Determining the amount of heat transferred to a refrigerator is a fundamental problem in thermodynamics, directly related to the operating efficiency of real household appliances and industrial installations. Understanding this process allows engineers to optimize cooling cycles and gives students and enthusiasts greater insight into the physics that governs energy transfer. The calculations are based on the first law of thermodynamics, which connects work, internal energy and heat into a single system of equations.
In contrast to the intuitive idea of “cold” as the absence of heat, physics considers the cooling process as the active selection of energy from one body and transferring it to another. Refrigerator, be it a laboratory installation or a kitchen unit, works like a heat pump pumping heat energy from a less heated body to a more heated one, while expending external work. It is the amount of energy released into the environment (heater) that is the key parameter that we will learn to find.
For correct calculations, it is necessary to clearly understand the structure of the thermal cycle and distinguish between the concepts of supplied and rejected heat. Errors in signs or confusion between the operation of the compressor and the heat flow can lead to incorrect results, so in this article we will analyze the calculation algorithms step by step, relying on strict physical definitions and time-tested formulas.
The physical essence of the thermal cycle
Any refrigerator operates in a closed cycle in which the working fluid (refrigerant) periodically changes its state. Ideal gas or real freon, passing through various parts of the cycle, either absorbs or releases energy. The amount of heat transferred to the refrigerator (heater) is denoted as $Q_2$ or $Q_{hot}$. This is the energy that is “ejected” into the atmosphere through the radiator on the back wall of the device.
According to the first law of thermodynamics, energy does not appear out of nowhere and does not disappear without a trace. In a closed system of a refrigeration machine, the energy balance is as follows: the work done on the working fluid by external forces (compressor), plus the heat taken from the object being cooled, is equal to the heat given to the heater. Mathematically, this is expressed by the equality $A + Q_1 = Q_2$, where $A$ is work, $Q_1$ is heat received from a cold body.
It is important to understand that the process of heat transfer always occurs spontaneously from hot to cold. To “force” heat to go in the opposite direction - from the cold contents of the chamber to the warm room - it is necessary to expend energy. Efficiency factor The (efficiency) of a refrigeration machine, often called the coefficient of performance, shows the effectiveness of this process.
Basic formulas for calculating heat
To find the required value $Q_2$ (heat given to the heater), two main approaches are most often used, depending on the initial data of the problem. If you know the work of the compressor $A$ and the heat received from the cooled body $Q_1$, then the calculation comes down to simple summation. The formula looks like this:
Q_2 = Q_1 + AThis formula is a direct consequence of the law of conservation of energy. It is applicable to any refrigeration machines, regardless of whether they contain ideal or real gas, and what kind of refrigerant is used. The main condition is that the system must complete a full cycle and return to its original state.
The second approach is necessary when the temperatures of the heater ($T_2$) and refrigerator ($T_1$) are given, and one of the heats or work is also known. For an ideal refrigeration machine operating according to the Carnot cycle, the heat ratio is equal to the temperature ratio:
Q_1 / Q_2 = T_1 / T_2From here we can express the desired value: $Q_2 = Q_1 \cdot (T_2 / T_1)$. Please note that temperatures here must be converted to the absolute scale (Kelvins). Using degrees Celsius in this relationship will lead to a gross physical error, since the Celsius scale does not start from absolute zero.
Why are absolute temperatures used in Carnot's formulas?
Temperature in thermodynamics is a measure of the average kinetic energy of molecules. Absolute zero (-273.15°C) is the state where thermal movement ceases. The relationships of physical quantities are correct only when the zero of the scale corresponds to the complete absence of the measured property.
Algorithm for solving problems in thermodynamics
To successfully find the amount of heat given to the refrigerator, it is necessary to act systematically. Chaotic substitution of numbers into formulas often leads to errors, especially in problems with complex conditions or compound cycles. Following a clear algorithm minimizes risks.
The first step is always to analyze the condition. Write down all known quantities: temperatures, work, efficiency, volumes or pressures, if given. Determine what exactly you need to find. Often, in a condition, the problem can be formulated indirectly, for example, “by how much has the internal energy increased” or “how much work has been done by the gas.”
The second step is the conversion of units of measurement to the SI system. This is a critical stage. Pressure is often given in atmospheres or millimeters of mercury, volumes in liters, and temperatures in degrees Celsius. Conversion all data in Pascals, cubic meters and Kelvins is the key to the correct numerical answer.
☑️ Algorithm solving the problem
The third step is choosing a formula. If the cycle is ideal (Carnot), we use temperature ratios. If the cycle is real or specific values of work and heat are given, we use the law of conservation of energy. In some cases, it may be necessary to calculate the work through the area of the figure on the graph in P-V coordinates.
Calculation through efficiency and the Carnot cycle
The Carnot cycle is an ideal reversible process consisting of two isotherms and two adiabats. It has the highest possible efficiency for these temperature limits. Although such refrigerators do not exist in reality, this cycle serves as a standard for assessing the efficiency of real devices.
The refrigeration coefficient (refrigerator efficiency) for the Carnot cycle is determined by the formula $\epsilon = T_1 / (T_2 - T_1)$. Knowing this coefficient and the work $A$ expended, we can find the heat removed from the cold body: $Q_1 = \epsilon \cdot A$. After this, using the first law of thermodynamics, we easily find the desired value: $Q_2 = Q_1 + A$.
There is also a direct connection between efficiency and heat: $\epsilon = Q_1 / A = Q_1 / (Q_2 - Q_1)$. By transforming this expression, we can obtain a formula for $Q_2$ directly in terms of efficiency and $Q_1$: $Q_2 = Q_1 \cdot (1 + 1/\epsilon)$. This allows you to solve problems without the step of calculating the work, if the coefficient of refrigeration is known.
Practical examples of calculations
Consider a specific problem for fixing the material. Let an ideal refrigerator operate according to the Carnot cycle. The freezer (refrigerator) temperature is -18°C and the room (heater) temperature is +27°C. The compressor performed 20 kJ of work. You need to find the amount of heat given to the air in the room.
First, we convert the temperatures into Kelvins: $T_1 = -18 + 273 = 255$ K, $T_2 = 27 + 273 = 300$ K. Find the temperature ratio: $T_1 / T_2 = 255 / 300 = 0.85$. This means that $Q_1 = 0.85 \cdot Q_2$. We substitute into the energy balance equation: $A = Q_2 - Q_1 = Q_2 - 0.85 \cdot Q_2 = 0.15 \cdot Q_2$. Hence $Q_2 = A / 0.15 = 20 / 0.15 \approx 133.3$ kJ.
In another example, pressures and volumes can be given. If a gas is isothermally compressed, giving off heat, the amount of heat is equal to the work done by external forces during this compression. For an ideal gas on the isotherm $Q = A = \nu R T \ln(V_1/V_2)$. Here it is important to correctly determine the signs: during compression, the work is positive, the gas releases energy.
Analysis of graphs and process diagrams
Often the conditions of problems are given in graphical form, for example, in $P-V$ (pressure-volume) or $T-S$ (temperature-entropy) coordinates. The ability to “read” such graphs is necessary to determine the signs of heat and work. The area under the curve on the $P-V$ graph is numerically equal to the work done by the gas.
If the cycle goes clockwise, the gas does positive work (heat engine). If counterclockwise, work is done on the gas (refrigerator or heat pump). In a refrigeration cycle, sections where the volume decreases (compression) usually correspond to the release of heat, if the process is not adiabatic.
On the Carnot cycle diagram, isothermal compression corresponds to a horizontal section (at constant temperature) where the gas releases heat $Q_2$. Adiabatic compression occurs without heat transfer. Knowing the coordinates of the break points of the graph, you can calculate the parameters of the state of the gas at each point and then calculate the heat flows.
Typical errors and nuances of calculations
One of the most common errors is the confusion between the heat received by the gas and the heat given off by the gas. In formulas, the first law of thermodynamics is often written as $Q = \Delta U + A_{gas}$. Here $Q$ is all the heat received by the system. If the gas gives off heat, $Q$ will be negative. When calculating the “heat given off,” we usually look for the modulus of this value.
Another nuance concerns units of measurement. In engineering, kilocalories or British thermal units (BTU) are often used. Conversion to Joules is required: 1 kcal $\approx$ 4184 J. It is also worth remembering that the power of the refrigerator (in Watts) is work per unit of time. To find heat, you need to multiply the power by the operating time of the cycle.
| Parameter | Designation | Unit of measurement (SI) | Communication formula |
|---|---|---|---|
| Heat | Q | Joule (J) | Q = cmΔT |
| Work | A | Joule (J) | A = pΔV |
| Temperature | T | Kelvin (K) | T = t + 273.15 |
| Refrigeration coefficient | ε | Dimensionless | ε = Q1 / A |
Can the efficiency of a refrigerator be greater than 1?
Yes, the coefficient of performance (COP) often exceeds one. This does not violate the laws of physics, since we do not create energy, but only transfer heat. For 1 Joule of consumed electricity, you can pump 3-5 Joules of heat from the chamber.
Real devices versus ideal models
In real life, refrigerators do not operate according to the ideal Carnot cycle. There are heat losses through insulation, friction in the compressor, incomplete evaporation of the refrigerant and other imperfections. Therefore, the actual amount of heat given off by the condenser will always be greater than the calculated one for an ideal cycle at the same performance.
Efficiency real compressors is 60-80% of the theoretical one. This means that part of the engine’s work does not turn into pumping heat, but into useless heating of the compressor itself and the environment. This additional heating also goes into the final balance of $Q_2$, increasing the overall output to the room.
⚠️ Attention: When calculating for real household refrigerators, you cannot use the Carnot formula directly to determine power consumption unless the real COP of the device is known. The data in the device passport (energy consumption in kWh/year) is a more reliable source than theoretical calculations.
Modern inverter compressors allow you to adjust the engine rotation speed, bringing the system operation closer to the optimal mode in different conditions. This reduces energy losses and makes the actual heat balance closer to the calculated one, but the principle of conservation of energy $Q_2 = Q_1 + A$ remains unchanged.
What is the difference between heat and temperature in these calculations?
Temperature is a measure of the average kinetic energy of molecules (degrees), and heat is the amount of energy transferred from one body to another (Joules). In refrigerator formulas, temperatures determine the direction and potential of flow, and heat is a quantitative measure of transfer.
Why is the back wall of the refrigerator hot?
This is where the condenser is located, where the refrigerant releases the accumulated heat ($Q_2$) to the environment. According to the law of conservation of energy, both heat from the chamber and heat equivalent to the operation of the compressor enters there.
Is it possible to cool a room by opening the door of a working refrigerator?
No, the room will heat up. The refrigerator will release more heat into the room ($Q_2$) than it will take out of it ($Q_1$), due to the addition of heat from the operation of the electric motor ($A$). Total balance: $Q_{room} = Q_2 - Q_1 = A > 0$.
How does the temperature in the room affect energy consumption?
The higher the heater temperature ($T_2$), the greater the temperature difference $(T_2 - T_1)$ and the more work $A$ needs to be spent to pump the same amount of heat $Q_1$. Therefore, in summer, refrigerators consume more electricity.
⚠️ Attention: If you are conducting experiments to measure heat flow, make sure that the refrigerator is installed at least 10 cm from the wall. Violation of the heat exchange of the condenser can lead to overheating of the compressor and failure of the equipment.