The question of what is the equivalent of coefficient of performance (efficiency) a heat engine with given parameters is often found in educational problems in thermodynamics and physics. However, to get the correct answer, it is not enough to simply substitute numbers into the formula. It is critical to understand that thermodynamic calculations require the use of an absolute temperature scale, rather than the familiar Celsius scale. In the case under consideration, when the temperature of the heater is 17°C and the refrigerator is 27°C, we are faced with a unique and physically paradoxical situation that requires a detailed explanation.
Before moving on to calculations, it is necessary to clearly define the roles of the participants in the process. In any heat engine, heater gives energy to the working fluid, and refrigerator receives the remaining heat. Logic dictates that heat spontaneously transfers from a hotter body to a colder one. If the problem statement says that the heater has a temperature of 17°C and the refrigerator has a temperature of 27°C, then this violates the basic principles of operation of a heat engine in a direct cycle. Let's figure out why this happens and how to correctly interpret such data.
First, let's conduct a primary analysis of the input data. Temperature is a measure of the average kinetic energy of particles. In classical thermodynamics, the Carnot cycle, which determines the maximum possible efficiency, operates only when the temperature of the heat source ($T_1$) is higher than the temperature of the sink ($T_2$). In our case, an inversion is observed: $t_1 = 17^\circ\text{C}$ and $t_2 = 27^\circ\text{C}$. This means that the so-called “heater” is colder than the “refrigerator”. This configuration is impossible for an engine that produces work, but it is quite possible for other devices, which we will talk about later.
⚠️ Attention: If in a physics problem the temperatures are reversed (the heater is colder than the refrigerator), then the classic heat engine will not work. To perform work under such conditions requires the expenditure of external energy, which is typical for refrigeration machines or heat pumps, and not for engines.
Converting temperatures to the absolute scale
The first and most important step in solving any thermodynamic problem is converting degrees Celsius to Kelvin. The conversion formula is simple: you need to add 273.15 to the value in degrees Celsius (in school problems it is often rounded to 273). The use of absolute temperature is necessary because the Celsius scale is relative, and zero on it does not mean the absence of thermal movement of molecules. Absolute zero is the point where thermal movement stops, and it is from this that efficiency calculations are made.
Let's perform the calculation for our values. The temperature of the heater $T_1$ will be $17 + 273 = 290$ K. The temperature of the refrigerator $T_2$ will be equal to $27 + 273 = 300$ K. The obtained values show that $T_1 < T_2$. The formula for the efficiency of an ideal heat engine (Carnot cycle) uses the ratio of the temperature difference to the heater temperature: $\eta = \frac{T_1 - T_2}{T_1}$. Substituting our numbers, we get a negative value, which physically means it is impossible to obtain work in such a cycle without external costs.
It is important to understand why negative efficiency or a value greater than one (if we were wrong in the formula) is impossible for the engine. Efficiency shows what fraction of the received heat is converted into useful work. If heat spontaneously flows from cold to hot (which is required for an engine to operate at these temperatures), it contradicts the second law of thermodynamics. Therefore, the correct answer in the context of the engine is that operation is impossible.
Analysis of the paradox: the heater is colder than the refrigerator
The situation described in the title, where the temperature of the heater is 17°C and the refrigerator is 27°C, is a classic “trap” for students. In a real heat engine, such as an internal combustion engine or steam turbine, the working fluid must expand by pushing a piston or blades. To expand, the gas must be heated. If the heat source (heater) has a temperature of 17°C, and the medium (refrigerator) has a temperature of 27°C, then the gas will not be able to receive heat from the heater, since heat is transferred only from hot to cold.
However, if we turn the problem around and consider the work heat pump or refrigeration machinethe picture changes. In these devices, the goal is not to obtain work, but to transfer heat from a less heated body to a more heated one. In our case, the machine could take heat from an environment with a temperature of 17°C and release it into an environment with a temperature of 27°C. But for this it will need external energy (electricity or mechanical work).
Thus, the answer to the question “what is the efficiency” depends on what exactly we consider. If we are talking about a heat engine, then its efficiency in this case is zero or the problem has no physical meaning in a direct cycle. If we consider the efficiency of heat transfer (although the coefficient of performance is used for refrigeration machines), then we are talking about a completely different device. The second law of thermodynamics says that a circular process is impossible, the only result of which would be the transfer of heat from a less heated body to a more heated one.
Why efficiency cannot be more than 1?
Efficiency greater than 1 would mean the creation of energy from nothing (perpetual motion machine of the first kind) or the complete conversion of heat into work without loss (perpetual motion machine of the second kind), which is prohibited by the laws of physics. The maximum theoretical efficiency is always less than 100%.
Carnot formula and theoretical limit
To calculate the maximum possible efficiency of any heat engine, the formula proposed by Sadi Carnot is used. It states that the efficiency of an ideal cycle depends only on the temperatures of the heater and refrigerator and does not depend on the design of the machine or the type of working fluid. The formula is as follows:
η = (T1 - T2) / T1Where $T_1$ is the absolute temperature of the heater, and $T_2$ is the absolute temperature of the refrigerator. As we have already found out, when substituting the values $T_1 = 290$ K and $T_2 = 300$ K, the numerator of the fraction becomes negative ($-10$). This is mathematical confirmation that the process will not go in the forward direction (the engine). The machine will not be able to do work based on such a temperature difference.
If in the problem conditions the temperatures were mixed up, and the heater had 27°C (300 K) and the refrigerator had 17°C (290 K), then the calculation would be as follows: $\eta = (300 - 290) / 300 = 10 / 300 \approx 0.033$. This would be only 3.3%. This low efficiency is explained by the small temperature difference. Real heat engines operate at much higher heater temperatures (hundreds of degrees) to achieve acceptable efficiency of 30-40%.
Practical application: refrigeration machines
Although the “heater 17°C, cooler 27°C” configuration is not possible for an engine, it ideally describes the operation of a household one refrigerator or air conditioner in a certain mode. Imagine that you need to cool a room (or chamber) where the temperature has dropped to 17°C, and the heat needs to be released into an atmosphere with a temperature of 27°C. In this case, the device works like a heat pump.
The efficiency of such devices is assessed not through efficiency, but through refrigeration coefficient ($\varepsilon$). It shows the ratio of the amount of heat taken from the cooled body to the work expended. The formula for an ideal Carnot cycle in refrigeration machine mode looks like this:
ε = T2 / (T1 - T2)Here $T_2$ is the temperature of the object being cooled (17°C or 290 K), and $T_1$ is the temperature of the medium where the heat is discharged (27°C or 300 K). Substituting the values, we get: $\varepsilon = 290 / (300 - 290) = 290 / 10 = $29. This means that for every unit of electricity expended (compressor work), we transfer 29 units of heat. This is a very high efficiency indicator, which explains why heat pumps are more economical than conventional electric heaters.
It is important to note that in real devices, due to friction, heat loss and imperfect refrigerants, the actual coefficient will be significantly lower than the theoretical one. However, the principle remains the same: the small temperature difference between the sources allows for high efficiency of heat transfer, as opposed to the generation of work.
☑️ Checking the operating conditions of the heat engine
Comparison of characteristics of heat engines
To better understand the difference between an engine and a refrigeration machine, as well as the effect of temperatures on their efficiency, consider the comparison table. It will help systematize knowledge about how systems behave under various temperature differences.
| Parameter | Heat engine | Refrigerator | Heat pump |
|---|---|---|---|
| Purpose work | Obtaining mechanical work | Cooling an object | Heating an object |
| Direction of heat flow | From hot to cold | From cold to hot (forced) | From cold to hot (forced) |
| Energy consumption | Heat from the heater | External work (electricity) | External work (electricity) |
| Basic formula (ideal) | $(T_1 - T_2) / T_1$ | $T_2 / (T_1 - T_2)$ | $T_1 / (T_1 - T_2)$ |
The table shows that with a small temperature difference ($T_1 - T_2$), the denominator in the formulas for refrigeration machines and heat pumps becomes small, which dramatically increases their efficiency. At the same time, for a heat engine, a small temperature difference leads to negligible efficiency. That is why for engines they strive to maximize the temperature of fuel combustion.
Typical errors when solving problems
When solving problems on the topic of efficiency, students often make a number of systematic errors that lead to an incorrect answer. The most common of these is the use of degrees Celsius instead of Kelvin. If you substitute 17 and 27 directly into the formula, the result will be radically different from the truth and will have no physical meaning. Always remember that thermodynamic potentials are related to absolute temperature.
The second mistake is the confusion between $T_1$ and $T_2$. In the Carnot formula, the denominator always contains the temperature of the heater ($T_1$), that is, the hotter body in the engine cycle. If you mix them up, you can get an efficiency value greater than 100%, which is an error signal. Efficiency cannot exceed one.
⚠️ Attention: When making calculations, always check the dimension of the quantities. If the answer yields a negative efficiency for the motor, this means that the process cannot proceed spontaneously, and the device must consume energy from the outside.
The third mistake is ignoring the conditions of the problem. If the condition says that the machine is thermal, but the temperatures are given in the order “fridge hotter than heater,” this is a test of attentiveness. The correct answer in this case should contain an analysis of the impossibility of the process, and not just a substitution of numbers. Physics requires a logical understanding of the phenomenon, and not blindly following an algorithm.
Conclusion and conclusions
Summarizing the consideration of the question of what the efficiency of a heat engine is equal to at a heater temperature of 17°C and a refrigerator of 27°C, we can draw an unambiguous conclusion: such a machine will not work in heat engine mode. The temperature difference is directed in the direction opposite to that required to obtain work. Heat cannot spontaneously flow from 17°C to 27°C, doing useful work.
However, if we consider this system as a refrigeration unit or heat pump, then it will function very efficiently, consuming external energy. The theoretical coefficient of refrigeration for such conditions is 29, which indicates high efficiency of heat transfer at a small temperature delta. Understanding this difference is key to a deep understanding of thermodynamics.
When solving such problems, always start by converting to Kelvin and analyzing the directions of heat flows. This will save you from gross mistakes and help you correctly interpret the physical meaning of the numbers obtained. Remember that the laws of physics are universal and do not depend on what numbers are given in the problem book.
Can the efficiency be 100%?
According to the second law of thermodynamics, the efficiency of a heat engine can never reach 100%. Part of the heat must be given to the refrigerator. Efficiency = 100% is possible only at absolute zero in the refrigerator, which is unattainable.
Why can’t you use degrees Celsius in the efficiency formula?
The Celsius scale is relative, its zero is chosen arbitrarily (the freezing point of water). Thermodynamic processes depend on the absolute energy of particles, which is zero only at absolute zero (-273.15°C). Using Celsius will violate proportionality and give an incorrect physical result.
What is the Carnot cycle?
The Carnot cycle is an ideal circular thermodynamic process consisting of two isotherms and two adiabants. It has the highest possible efficiency for any heat engines operating between the set temperatures of the heater and refrigerator. Real cycles are always less efficient.
Can efficiency be negative?
For a heat engine, negative calculated efficiency means that the engine does not operate in a direct cycle, but requires a supply of work (pump mode). Physically, “negative efficiency” as a property of a machine does not exist; it is only a mathematical signal about the wrong direction of the process for the selected mode.
How to increase the efficiency of a real engine?
To increase efficiency, it is necessary to maximize the temperature of the heater (using heat-resistant materials) and reduce the temperature of the refrigerator (by improving the cooling system). Reducing friction and heat loss to the environment also helps.