Heating power of a room affecting object of different temperature

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.

• The main mechanism of heat transfer in this case is thermal convection, so the answer depends on the geometry of the object. – Alex Trounev Feb 2 at 23:14
• The object is a 4x4cm plate with a negligible height. – V.Iron Feb 2 at 23:19
• 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. – V.Iron Feb 2 at 23:25
• Is the plate hanging or lying on the table? – Alex Trounev Feb 2 at 23:26
• Laying - only one side is exposed – V.Iron Feb 2 at 23:26

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:

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

• I'm aware of Newton's cooling law but I don't know the heat transfer coefficient to plug into the equation – V.Iron Feb 2 at 23:34
• $h$ is usually determined experimentally. You can consult engineering pages for it. For alumina/air some values should be available. Google is your friend. – Gert Feb 2 at 23:37
• Thanks for the upvote. An edit has also been made. – Gert Feb 3 at 16:34
• @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. – JMac Feb 4 at 20:59
• @JMac: well, you need both: a credible value for $h$ and a workable model. – Gert Feb 5 at 15:19