I have a object with a surface area of about $16 \, \mathrm{cm}^2 ,$ and I'm trying to calculate the rate at which it will be heating up if placed into a warmer room, so I can apply the same amount of cooling to make the temperature constantly lower than the room temperature.

For instance my room temperature is $20 \sideset{^{\circ}}{}{\mathrm{C}}$ and the object's temperature is $0 \sideset{^{\circ}}{}{\mathrm{C}} .$ If it helps, the object is made out of a ceramic material $\left(\text{Al}_2 \text{O}_3 \right) .$

I know it can be somehow calculated with Newton's law of cooling, but I can't seem to get past the heat transfer coefficient.

  • $\begingroup$ The main mechanism of heat transfer in this case is thermal convection, so the answer depends on the geometry of the object. $\endgroup$ Feb 2, 2019 at 23:14
  • $\begingroup$ The object is a 4x4cm plate with a negligible height. $\endgroup$
    – V.Iron
    Feb 2, 2019 at 23:19
  • $\begingroup$ I was trying to keep it general but to make it clear, my object is a peltier module and I'm trying to find out what heat load will it have in a room with constant temperatur so that I can figure out the deltaT of it's sides. However I can't crack, how much will the room heat the module. $\endgroup$
    – V.Iron
    Feb 2, 2019 at 23:25
  • $\begingroup$ Is the plate hanging or lying on the table? $\endgroup$ Feb 2, 2019 at 23:26
  • $\begingroup$ Laying - only one side is exposed $\endgroup$
    – V.Iron
    Feb 2, 2019 at 23:26

1 Answer 1


Assume the room's temperature to be $T_0$ and constant, the object's temperature $T$ (considered uniform), its surface area $A$, its mass $m$, its heat capacity $c_p$ and $h$ the heat transfer coefficient.

With Newton's Cooling/Heating Law we get:


For an infinitesimal increase in temparature of the object $dT$, then:


Combined we have:

$$\frac{mc_p dT}{dt}=hA(T_0-T)$$

$$\frac{dT}{T_0-T}=\alpha dt$$

Where $\alpha=\frac{hA}{mc_p}$.

Integrated we get, with $T_i$ the initial temperature of the object:

$$-\ln\frac{T_0-T(t)}{T_0-T_i}=\alpha t$$

So that:

$$T(t)=T_0-(T_0-T_i)e^{-\alpha t}$$

So the rate of change of $T$ isn't linear, it's exponential. 'On paper' it takes $t\to +\infty$ for the object's temperature to reach $T_0$.


The OP revealed in the comments that the object is laying on the floor. So it is partially exposed to the floor (through surface area, say $A_1$) and partially to the air in the room (through surface area, say $A_2$). If we assume floor and air to be at the same constant temperature $T_0$ then we have two heat flows:


The rest of the derivation is then the same as above, except:


  • $\begingroup$ I'm aware of Newton's cooling law but I don't know the heat transfer coefficient to plug into the equation $\endgroup$
    – V.Iron
    Feb 2, 2019 at 23:34
  • 1
    $\begingroup$ $h$ is usually determined experimentally. You can consult engineering pages for it. For alumina/air some values should be available. Google is your friend. $\endgroup$
    – Gert
    Feb 2, 2019 at 23:37
  • $\begingroup$ Thanks for the upvote. An edit has also been made. $\endgroup$
    – Gert
    Feb 3, 2019 at 16:34
  • $\begingroup$ @Gert The problem is generally finding an appropriate h to use. There are likely a huge number of values available, but the actual value depends on geometry and surroundings along with materials. $\endgroup$
    – JMac
    Feb 4, 2019 at 20:59
  • $\begingroup$ @JMac: well, you need both: a credible value for $h$ and a workable model. $\endgroup$
    – Gert
    Feb 5, 2019 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.