Finding heat dissipation from resistor surface temperatures

[DJJB]

So I am an engineering student working on co-op and I ran into a problem.

I have a resistor like the image above. I placed this resistor in free space, leads in vertical position (Ambient: 296K ) and I placed 10W of electrical power into it. What I need to find is the thermal coefficient (How much thermal energy is transmitted to the air from the aluminium outer casing)

I placed 6 thermocouples, one on each side and recorded the following temperatures:

Front: 442K
Back: 486K
Sides x 2: 462K
Bottom: 453K

Is anyone aware of a method to convert surface temperature of a given solid to power? So far I have only come across thermal conductivity which is dependent on thickness. Hopefully there is an easier solution which one of you may know. Thanks.

 

[NobodySpecial]

You know the power radiated from it - it has to equal the electrical power you put into it.

The thermal coeff is normally given in terms of temperature difference per power (or the other way around)

 

[DJJB]

You know the power radiated from it - it has to equal the electrical power you put into it.

The thermal coeff is normally given in terms of temperature difference per power (or the other way around)

Ideally, Yes. But in the real world there are a lot of losses associated with this such as sound, vibration, thermal loss in the wires and leads and so on. Nothing is 100% efficient.

 

[NobodySpecial]

All of which must end up as thermal energy anyway - but that doesn't matter

The thermal coef of a resistor is power/temp-difference (or the other way up)

Nothing is 100% efficient.

No machine is 100% - whether this applies to a heater is a surprisingly tricky question