Calcylator
Room heating energy

Room heating energy:
how much heat a temperature rise takes

The heat needed to warm anything is its mass, its specific heat and the temperature rise multiplied together, and air and water differ enormously.

Calcylator Editorial Team

Updated · 4 min read

The three-factor sum

Warming a substance takes heat in proportion to how much of it there is, how much it resists warming and how far the temperature has to go. Those three ideas are mass, specific heat and temperature change, and multiplying them gives the energy.

Heat required =Q = m × c × ΔT
Q:
energy, in kJ
m:
mass, in kg
c:
specific heat, in kJ/(kg·K)
ΔT:
temperature rise, in K or °C
A temperature difference in °C has the same size as in K. 3,600 kJ equals 1 kWh.
  • Mass, m

    100 kg of water

  • Specific heat, c

    4.18 kJ/(kg·K)

  • Temperature rise, ΔT

    10 °C

Heat required

4,180 kJ

100 × 4.18 × 10 = 4,180 kJ, which is 4,180 ÷ 3,600 = 1.16 kWh.

Water has one of the highest specific heats of common materials, which is why it stores and carries heat so well in tanks, radiators and underfloor systems.

Heating the air in a room

To warm the air in a room, first convert the volume into mass using the density of air, around 1.2 kg/m³ at typical room conditions. The specific heat of air is roughly 1.005 kJ/(kg·K), a quarter of that of water.

  • Room volume

    150 m³ (10 × 5 × 3 m)

  • Air density

    1.2 kg/m³

  • Specific heat

    1.005 kJ/(kg·K)

  • Rise

    10 °C

Heat required

about 1,809 kJ

Mass = 150 × 1.2 = 180 kg; 180 × 1.005 × 10 = 1,809 kJ = 0.50 kWh.

That is surprisingly little. A 2 kW heater, 2 kJ per second, would take 1,809 ÷ 2 = about 905 seconds, a little over 15 minutes, to supply it with no losses at all.

The air in a room is only a small part of what needs to be warmed, and a real room takes far longer, which the next section explains.

Why real rooms take longer

Walls, floors, furniture and ceilings hold far more heat than the air, and they absorb a large share of what the heater supplies. A concrete floor and brick walls may hold hundreds of times the heat capacity of the air. Meanwhile heat leaks out through walls, windows and ventilation while you warm the room.

What soaks up and loses the heat
Item in the roomTypical role in the heat balance
AirSmall mass, warms quickly, leaks out with ventilation
Walls and floorsVery large heat store, slow to warm, give the room its feel
Furniture and contentsModerate store, takes heat gradually
Windows and gapsMain routes for heat to escape

For that reason, the m × c × ΔT figure for the air is a minimum, and a practical estimate uses heat-loss figures and a heater rated against the losses instead.

Specific heat of common materials

The value of c differs a great deal between materials, which explains why some things warm up quickly and others hold heat for hours. Typical figures are listed below; use data for your own material where precision matters, because composition changes the number.

Illustrative values for the same mass and rise
MaterialSpecific heat, kJ/(kg·K)Energy for 100 kg over 10 °C
Water4.184,180 kJ
Air1.0051,005 kJ
Concrete (typical)0.88880 kJ
Brick (typical)0.84840 kJ
Steel0.49490 kJ

Per kilogram, water takes more than four times the energy of air for the same rise. Per cubic metre the difference is far larger, because water is about 800 times denser than air, and that is why a small tank can carry as much heat as an enormous volume of air.

From kJ to kWh and running cost

Electricity is billed in kilowatt-hours, so convert kilojoules by dividing by 3,600. The 1,809 kJ for the air is 0.50 kWh, and the 4,180 kJ for the water is 1.16 kWh. Multiply by your unit price to cost it.

  • Energy

    4,180 kJ

  • Heater

    2 kW

  • Efficiency

    100% (electric resistance)

Run time

about 35 min

4,180 ÷ 2 = 2,090 s = 34.8 minutes at 2 kW, and 1.16 kWh of electricity.

An appliance cost tool applies a tariff to the kWh figure, and an energy tool converts watts and hours into kWh, so they are practical for the final step. A heat pump or a heat-recovery system changes the electricity needed per unit of heat, which is why coefficient of performance matters.

Choosing a heater for the job

The power of the heater sets how fast the energy arrives, and the losses set how much power is needed just to hold the temperature. A practical selection therefore needs two numbers: a steady-state loss at the design temperature difference, and a warm-up margin to bring the room up in the time you accept.

  1. Estimate the steady loss in watts through walls, roof, floor, glazing and ventilation.
  2. Add the warm-up energy for the air and for the fabric, divided by the time you allow.
  3. Choose a heater rated at or above the sum, rounding up to a standard size.
  4. Check the electrical supply and the cost of running it for the hours you will use it.

A quick check with the sum above shows whether the heater is in the right order of magnitude, but the final selection should follow a heat-loss calculation for the actual building and a recognised design method.

Assumptions worth stating

The formula assumes a constant specific heat, no change of phase and no heat loss. Over a narrow range of temperature these are fair for water and air. Moisture in the air, humidification or boiling would need extra terms.

  • Specific heat values are approximate and shift slightly with temperature.
  • Air density depends on temperature, altitude and humidity.
  • The sum gives energy delivered to the material, not the energy bought, which depends on the efficiency of the heater.
  • Heat loss over the heating period is additional and often larger than the sum itself.

Slips to watch for

Mistakes in heating sums tend to come from units and from forgetting what has been left out. Kilojoules and kilowatt-hours differ by a factor of 3,600, and degrees Celsius and kelvin share the same step size, so confusing them in a difference does no harm, but using kelvin values in place of a difference does.

Another slip is to treat the air-only figure as the heater's job. The result of about 1,809 kJ for a 150 m³ room looks tiny, yet it ignores the thermal mass of the building and the continuing losses. If a heater of the apparent size struggles in practice, the explanation is almost always the fabric and the leakage rather than a fault in the heater.

Common questions

How do I calculate the energy needed to heat something?

Multiply its mass in kg by its specific heat in kJ/(kg·K) and by the temperature rise. For 100 kg of water with a 10 °C rise, 100 × 4.18 × 10 gives 4,180 kJ, or 1.16 kWh after dividing by 3,600.

How much energy does it take to heat the air in a room?

Convert volume to mass with air density near 1.2 kg/m³, then use m × 1.005 × ΔT. A 150 m³ room is 180 kg; a 10 °C rise needs about 1,809 kJ, or 0.5 kWh, ignoring all losses and the heat taken by the walls.

Why does a room take so long to heat if the air needs so little?

The walls, floor and furniture hold much more heat than the air, and heat leaks out through windows and gaps. The heater has to warm all of that as well, so the real time is many times the air-only figure.

How do I convert kJ to kWh?

Divide the number of kilojoules by 3,600, because 1 kWh equals 3,600 kJ. So 4,180 kJ is 1.16 kWh. Multiply by your electricity price to find the running cost, and account for the heater's efficiency or COP.

Was this guide helpful?

Continue reading

View all blogs