The energy loss caused by Thermal Resistance

Question

The absolute thermal resistance can be expressed in the following formula:

$$R = \frac{\Delta T}{P} \left[\frac{K}{W}\right]$$

It shows how big is the temperature difference between two points in the object per watt of applied energy. So, for known R = 3 K/W and P = 2 W we will get 6 K drop.

What happens with this power loss (temperature)? In case of electric analogy the lost current voltage drop (energy loss) goes into heat, but it is not clear for me where this energy goes to. Or is it only mathematical model which is used only in calculations?

Answer 1

I'd say it's a bit more than a mathematical model, but you are taking the analogy to a place where it can't be applied.

First to clear up some terminology in your question about the analogy. The temperature loss is equivalent to a voltage loss when going through a resistor, not the power lost from the circuit. In the electrical analogy for a circuit, there is no lost current; the current has to be constant across for a single loop/line.

Current is also analogous to heat transfer rate, and that's where the analogy starts to fall apart. Heat transfer rate already measures where the energy is going. That's the point, it measures the thermal energy transfer per unit time. You're already measuring where the energy goes (and it's rate) when you measure $\dot Q$ (or $P$ in your equation).

So it basically falls apart because an electric circuit is only telling you where the charge goes; which doesn't directly measure energy. But the heat transfer equations tell you where the heat goes, which is directly telling you about the energy.