spent on heat transfer;
• T₁ and T₂ —temperature of the heater (ambient) and refrigerator (freezer).

For our example (T₁ = 900 K, T₂ = 300 K):

ε = 300 / (900 - 300) = 0,5

This means that for 1 J of consumed electricity the refrigerator will transfer 0.5 J of heat from the chamber. Real refrigerators have ε ≈ 2–5 due to optimization of cycles and thermal insulation.

7. How to improve the efficiency of real heat engines

Although the Carnot efficiency is a theoretical limit, engineers are constantly looking for ways to get closer. Here are the key approaches:

• 🔥 Increasing the heater temperature (T₁). For example, in gas turbines, heat-resistant materials are used to increase the temperature of fuel combustion.
• ❄️ Reducing the temperature of the refrigerator (T₂). In cryogenic engines, liquid nitrogen (-196°C) can serve as a refrigerator.
• ♻️ Reducing heat losses. Thermal insulation of pipelines, heat recovery (for example, in cogeneration plants).
• ⚙️ Reducing mechanical losses. Use of high-quality lubricants, bearings, optimization of aerodynamics (for turbines).

In household refrigerators, to increase ε they use:

• Multilayer thermal insulation of walls;
• Inverter compressors (stepless power control);
• Systems No Frost for uniform distribution of cold.

⚠️ Attention: Some countries have minimum efficiency standards for household appliances (for example, Energy Star in the USA, when buying a refrigerator or air conditioner, pay attention to the energy efficiency class (from A+++ to D).

8. Problems for independent solution

Secure the material by solving similar problems:

1. The temperature of the heater heater is 1200 K, the refrigerator - 400 K. Find the maximum efficiency.
2. The efficiency of an ideal heat engine is 40%. The temperature of the refrigerator is 300 K. Find the temperature of the heater.
3. The refrigerator operates on a reverse Carnot cycle with a coefficient of performance of 4. The air temperature in the room is 27°C. Find the temperature in the freezer.

Answers:

1. η_max = 1 - (400/1200) ≈ 66,67%;
2. 0.4 = 1 - (300/T₁) → T₁ = 500 K;
3. 4 = T₂ / (300 - T₂) → T₂ ≈ 240 K (-33°C).

FAQ: Frequently asked questions about the efficiency of heat engines

Can the efficiency of a heat engine be greater? 100%. understanding.

No. Efficiency shows the share useful work from the supplied heat. According to the first law of thermodynamics, energy cannot arise from nothing, so η is always ≤ 100%. Used in refrigerators coefficient of performance (COP), which may be >1, but this is not efficiency in the classical sense.

Why is the efficiency so low in internal combustion engines?

Main reasons:

• Incomplete fuel combustion;
• Heat loss through the exhaust system and radiator;
• Mechanical friction losses in pistons, bearings;
• Energy costs for auxiliary systems (generator, pumps).

How does ambient temperature affect on efficiency?

Ambient temperature (T₂) is the temperature of the refrigerator for most heat engines. The lower it is, the higher it is maximum. possible efficiency (all other things being equal), thermal power plants can operate more efficiently in cold climates than in hot climates.

Is it possible to create a heat engine with an efficiency of 100%?

No, this is prohibited. second law of thermodynamics. Even an ideal Carnot machine has η < 100%, since part of the heat is always transferred to the refrigerator.

What is the difference between efficiency and coefficient of Performance (COP)?

Parameter	Efficiency (η)	COP
Definition	The fraction of heat converted into work	The amount of heat transferred per 1 J of work expended
Value range	0 ≤ η < 1	COP > 1 (for refrigerators)
Application	Heat engines (engines, turbines)	Refrigerators, air conditioners, heat pumps