I have a basic system, let's say a cube of air of size $1m \times 1m \times 1m$. I place a metal heater next to it, for simplicity its size will be $0.1m \times 0.1m \times 0.1m$ (so the contact area would be $0.1m \times 0.1m$). How can we calculate the change of air's temperature?

Some important rules:

  • Heater's Work is a constant and known amount.
  • Air's volume is $1m \times 1m \times 1m$.
  • Contact area is $0.1m \times 0.1m$.
  • Heat doesn't escape nor come any other way than from the heater.
  • Heater is a metal, eg. alluminium. Air is an actual Earth air and not a perfect gas.

We can use the first law of thermodynamics to calculate heater's heat:

$$Q = ΔU + W, \quad \text{can we say} \quad \Delta U = 0?$$

For air temperature change we can use this formula:

$$\Delta t = Q/c_h m$$


  • $c_h$ - specific heat
  • $m$ - air mass

Saying we know $c_h$ and $m$ we now need $Q$. How can we calculate the $Q$ knowing $Q$ of heater?

PS. (or $W$ if we don't know $Q$). I'm sorry for my incomprehension and any mistakes, as I am a beginner.

  • $\begingroup$ "We can neglect the fact of convenction and focus on heat that is transferred to air as whole." This doesn't make any sense whatsoever. Heat transfer will be primarily through convection and radiation. $\endgroup$
    – Gert
    Commented Mar 17, 2019 at 22:45
  • $\begingroup$ Alright, I will cut it out if this doesn't make sense, thank you @Gert $\endgroup$
    – Elgirhath
    Commented Mar 17, 2019 at 22:46
  • $\begingroup$ Q is the power supplied by the heater times the time the heater is operating, neglecting the heat required to raise the temperature of the heater metal. $\endgroup$ Commented Mar 17, 2019 at 23:15
  • $\begingroup$ @ChesterMiller right, but since P = W/t, isn't P*t = W <=> Q = W? $\endgroup$
    – Elgirhath
    Commented Mar 17, 2019 at 23:18
  • $\begingroup$ Sure, if you neglect the change in internal energy of the heater metal. $\endgroup$ Commented Mar 17, 2019 at 23:51

1 Answer 1


A first approximation can be obtained by means Lumped Systems Analysis, with the following assumptions:

  • heater operates at constant temperature $T_H$
  • air temperature $T(t)$ increases over time but is uniform over the $1\ \mathrm{m} \times 1\ \mathrm{m} \times 1\ \mathrm{m}$ domain. There are no temperature gradients in the domain
  • heat transfer from heater to air proceeds via convection and radiative transfer only

The convective heat ($Q_C$) flow from the heating element is given by Newton's cooling law:


where $h$ is the (convective) heat transfer coefficient, $A$ the surface area of the heater and $T(t)$ the air's temperature (as a function of time).

The radiative heat ($Q_R$) flow from the heating element is given by Stefan-Boltzmann Law:

$$\frac{\mathrm{d}Q_R}{\mathrm{d}t}=A\sigma\epsilon[T_H^4-T(t)^4]\tag{2}$$ The total heat ($Q$) flowing from the heating element is obtained by combining $(1)$ and $(2)$:




where $c_v$ is the air's heat capacity (at constant volume) and $m$ is the total mass of air.

With $(4)$ into $(3)$:


$(5)$ is a first order differential equation (DE), which is separable in the variables $T(t)$ and $t$.

If $T_0$ is the initial temperature of the air (at $t=0$) then integrate the DE between $(0,T_0)$ and $(t,T(t))$ to get the temperature evolution in time, $T(t)$.

  • $\begingroup$ This is great! What emmisivity would be proper for eg. alluminium heater, painted white? This article you attached states that alluminium emmisivity is 0.03 and paint's emmisivity is 0.9 (That seems like a huge difference). I'm trying to approximate that value for just a regular home radiator. $\endgroup$
    – Elgirhath
    Commented Mar 18, 2019 at 10:43
  • $\begingroup$ Your best bet is black, matt paint. Its emmisivity is close to $1$. Thanks for the upvote! $\endgroup$
    – Gert
    Commented Mar 18, 2019 at 13:59

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