Describes how much heat energy is transferred during the whole process. kJ and kWh are not different quantities, but different units of the same energy. The kilojoule (kJ) is convenient in thermodynamic calculations because specific heat capacities are often expressed in kilojoules. The kilowatt-hour (kWh) is practical when energy is linked to equipment power, operating time or electricity consumption. 1 kWh = 3,600 kJ.
Describes how fast energy is transferred, i.e. how much energy is transferred in a given time. 1 kW = 1 kJ/s. For example, an ideal 2 kW heater transfers 2 kJ of energy every second. The same amount of energy can therefore be transferred slowly with low power or quickly with high power. Power alone does not tell the total amount of energy – time is also needed. For example, 2 kW × 3 h = 6 kWh.
Kelvin is the absolute temperature scale used in thermodynamics. Its zero point is absolute zero: 0 K = −273.15 °C. Kelvin is written without a degree sign: 273.15 K, not °K. For temperature differences, a change of 1 K is equal in size to a change of 1 °C. The conversion is T(K) = t(°C) + 273.15. Absolute temperature is required in many thermodynamic calculations.
Why do different materials warm up differently?
Specific heat capacity – what does it mean?
Specific heat capacity tells how much energy is needed to raise the temperature of one kilogram of a material by one degree. The higher the value, the more energy the material can absorb before its temperature rises.
Think of it this way
If 1 kg of water and 1 kg of copper receive the same amount of heat energy, the temperature of the copper rises much more. Water needs about 4.19 kJ to heat one kilogram by one degree, while copper needs only about 0.39 kJ.
For a temperature difference, K = kelvin: a change of one kelvin is the same size as a change of one degree Celsius. Therefore kJ/(kg·K) directly expresses the energy needed per kilogram and per degree of temperature change.
| Material | c, approx. kJ/(kg·K) | What does the value indicate? |
|---|---|---|
| Water | 4,19 | Stores a lot of heat |
| Potato | approx. 3.6 | High water content → heats fairly slowly |
| Air | 1,01 | Per unit mass; air has low density |
| Aluminium | 0,90 | Requires much less energy than water |
| Iron / steel | approx. 0.45–0.50 | Temperature changes more readily than water |
| Copper | 0,39 | Low specific heat capacity |
| Concrete | approx. 0.88 | Structures can store much heat because of their large mass |
| Ice | approx. 2.1 | Below 0 °C, before melting |
The values are approximate values suitable for teaching. Specific heat capacity can vary slightly with temperature and material composition.
Three things determine the required heat energy: the amount of material m, its specific heat capacity c and the temperature change ΔT. Therefore the equation is Q = m · c · ΔT.
Energy to heat a material
Q = m · c · ΔT
10 kg of water, temperature rises by 20 K: Q = 10 × 4.19 × 20 = 838 kJ
Energy from power and time
E = P · t
A 2 kW heater runs for 3 hours: E = 2 kW × 3 h = 6 kWh
How long does heating take?
t = E / P
6 kWh of energy, power 3 kW: t = 6 / 3 = 2 h
The same amount of energy can be transferred quickly with high power or slowly with low power.
Water stores a lot of heat
Heating water
The specific heat capacity of water is about 4.19 kJ/(kg·K). This means that heating one kilogram of water by one degree requires 4.19 kJ of energy.
Before:10 kg water, +10 °C
After: same water,
Try heating water
There are 10 kg of water and the starting temperature is always +10 °C. Change the final water temperature.
m = 10 kgAlkutila = +10 °Cc = 4,19 kJ/(kg·K)
Temperature difference ΔT50 K
Heat energy2 095 kJ
Same energy0.58 kWh
Time at 2 kW17.5 min
As the final temperature rises, the temperature difference increases and more energy is required. The calculation does not include heating the vessel or heat losses to the surroundings.
Try it yourself
How do material, amount and power affect heating?
Choose a material and change the amount, starting and final temperatures, and heating power. You will immediately see how much energy is required and how long heating theoretically takes.
c ≈ 4.19 kJ/(kg·K)
1 kg1000 kg
0,5 kW20 kW
Selected materialWater
Temperature difference ΔT50 K
Required heat energy
0.58 kWh
2,095 kJ
Theoretical heating time
17 min 28 s
Q = 10 kg · 4.19 kJ/(kg·K) · 50 K = 2,095 kJ
Water requires a large amount of energy to change its temperature because its specific heat capacity is high.
The calculation uses the approximate specific heat capacity from the table and assumes that all heating power is transferred to the material. Heat losses, heating of the container and phase changes are not included.
Flowing air
Heating and cooling power for air
For air, a volume-based heat capacity is used here so that every calculation step and unit remains visible.
Volumetric heat capacity of air ≈ 1.2 kJ/(m³·K)
Changing the temperature of one cubic metre of air by one degree requires about 1.2 kJ of energy. This is a practical approximation for ordinary conditions.
Try cooling power
Airflow is 1.0 m³/s and the air leaving the coil is +15 °C. Change the incoming air conditions.
V̇ = 1,0 m³/sLeaving air = +15 °C
Sensible capacity18,1 kW
Latent capacity8,5 kW
Total capacity26,6 kW
Condensate11,5 kg/h
The air cools at the coil and moisture condenses from it. Total capacity is therefore higher than sensible capacity alone.
Example 3
500 m³/h of air is heated from −10 to +20 °C
5,0 kW
V̇ = 500 m³/hΔT = 30 K
1,2 kJ/(m³·K) · 30 K = 36 kJ/m³
36 kJ/m³ · 500 m³/h = 18 000 kJ/h
18 000 kJ/h ÷ 3 600 s/h = 5,0 kJ/s = 5,0 kW
Example 4
1,000 m³/h of air is cooled from +30 to +18 °C
4,0 kW
V̇ = 1 000 m³/hΔT = 12 K
1,2 kJ/(m³·K) · 12 K = 14,4 kJ/m³
14,4 kJ/m³ · 1 000 m³/h = 14 400 kJ/h
14 400 kJ/h ÷ 3 600 s/h = 4,0 kJ/s = 4,0 kW
Heat moves in three ways
Conduction, convection and radiation
In refrigeration equipment these phenomena often occur at the same time, even though they can be examined one at a time.
1
Conduction
Heat transfers within a material or between materials in contact. For example, heat conducts through the wall and insulation of a cold room.
2
Convection
Heat is transferred by a moving liquid or gas. A fan increases heat transfer between air and an evaporator or condenser.
3
Radiation
Heat transfers as electromagnetic radiation without contact. The sun and warm surfaces radiate heat toward colder surfaces.