Heat engines surround us everywhere: from refrigerators and air conditioners to car engines and power plants. Their efficiency is measured coefficient of efficiency (efficiency) —a key parameter showing what proportion of the received heat the machine converts into useful work. But how to calculate the efficiency of a heat engine if the temperatures of the heater ( ideal heat engine, if the heater temperatures are known (140°C) and refrigerator (17°C) are known? The answer lies in the laws of thermodynamics and the Carnot cycle - a theoretical model that sets the efficiency limit for any real devices.
In this article we will not only derive the exact formula and carry out calculations, but also analyze the physical the meaning of each parameter. You will learn why temperatures need to be converted to kelvins, how to avoid common errors in calculations, and why even an ideal machine cannot have 100% efficiency. And for those who plan to apply this knowledge in practice - for example, when diagnosing refrigeration equipment or optimizing heating systems - we have prepared a checklist for checking calculations and an FAQ with an analysis of complex cases.
What is an ideal heat engine and the Carnot cycle
An ideal heat engine is an abstract a model that works according to Carnot cycleproposed by the French physicist Sadi Carnot in 1824. In reality, such machines do not exist, but they serve a standard of the highest possible efficiency for any heat engines operating in a given temperature range.
The Carnot cycle consists of four reversible processes:
- 🔥 Isothermal expansion (gas receives heat
Q₁from the heater at temperatureT₁). - ↗️ Adiabatic expansion (gas expands without heat exchange, temperature drops to
T₂). - ❄️ Isothermal compression (gas gives off heat
Q₂to the refrigerator at temperatureT₂). - ↘️ Adiabatic compression (gas is compressed without heat exchange, the temperature returns to
T₁).
The key advantage of the Carnot cycle is it reversibility: all processes can be carried out in the reverse order without loss of energy. It is this property that allows you to achieve maximum efficiency among all heat engines operating between two heat reservoirs with temperatures T₁ and T₂.
Efficiency formula for an ideal heat engine
Efficiency (η) of the Carnot cycle is determined only by the temperatures of the heater (T₁) and refrigerator (T₂) and is calculated by the formula:
η = (T₁ - T₂) / T₁ = 1 - (T₂ / T₁)Where:
T₁— absolute temperature of the heater (in kelvins).T₂—the absolute temperature of the refrigerator (in kelvins).η—efficiency (dimensionless value, often expressed as a percentage).
Please note: the formula uses absolute temperaturesand not degrees Celsius This is! is critical, since the Kelvin scale starts from absolute zero (0 K = -273.15°C), and only in it the temperature ratio correctly reflects physical processes.
Calculation of efficiency for T₁=140°C and T₂=17°C: step-by-step instructions
Now we apply the formula to the given conditions. Temperatures are given in degrees Celsius, so the first step is to convert them to Kelvin:
- Convert the heater temperature:
T₁ = 140°C + 273.15 = 413.15 K. - Convert the temperature of the refrigerator:
T₂ = 17°C + 273.15 = 290.15 K. - Substitute the efficiency into the formula:
η = 1 - (290.15 / 413.15) ≈ 1 - 0.702 ≈ 0.298. - Convert into percentages:
η ≈ 0.298 × 100% ≈ 29.8%.
Thus the maximum possible efficiency of an ideal heat engine at heater temperatures 140°C i refrigerator 17°C is 29.8%This means that even under ideal conditions, only 29.8% of the heat received from the heater is converted into useful work, and the remaining 70.2% is transferred to the refrigerator.
☑️ Checking efficiency calculations
Why efficiency cannot be 100%: the second law of thermodynamics
Many people wonder: why even an ideal machine is not able to convert all the heat into work? The answer is given the second law of thermodynamics, which can be formulated as follows:
⚠️ Attention: It is impossible to create a heat engine that would have an efficiency of 100%, even if there is no refrigerator (that is, T₂ = 0 KIn practice, achieving absolute zero is impossible, but any real machine always has). losses.
The physical meaning of this law is that entropy (the measure of chaos in the system) never decreases spontaneously. Part of the heat required must be transferred to the refrigerator to compensate for the increase in entropy during the transfer of energy from the heater. therefore, the efficiency of the Carnot cycle is always less than unity:
- 🔥 If
T₂ > 0 K, thenη < 1. - ❄️ When
T₂ = T₁(same temperatures)η = 0—the machine does not do work. - 📈 The greater the difference
T₁ - T₂, the higher the efficiency, but it never reaches 100%.
For example, if a refrigerator had temperature 0 K (which is impossible), the efficiency would tend to 100%. But in reality, even in space, where the temperature is close to 2.7 K (relict radiation), it is unrealistic for technical systems to achieve such cold.
Comparison with real heat engines
The ideal Carnot cycle serves as a benchmark for assessing the efficiency of real devices. However, in practice, the efficiency is always lower due to:
| Machine type | Theoretical efficiency (Carnot) | Real efficiency | Reasons losses |
|---|---|---|---|
| Steam turbines (TPP) | ~40–50% | 30–35% | Friction, heat loss, non-ideal cycle |
| ICE (gasoline) | ~50–60% | 20–30% | Incomplete combustion, heat transfer, mechanical losses |
| Refrigerators | ~10–20% | 3–8% | Refrigerant leaks, heat exchange with the environment |
| Heat pumps | ~300–500% | 200–400% | Electrical losses, imperfect compressor |
As can be seen from the table, real machines lose a significant part For example, gasoline engine with theoretical efficiency in 50% in practice it produces only 25% due to:
- 🔥 Incomplete combustion of fuel (part of the energy is lost with the exhaust gases).
- 🛢️ Friction in moving parts (pistons, crankshaft).
- 🌡️ Heat loss through the engine housing.
For refrigerators the situation is even more complicated: their efficiency is often assessed through efficiency coefficient (COP), which may exceed 100% (since they do not produce work, but “pump” heat). But even here, the real values are far from ideal.
Why is the heat pump efficiency >100% in the table?
Heat pumps do not produce energy, but transfer heat from a cold environment to a warm one, consuming electricity. COP shows how much heat is transferred per 1 kW of consumed electricity. For example, COP=4 means that 4 kW of heat is transferred per 1 kW of electricity.
Typical errors when calculating efficiency
Even in a simple Carnot formula, it is easy to make mistakes that will lead to incorrect results. Here are the most common of them:
- Using Celsius instead of Kelvin:
If you substitute
140°Cand17°Cdirectly, you getη = 1 - (17/140) ≈ 87.9%— absurdly high resultsince temperatures are not reduced to absolute scale. - Mixed temperatures:
If you accidentally swap
T₁andT₂, the efficiency will become negative, which is physically impossible. - Ignoring the temperature sign:
Absolute temperature is always positive. An error in the sign (for example,
-273°Cinstead of0 K) will lead to incorrect calculations. - Not taking into account units of measurement:
If temperatures are given in Fahrenheit, they must first be converted to Celsius, and then to Kelvin.
⚠️ Attention: When calculating for refrigeration units (for example, in air conditioning systems), reverse Carnot cycleis sometimes used. In this case, the efficiency formula changes toCOP = T₂ / (T₁ - T₂), whereCOPis the efficiency coefficient. Do not confuse these values!
Practical application of efficiency calculations
Knowledge of theoretical efficiency helps in several practical scenarios:
- Diagnostics of refrigeration equipment:
If the actual efficiency of your refrigerator or air conditioner is significantly lower than the theoretical one (for example,
5%instead of15%), this may indicate:- 🛠️ Leakage refrigerant.
- 🧊 Heat exchanger contamination.
- 🔧 Compressor malfunction.
Heat pumps are selected based on COP, which depends on the temperature difference between the heat source (for example, soil or air) and the heated room. The smaller this difference, the more efficient the work.
Oils with low viscosity reduce friction in the engine, bringing its efficiency closer to the theoretical maximum.
For example, if you select inverter air conditioner, it COP should be close to theoretical for your climate zone. In Moscow (where in summer T₁ ≈ 30°C, and in winter T₂ ≈ -10°C), the ideal COP for heating will be:
COP_ideal = T₁ / (T₁ - T₂) = 273.15 / (273.15 - 263.15) ≈ 27.3The actual values will be 2-3 times lower, but this provides a guideline for comparing models.
FAQ: Frequently asked questions about the efficiency of heat engines
Can the efficiency of a heat engine exceed 100%?
No, for heat engines engines (machines that convert heat into work) efficiency is always less than 100%. However, thermal pumps and refrigerators can have an efficiency coefficient (COP) greater than 100%, since they do not produce energy, but transfer heat at a lower cost.
Why is the temperature difference used in the Carnot formula?
The difference (T₁ - T₂) reflects maximum possible workwhich can be obtained by transferring heat from the heater to the refrigerator. The greater the temperature difference, the more energy can be converted into work (but never all).
How does the temperature of the refrigerator affect the efficiency?
The lower refrigerator temperature (T₂), the higher the efficiency, since the proportion of heat transferred to the refrigerator decreases. However, in practice, excessive cooling requires additional energy (for example, in cryogenic systems).
Can the Carnot formula be used for real engines?
The Carnot formula gives theoretical maximum, but real engines operate on different cycles (for example Otto for internal combustion engines or Rankine for steam turbines). Their efficiency is calculated taking into account the specifics of the processes, but always remains below the Karnoff limit.
What heater temperature is considered optimal for maximum efficiency?
The optimal temperature depends on technical limitations. For example, in steam turbines T₁ it is limited by the strength of materials (usually 500–600°C). In ICE the combustion temperature reaches 2000°C, but the efficiency is limited by mechanical losses.