Solving problems in thermodynamics often causes difficulties for students and schoolchildren, especially when it comes to heat engines. In this material we will analyze in detail the classic problem where The working fluid of a heat engine with an efficiency of 15 delivers 340 J to the refrigerator per operating cycle., and we need to find useful work. Understanding these processes is critical for successfully passing exams and in-depth study of physics.
A heat engine is a device that converts the internal energy of a fuel or other source into mechanical work. The key elements here are the heater, the working fluid and the refrigerator. In our task, we will rely on the laws of conservation of energy and the determination of the coefficient of efficiency (efficiency) to find the desired value.
Physical essence of a heat engine
Any heat engine operates in a closed cycle. Working fluid (for example, gas or steam) receives an amount of heat $Q_1$ from the heater. Part of this energy is converted into mechanical work $A$, and the remaining part $Q_2$ is given to the refrigerator. It is this process of energy release that is described in the conditions of our problem.
A refrigerator in thermodynamics is any object or medium whose temperature is lower than the temperature of the working fluid at the moment of heat release. In real internal combustion engines this can be atmospheric air, and in steam turbines it can be a condenser with running water. It is important to understand that Efficiency 15% means that only 15% of the energy received is used for useful purposes.
The efficiency of the engine directly depends on the temperature difference between the heater and refrigerator. The higher this difference, the more work can be obtained from the same amount of heat. However, in real conditions, we are limited by materials and environmental standards.
⚠️ Attention: In problems, the amount of heat received from the heater ($Q_1$) and the amount of heat given to the refrigerator ($Q_2$) are often confused. Read the condition carefully: the phrase “gives to the refrigerator” always refers to $Q_2$.
Mathematical apparatus and formulas
To solve the problem, we need a basic set of formulas that describe the operation of a heat engine. The efficiency coefficient ($\eta$) is defined as the ratio of the work done $A$ to the amount of heat received $Q_1$.
The efficiency formula is as follows:
η = A / Q1
Also, the work is equal to the difference between the received and released heat:
A = Q1 - Q2
Substituting the second expression into the first, we obtain a universal formula for calculation:
η = (Q1 - Q2) / Q1 = 1 - (Q2 / Q1)Where:
- 🔥 Q1 — amount of heat received from the heater (J);
- ❄️ Q2 — amount of heat given to the refrigerator (J);
- ⚙️ A — perfect useful work (J);
- 📊 η — efficiency factor (in fractions of unity).
In our case The efficiency is 0.15 (15%), and the heat given off is 340 J. These data are sufficient to find all the unknown parameters of the cycle.
Step-by-step algorithm for solving the problem
Let's start the solution with data analysis. We are given: $\eta = 15\% = 0.15$ and $Q_2 = 340$ J. We need to find job $A$. The first step is to express the unknown $Q_1$ (heat received from the heater) from the efficiency formula.
Transform the formula $\eta = 1 - (Q_2 / Q_1)$ relative to $Q_1$:
Q2 / Q1 = 1 - η
Q1 = Q2 / (1 - η)
Substitute the numerical values:
Q1 = 340 / (1 - 0.15) = 340 / 0.85 = 400 J
Now, knowing $Q_1$, let's find a job. The work is equal to the heat difference:
A = Q1 - Q2 = 400 - 340 = 60 J
Thus, the required work is 60 Joules. This value shows how much energy the machine actually used to move the piston or rotate the shaft.
☑️ Algorithm for solving efficiency problems
Analysis of energy losses
Why, when receiving 400 J of energy, the machine did only 60 J of work? The answer lies in the nature of thermal processes. Most of the energy (340 J) is inevitably dissipated in the environment. This is a fundamental limitation described by the second law of thermodynamics.
Energy losses in heat engines occurs for several reasons:
- 🌡️ Incomplete combustion of fuel in the combustion chamber;
- 🔩 Friction of moving parts mechanism;
- 💨 Heat exchange with the outer walls of the cylinder;
- 🌪️ Heat loss with exhaust gases.
An efficiency of 15% is a rather low indicator for modern engines, but quite realistic for old steam engines or Stirling engines with suboptimal parameters. Modern diesel engines can achieve 40-50% efficiency, but they also give off more than half of the energy to heat.
⚠️ Attention: Do not try to create an engine with 100% efficiency. These are the laws of physics (perpetual motion machine of the second kind). Part of the heat obliged goes into the refrigerator.
Why can’t all the heat be used?
Complete conversion of heat into work is possible only if the refrigerator has a temperature of absolute zero (-273.15°C), which is unattainable in real conditions. Therefore, part of the energy is always lost.
Comparative table of cycle parameters
For clarity, we will summarize all the calculated and initial parameters into a single table. This will help you better see the relationship between the energy received, useful work and losses.
| Parameter | Designation | Value | Share in balance |
|---|---|---|---|
| Heat from heater | Q1 | 400 J | 100% (base) |
| Useful work | A | 60 J | 15% (efficiency) |
| Heat to the refrigerator | Q2 | 340 J | 85% (losses) |
| Efficiency coefficient | η | 0.15 | - |
As can be seen from the table, the lion's share of energy (85%) does not do useful work. Engineers are constantly striving to reduce this figure using turbocharging, heat recovery and improved materials.
Practical significance of calculations
Why do we need to be able to solve such problems? Calculation of the energy balance is necessary when designing any power plants. Whether it is nuclear power planta car engine or an airplane jet engine, these principles apply everywhere.
Knowing that the working fluid delivers 340 J, engineers can calculate the required power of the cooling system. If heat dissipation is insufficient, the engine will overheat and fail. Therefore, calculating $Q_2$ is not just a training exercise, but a vital part of engineering practice.
In addition, understanding the efficiency helps to assess the economic efficiency of equipment. A low-efficiency engine uses more fuel to do the same job, increasing operating costs and emissions.
⚠️ Attention: Always check the units when solving problems. If heat is given in kilojoules (kJ), and work is required in joules (J), do not forget to convert: 1 kJ = 1000 J.
Frequently asked questions (FAQ)
What is a working fluid in thermal machine?
A working fluid is a substance (gas, steam, liquid) that cyclically changes its state (heats, expands, cools, contracts), while performing mechanical work. In an internal combustion engine it is a mixture of gases; in a steam turbine it is water vapor.
Can the efficiency of a heat engine be greater than 1?
No, it cannot. This would contradict the law of conservation of energy. Efficiency is always less than unity (or 100%), since part of the energy is inevitably dissipated in the form of heat without being converted into work.
Why is it given in the problem that the heat given is given, and not the heat received?
This is a standard technique for complicating problems. By giving $Q_2$ (heat transferred) and efficiency, the compilers test the student’s ability to operate with the formula $\eta = 1 - Q_2/Q_1$, and not simply substitute numbers into $\eta = A/Q_1$.
What does the maximum theoretical efficiency depend on?
The maximum efficiency (Carnot cycle efficiency) depends only on the temperatures of the heater ($T_1$) and refrigerator ($T_2$) and is calculated by the formula: $\eta = 1 - T_2/T_1$. Temperatures must be expressed in Kelvin.