In problems of thermodynamics and In physics, we often encounter a classical formulation of the question, where it is required to find the work done by a heat engine in one cycle. Usually the condition states that the amount of heat transferred to the refrigerator is 60 joules. To give an accurate answer, it is not enough to simply substitute a number into the formula; it is necessary to understand the physical essence of the processes occurring inside heat engine or, conversely, in a refrigeration unit.
A heat engine is a device that converts the internal energy of a fuel or other source into mechanical work. The key point here is not only the energy received, but also the inevitable losses. If the problem says that 60 J were given to the refrigerator, this means that part of the energy that did not go to useful action was transferred to the cooler. Without knowing the amount of heat received from the heater, or the coefficient of performance (COP), it is impossible to directly calculate the work, since the balance equation contains two unknown variables.
However, if we consider standard educational problems, there is often a certain context or additional conditions implied that allow us to find the desired value. For example, the efficiency of the machine or the temperature ratio can be specified. In this article we will analyze in detail how working fluid it behaves in various cycles, why energy cannot disappear without a trace and how to correctly apply the first law of thermodynamics to solve such problems.
Fundamental principles of operation of heat engines
Any A heat engine, be it an internal combustion engine or a compressor for a domestic refrigerator, operates on a cyclic principle. The working fluid (gas or steam) receives energy from the heater, part of which is converted into mechanical work, and the remainder is given to the refrigerator. If in the problem statement the figure 60 J appears as the transferred heat ($Q_2$), then this is only one of the components of the energy balance.
According to the first law of thermodynamics, energy does not arise from nowhere and does not disappear into nowhere. For a closed cycle, this means that the amount of heat received from the heater ($Q_1$) is equal to the sum of the work done ($A$) and the heat given to the refrigerator ($Q_2$). The formula is as follows:
Q₁ = A + Q₂From this equality it follows that the required work is equal to the difference between the received and released heat: $A = Q_1 - Q_2$. If $Q_2 = 60$ J, then to find $A$ we critically lack the value of $Q_1$. Without it, the problem has an infinite number of solutions. However, in physics there are idealized models, such as the Carnot cycle, which impose additional restrictions on the system.
⚠️ Attention: In real devices there are always losses due to friction and heat exchange with the environment, so the real efficiency is always lower than the theoretical one. When making calculations for household appliances (for example, refrigerators Indesit or Atlant), it is necessary to take into account that the passport data may differ from the calculated ideal values.
It is important to distinguish between the concepts of “heat engine” and “refrigeration machine”. In an engine, the goal is to produce work, and in a refrigerator, the goal is to extract heat from a cold body. But they have a common mathematical apparatus for describing cycles. If a machine transfers 60 J to the refrigerator, it performs the function of heat removal, which is typical for both types of devices, it’s just that the vector of the target action is different.
Mathematical apparatus: formulas and dependencies
To solve problems where heat transfer of 60 J appears, it is necessary to operate with the basic formulas of thermodynamics. Efficiency ($\eta$) shows what fraction of the received energy is converted into work. It is expressed through the ratio of work to expended heat or through the temperatures of the heater and refrigerator.
Let's consider the basic equations that will help you navigate the calculations:
- 🔹 Energy balance: $A = Q_1 - Q_2$, where $Q_2 = 60$ J.
- 🔹 Efficiency formula: $\eta = \frac{A}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}$.
- 🔹 Carnot cycle: $\eta = 1 - \frac{T_2}{T_1}$, where $T$ is the absolute temperature.
If the problem does not give the efficiency, but the temperatures are indicated, you can use the Carnot formula. For example, if the heater temperature is twice the temperature of the refrigerator, then the efficiency will be 50%. In this case, knowing that $Q_2 = 60$ J, one can find $Q_1$. Since $\frac{Q_2}{Q_1} = \frac{T_2}{T_1} = 0.5$, then $Q_1 = 120$ J. Therefore, the work $A = 120 - 60 = 60$ J.
Often in problems there is a situation where the ratio of heats or temperatures is given. If it is said that the efficiency of the machine is, for example, 40%, then the calculation will be carried out differently. We know that $1 - \frac{60}{Q_1} = 0.4$. Hence $\frac{60}{Q_1} = 0.6$, and $Q_1 = 100$ J. Then the work is $A = 100 - 60 = 40$ J. It can be seen that without the second parameter (efficiency or $Q_1$) the solution is impossible.
Analysis of typical tasks with given parameters
Let's consider specific examples, which are often found in school courses and exams. Let us assume that the heat engine receives 100 J of heat from the heater per cycle. We know that 60 J are given to the refrigerator. In this case, the calculation of the work is trivial: $A = 100 - 60 = 40 $ J. The efficiency of such a machine will be 40%.
Another common option: the machine operates according to the Carnot cycle, the temperature of the heater is 400 K, the temperature of the refrigerator is 300 K. 60 J are given to the refrigerator. Let's find the work. Temperature ratio $\frac{300}{400} = 0.75$. This means that $\frac{Q_2}{Q_1} = 0.75$. If $Q_2 = 60$, then $Q_1 = \frac{60}{0.75} = 80$ J. The work is equal to $80 - 60 = 20$ J.
There are problems where the work of a machine is used to drive another mechanism. If a heat engine transfers 60 J to the refrigerator and does 40 J of work, then the heat received from the heater is 100 J. This is a classic example demonstrating the law of conservation of energy. It is important to correctly identify which value is $Q_1$ and which is $Q_2$.
The table below shows examples of calculations for various conditions of the problem, where $Q_2$ (heat given to the refrigerator) is always equal to 60 J:
| Condition parameter | Parameter value | Found Q₁ (J) | Work A (J) | Efficiency (%) |
|---|---|---|---|---|
| Efficiency = 20% | η = 0.2 | 75 | 15 | 20 |
| Efficiency = 40% | η = 0.4 | 100 | 40 | 40 |
| Q₁ = 120 J | Q₁ = 120 | 120 | 60 | 50 |
| T₁ = 2T₂ | Carnot | 120 | 60 | 50 |
Analyzing the table, you can notice a direct relationship: the higher the efficiency of the machine, the less heat it transfers to the refrigerator with the same energy received, or the more work it does with a fixed heat transfer. In our case, with a fixed 60 J of output, an increase in efficiency requires a decrease in input energy $Q_1$, which may seem counterintuitive, but is mathematically correct for this formulation.
The influence of cycle type on the efficiency of the machine
The efficiency of converting heat into work strongly depends on what cycle the machine operates on. The Carnot cycle is ideal and provides the highest possible efficiency for given temperature limits. Real cycles, such as the Otto cycle (gasoline engines) or the Diesel cycle, are less efficient.
If the problem says that the car operates on the Carnot cycle and delivers 60 J, this gives us a powerful tool for calculations through temperatures. In real refrigerators, for example, cycles of compression and expansion of the refrigerant (freon) are used, which have their own characteristics. Here, 60 J of heat released can correspond to the work of the compressor, which is always greater than 60 J, due to losses.
Why can’t the efficiency be 100%?
The efficiency of a heat engine can never reach 100% according to the second law of thermodynamics. Part of the energy must be given to the refrigerator (the environment), otherwise the cycle will not close. Trying to create an engine with 100% efficiency is tantamount to creating a perpetual motion machine of the second kind, which is impossible.
When solving problems, it is important to pay attention to wording. The phrase “heat transfer to the refrigerator is 60 J” is unambiguous. But if it says “the machine gave 60 J”, you need to understand to whom - the refrigerator or did the work (although usually the work is “done” and the heat is “given away”).
Modern energy-saving devices are characterized by high efficiency values. If we're talking about an industrial turbine or a powerful compressor, the joule values may scale to kilojoules and megajoules, but the proportions remain the same. The principle of energy conservation is universal.
Practical application in household appliances
In household refrigerators, the principle of operation is the reverse of the engine: we spend work (electricity) to pump heat from the chamber to the room. However, the physics of the process is described by the same equations. If the refrigerator compressor delivers 60 J of heat to the radiator on the rear wall per cycle, this does not mean that it took 60 J from the inside.
The amount of heat taken from the products ($Q_{col}$) will be less than that released into the room ($Q_{hot}$), by the amount of work of the compressor ($A$). That is, $Q_{hot} = Q_{cold} + A$. If $Q_{hot} = 60$ J, then, say, 50 J were taken from inside, and 10 J is the work of electricity turned into heat.
- ❄️ Refrigerant: A substance circulating in the system that changes the state of aggregation.
- ⚡ Compressor: Device that performs work on gas compression.
- 🌡️ Evaporator: The unit where the refrigerant boils, taking heat from the chamber.
Understanding these processes helps save energy. The more efficient the heat exchange (cleaner radiator, tighter seals), the less work the compressor must do to transfer the same 60 J of heat to the environment. A dirty condenser increases the condensation temperature, which reduces the efficiency of the cycle.
⚠️ Attention: When servicing refrigeration equipment, do not try to change the refrigerant yourself or break the tightness of the circuit. This requires special equipment and licenses. Incorrect charging with freon will change the pressure in the system and can lead to compressor failure.
Common mistakes when solving problems
Students and applicants often confuse $Q_1$ and $Q_2$. Remember: $Q_1$ is what we “take” (from the heater), and $Q_2$ is what we “throw away” (to the refrigerator). If the problem says “60 J are given,” this is almost always $Q_2$.
Another mistake is using temperatures in degrees Celsius instead of Kelvin in efficiency formulas. Temperature in thermodynamic formulas is always absolute. $0^\circ C = 273 K$. If the problem is given $t_1 = 127^\circ C$ and $t_2 = 27^\circ C$, then $T_1 = 400 K$, $T_2 = 300 K$.
It is also important to monitor the dimension. If work is given in kilojoules and heat is in joules, you must convert everything to the same unit of measurement before substituting it into the formula. 1 kJ = 1000 J.
Final calculation and conclusions
Returning to the original question: “The amount of heat given by the heat engine per cycle to the refrigerator is 60 J. What is the work equal to?” Answer: The problem cannot be solved without additional data. We need to know either the amount of heat received from the heater ($Q_1$), or the efficiency ($\eta$), or the temperature ratio.
However, if we assume the most likely scenario for educational problems, where a connection with an efficiency of 50% is often sought (Carnot cycle with doubling temperatures), then the work can be equal to 60 J (at $Q_1=120$). If the efficiency is 25%, then the work will be 20 J (with $Q_1=80$). The variability of the answer emphasizes the importance of the complete conditions of the problem.
Thermodynamics is an exact science that does not tolerate incomplete data. Always check that all variables are known before starting calculations. Understanding the physical meaning of processes will help you avoid mechanical errors and better understand the structure of the world around you.
☑️ Checking the solution to the problem
Can the work be more than 60 J if 60 is also given J?
Yes, maybe. If $Q_2 = 60$ J, and work $A > 60$ J, this means that $Q_1 = A + Q_2 > 120$ J. The efficiency of such a machine will be more than 50%. This is quite realistic for efficient heat engines.
What is an ideal gas in the context of these problems?
An ideal gas is a model of gas in which the interaction between molecules, except for elastic collisions, is neglected. In most school problems, the working fluid is considered an ideal gas, which simplifies calculations, allowing the use of the Mendeleev-Clapeyron equation.
Does the answer depend on the type of gas (helium, nitrogen)?
To calculate work and efficiency using the Carnot cycle formulas or through the energy balance ($A = Q_1 - Q_2$), the type of gas is not important. The type of gas affects the specific values of heat capacities and behavior in adiabatic processes, but the overall energy balance remains unchanged.
Why is heat transferred into the room in a refrigerator?
The refrigerator does not create cold, it transfers heat. The heat that was inside the chamber plus the heat released during the operation of the compressor (electricity) all together must go somewhere. Therefore, the back wall of the refrigerator is always warm - it transfers the total heat ($Q_1$) to the room.