# Heat Transfer and Heat added to a system

1. Oct 11, 2015

### tonyjk

Hello,
Let's say the temperature of an object is constant and equal to 20 °C. If we put this object to a surrounding which has a temperature of 25 °C, than there's heat transfer from the surrounding to object by convection. If q is the heat transfer (in Watt) than after a specific time (Δt) we will have Temperature of the object equal to 25 °C ( If we consider the temperature of the surrounding constant) and thus we will have a net heat transfer equal to Q = qxΔt if q is constant. My question is, if we consider a 1D heat transfer, the surface of the object exposed to the surrounding will get the heat and thus will receive each dt the amount of heat q( q is measured like a current, in a cross section of the object) then will gain as internal energy DU = q but we say that all the object gained the heat q not only the surface exposed to the surrounding. Can we say, first the surface of the object exposed will gain an internal energy equal to q (increasing its temperature) and thus "distributing" its energy "uniformly" to the other parts of the object so we can say that all the object gained DU as energy ?

Thank you

2. Oct 11, 2015

### CWatters

Clearly all the heat gained by the object must go through the surface. So if the object as a whole gains energy q as then that energy must have gone through the surface "first". However you can't say that after time t the surface layer will have gained an amount q because it also looses heat to the core of the object in the same time t. In other words after time t the energy in the surface layer will be less than q as some has been distributed.

The problem with your example is that you have a dynamic situation where the heat flowing changes with time. For example as the core temperature rises the heat flowing in reduces. The objects temperature would be asymptotic to 25C. If you want to know when it reaches 25C you would have to decide how close to 25C is close enough. I think it might be easier to read up and understand heat flow under steady state conditions before looking at the situation you describe. It's easier to see what's going on.

3. Oct 11, 2015

### tonyjk

Great so all the energy gained at the surface will be "distributed" to the object right?

4. Oct 11, 2015

### CWatters

Yes. Where else can it go?