How to correctly calculate the amount of heat given to a refrigerator: theory and practice

When it comes to the operation of a refrigerator, many people only imagine cooling food. However, from the point of view of thermodynamics, this process is much more complicated: the refrigerator does not just “take” heat, but moves it from the internal chamber into the environment. To understand exactly how much heat is given off to a refrigeration unit, you need to understand the basics of heat transfer, take into account the system parameters and apply the correct formulas.

In this article we will look in detail at how to calculate the amount of heat given off to a refrigerator, taking into account different scenarios: from everyday tasks (for example, defrosting) to engineering calculations. You will learn what physical laws underlie the process, how to use the heat balance equation, and where to find the necessary input data. And for those who prefer practice to theory, we will provide real examples of calculations with explanations.

Important: if you plan to use this knowledge to repair or modify a refrigerator, remember that incorrect actions can lead to violation of the tightness of the system or failure of the compressor. All theoretical calculations are applicable only if the equipment is in good working order.

Basic concepts: what is the “amount of heat” in the context of a refrigerator

In thermodynamics quantity of heat (Q) is the energy transferred from one body to another as a result of temperature differences. In the refrigerator, this process occurs in several zones:

• 🔹 Evaporator (internal radiator): here the refrigerant takes heat from the chamber, cooling the air and food.
• 🔹 Condenser (rear or lower grill): here the heat from refrigerant is released into the environment.
• 🔹 Compressor: heats up when the refrigerant is compressed, also releasing heat.

When they talk about “heat given to the refrigerator,” they usually mean total amount of energy, which the system must remove from the internal volume within a certain period of time. This value depends on:

• 📏 Volume of the refrigerator/freezer compartment.
• 🌡️ Temperature difference between inside and outside.
• ⚙️ Compressor power and type refrigerant.
• ⏱️ Operating time (on/off cycle).

For household calculations, a simplified model is often used, where the refrigerator is considered as heat pump. In this case, the amount of heat released can be expressed through efficiency coefficient (COP) and the consumed electrical power.

Physical laws underlying the calculations

To correctly calculate the amount of heat, it is necessary rely on two key principles:

1. First law of thermodynamics (law of conservation of energy):

   Energy does not appear or disappear, but only passes from one form to another. In the refrigerator, electrical energy is converted into heat, which is then distributed between the internal chamber and the environment.

2. Heat balance equation:

   The sum of the heat given and received by the system is zero. For a refrigerator, this means that the heat taken from the chamber (Qcold) is equal to the heat released into the environment (Qambient) plus the work of the compressor (A).

Mathematically, this is expressed formula:

Qambient = Qcold + A

where:

• Qambient — heat given off to the environment (the desired value).
• Qcold — heat, taken from the fridge compartment.
• A —the work done by the compressor (can be expressed in terms of power consumption).

For real refrigerators, the efficiency coefficient (COP) usually lies in the range 2–6. This means that per 1 kW of consumed electricity, the refrigerator transfers 2–6 kW heat.

Practical formulas for calculating heat

Depending on the available data, the amount of heat can be calculated in several ways. Let's consider the three most common approaches.

1. Through the compressor power and operating time

If known compressor power consumption (P, W) and its operating time (t, s), then the amount of heat given to the environment can be found by the formula:

Qokr = P × t × COP

Example: Compressor capacity 150 W worked 10 minutes (600 s) s COP = 4. Then:

Qokr = 150 × 600 × 4 = 360,000 J = 360 kJ

2. Through temperature change and heat capacity

If you need to calculate how much heat was given to the refrigerator when cooling a specific object (for example, water or product), use the formula:

Q = m × c × ΔT

where:

• m — mass of the object (kg).
• c — specific heat (J/(kg K)). For water c ≈ 4186 J/(kg K).
• ΔT —temperature difference before and after cooling (K).

Example: Cooling 1 kg of water s 20°C up to 5°C:

Q = 1 × 4186 × (20 - 5) = 62,790 J ≈ 62.8 kJ

3. Through heat flow and surface area

For engineering calculations (for example, when designing cooling systems), the heat transfer formula is used:

Q = k × S × ΔT × t

where:

• k — heat transfer coefficient (W/(m² K)).
• S — heat transfer surface area (m²).
• ΔT — temperature difference between the chamber and the environment (K).
• t — time (s).

This method requires knowledge of the technical characteristics of the refrigerator, which are usually indicated in the product passport.

Step-by-step instructions: how to calculate the heat for a household refrigerator

Suppose you want to find out how much heat your refrigerator gives off Atlant XM 4021-000 in one operating cycle. Follow this algorithm:

Take compressor power readings from the nameplate|Measure the operating time of the compressor (with a stopwatch or timer)|Check the COP coefficient (from the documentation or take the average value of 3.5)|Determine the temperature difference inside and outside (thermometer)

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Step 1. Find the compressor power.

Usually it is indicated on nameplate (metal plate) on the back wall of the refrigerator. For example, for Atlant XM 4021-000 the compressor power is 120 W.

Step 2. Measure the operating time of the compressor.

Turn on the stopwatch when the compressor starts and stop when it turns off. Let's say the cycle took 8 minutes (480 s).

Step 3. Determine COP.

If there is no data in the documentation, use the average value 3.5 for household refrigerators.

Step 4. Substitute the data into the formula.

Qokr = 120 × 480 × 3.5 = 201,600 J ≈ 201.6 kJ

This means that in one cycle the refrigerator gave up the environment is approximately 201.6 kJ heat.

Step 5 (optional). Compare with heat gain.

If you know how much heat entered the chamber (for example, from warm products placed there), you can check the efficiency of the system. For example, if you placed 2 kg of water at 25°Cin the refrigerator, and it cooled to 4°C, then:

Qinflow = 2 × 4186 × (25 - 4) ≈ 158,872 J ≈ 158.9 kJ

Comparing Qambient and Qinflow, you can evaluate how efficiently the refrigerator operates.

Table: Specific heat capacity of common products

For When calculating the amount of heat transferred to the refrigerator when cooling products, data on their specific heat capacity is required. Below is a table for the most commonly stored products:

Use this data for accurate calculations. For example, if you cool 1.5 kg of beef from 18°C to 2°C, then:

Q = 1.5 × 3350 × (18 - 2) ≈ 75,050 J ≈ 75 kJ

Typical errors in calculations and how to avoid them

Even when using the correct formulas it is easy to make errors that will distort the result. Here are the most common of them:

• ❌ Ignoring phase transitions:

  When freezing water or defrosting ice, it is necessary to take into account latent heat of melting/crystallizationFor example, for turning 1 kg water into ice. when 0°C it is required to remove 334 kJ heat - this is 5 times more than when cooling the same water by 20°C!

• ❌ Wrong choice of COP:

  The efficiency coefficient depends on the type of refrigerator For absorption models (for example, operating on gas) COP usually lower (0.5–1.2) than for compression (2–6).

• ❌ Neglect of heat inflows:

  Heat enters the chamber not only from the products, but also through the walls, when the door is opened, from lighting. In engineering calculations, this is taken into account as Qinflow.

To minimize errors:

1. Always check real conditions (ambient temperature, humidity, chamber load).
2. For accurate calculations, use data from technical data sheet refrigerator.
3. If phase transitions are possible (for example, frost formation), add the corresponding terms to the formula.

Examples of calculations for different scenarios

Let's consider three real situations in which it may be necessary to calculate the amount of heat.

Scenario 1: Cooling drinks before a party

Task: Placed in the refrigerator Samsung RB-30 J3200EF 10 bottles water on 0.5 l each (total 5 kg) with temperature 25°C. After how long will they cool down to 4°Cif the compressor power is 180 Wand COP = 4?

Solution:

1. Calculate the amount of heat that needs to be removed:

   Q = 5 × 4186 × (25 - 4) ≈ 439 530 J ≈ 440 kJ

2. Let's find the cooling power:

   Pcool = Pcompr × COP = 180 × 4 = 720 W

3. Let's determine the time:

   t = Q / Pcool = 440,000 / 720 ≈ 611 s ≈ 10 minutes

Answer: Drinks will cool in approximately 10–12 minutes (taking into account heat loss).

Scenario 2: Defrosting the freezer chambers

Task: In the freezer compartment Indesit DF 4180 W accumulated 1.2 kg ice at -18°CHow much heat will be required to melt the ice and heat the water to 20°C?

Solution:

The process consists of two stages:

1. Ice melting:

   Q1 = m × λ = 1.2 × 334,000 ≈ 400,800 J

   (where λ is the specific heat of melting of ice).

2. Water heating:

   Q2 = 1.2 × 4186 × (20 - 0) ≈ 100 464 J

Total:

Qtotal = Q1 + Q2 ≈ 501,264 J ≈ 501 kJ

Scenario 3: Assessment of heat release from the refrigerator to the room

Task: The refrigerator LG GA-B489 YDQZ operates in the 16 hours a day power mode 150 W. How much heat does it release into the room per month if COP = 3.8?

Solution:

1. Daily compressor operation:

   Qday = 150 × (16 × 3600) × 3.8 ≈ 33,177,600 J ≈ 33,178 kJ

2. Monthly heat release:

   Qmonth = 33,178 × 30 ≈ 1,000,000 kJ ≈ 278 kW h

Answer: In a month, the refrigerator will release about 278 kWh heat into the room - this is comparable to the operation of a heater with a power of 150 W.

When heat calculations are critical

Knowing the amount of heat given off by the refrigerator is necessary not only for educational purposes. There are several practical situations where this data becomes key:

• 🔧 Repair and modernization:

  When replacing a compressor or refrigerant, you need to make sure that the new part can cope with the thermal load. For example, if you install a more powerful compressor, but not If you change the capacitor, this can lead to overheating.

• ⚡ Optimizing energy consumption:

  Analyzing heat losses helps reduce energy consumption. For example, if it turns out that 30% heat enters through the door seal, replacing it can be considered. justified.

• ❄️ Design of cooling systems:

  For industrial refrigeration chambers, heat calculation is mandatory. An error in the choice of equipment can lead to cooling or excessive load to the compressor.

• 🏠 Indoor microclimate:

  If several refrigerators are running in a small kitchen, their total heat release can affect the temperature. This is important for ventilation and air conditioning systems.

In industry, specialized software is used for such calculations (for example, CoolProp or Refprop), but for domestic purposes the formulas given above are sufficient.

FAQ: Frequently asked questions about calculating the heat of a refrigerator

Qinflow ≈ 0.7 × V × ΔT × t

Q ≈ 0.7 × 0.2 × 20 × 30 ≈ 84 kJ

COP ≈ (Annual consumption, kWh) / (Chamber volume, l × 10)

COP ≈ 250 / (200 × 10) = 1.25