# Heat Transfer Conservation in a natural convective flow

In a initially stationary area, suppose I have a heat source of say 10 W.

Now if I consider a fictional cube (box)/ boundary of say volume 1 m3 around it. And I sum up the heat across the surfaces of this cube, would it be 10 W as well? The cube boundaries are not solid. Just a fictional volume to consider.

Due to the heat, the air near it moves and circulates due to convection. So my understanding was the some amount of heat energy is transferred to kinetic energy to move the air. So if someone considers a fictional cube, the heat transfer of it should be less than 10W, as some of it used to move the air molecules. Am I right in saying this?

the heat transfer of it should be less than 10W, as some of it used to move the air molecules. Am I right in saying this?

No. If you have a box dissipating 10 W, then at steady state all 10 W is transferred to the surroundings. Heating of the air is synonymous with increasing its internal energy and its kinetic energy; these can't be decoupled. If the air rises, as in natural convection, it expands and cools down. This does not take or sequester extra energy from the box; it's part of the 10 W. Does this make sense?

• Can you explain this more on why cant they be decoupled? - Heating of the air is synonymous with increasing its internal energy and its kinetic energy; these can't be decoupled. Also the box is just a fictional boundary I am talking about, not solid. Commented Dec 1, 2021 at 2:29
• I mean that the air is heated at a rate 10 W, the internal energy increases at a rate of 10 W, and the kinetic energy of the molecules increases at a rate of 10 W. It's all the same effect. Any region of heated air expands and becomes rarefied and then rises due to buoyancy, cooling down somewhat in the process while also diffusing heat to its own surroundings. Commented Dec 1, 2021 at 5:41

As the box' inside and walls heat up, the walls inevitably start losing heat convectively to the (presumed colder) surroundings.

This can be modelled fairly well using Newton's Law of cooling:

$$\dot{Q}=hA(T_{in}-T_{\infty})$$ where the LHS is the heat flux (loss, in $$\mathrm{Js^{-1}}$$), $$h$$ the heat transfer coefficient, $$A$$ the total surface area exposed to the surroundings, $$T_{in}$$ the temperature of the box and $$T_{\infty}$$ the temperature of the surroundings.

So as $$T_{in}$$ increases, so does $$\dot{Q}$$, until steady state is achieved (box temperature is constant):

$$P=hA(T_{SS}-T_{\infty})$$

with $$T_{SS}$$ the Steady State Temperature and $$P$$ the power of the heater.

You are talking about an initially transient situation. Initially, part of the 10 W will be used to increase the temperature of the air in the box, and part of it will go into the kinetic energy of the convection flow. So, certainly, at short times, not all the heat will be leaving the cube. But, eventually, the system will reach steady state, with the temperatures within the cube and the convection flow becoming constant. After that, all 10 W will leave the cube.