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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.

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  • $\begingroup$ The main mechanism of heat transfer in this case is thermal convection, so the answer depends on the geometry of the object. $\endgroup$ – Alex Trounev Feb 2 at 23:14
  • $\begingroup$ The object is a 4x4cm plate with a negligible height. $\endgroup$ – V.Iron Feb 2 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 at 23:25
  • $\begingroup$ Is the plate hanging or lying on the table? $\endgroup$ – Alex Trounev Feb 2 at 23:26
  • $\begingroup$ Laying - only one side is exposed $\endgroup$ – V.Iron Feb 2 at 23:26
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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:

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

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

$$dQ=mc_pdT$$

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 $T_0$.


Edit:

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:

$$\frac{dQ}{dt}=h_1A_1(T_0-T)+h_2A_2(T_0-T)=(h_1A_1+h_2A_2)(T_0-T)$$

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

$$\alpha=\frac{h_1A_1+h_2A_2}{mc_p}$$

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  • $\begingroup$ 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 at 23:34
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    $\begingroup$ $h$ 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 at 23:37
  • $\begingroup$ Thanks for the upvote. An edit has also been made. $\endgroup$ – Gert Feb 3 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 at 20:59
  • $\begingroup$ @JMac: well, you need both: a credible value for $h$ and a workable model. $\endgroup$ – Gert Feb 5 at 15:19

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