I am also confused as to what power consumption has to do with GPU temperature.
I could make an HD5450 run at 100c if I wanted, and cool a GTX295 down to 60c, tell me which is using more power. Heat has nothing much to do with power, and everything to do with the cooling system used.
hmm heat and temperature are related... but different things. Temperature is a measure of energy within 1 system (ie. internal energy). Heat is a measure of kinetic energy transferred between 2 systems.
If everything stays the same, then a system with higher power consumption will have a higher temperature.
But taking your example above, the HD5450 runs at 100C and GTX295 at 60C... you ask which system is consuming more power.. and we may (wrongly) conclude that the HD5450 consumes more power because it has a higher temperature. However, it is wrong to conclude like so because the 2 systems are different, so we cannot do a direct comparison without involving entropy into the equation.
The equation for temperature (T) for a system at equilibrium is :
T = (change in energy) / (change in entropy)
In order for the GTX295 system to run at such a relatively low temperature, there has to be a big big cooler be coupled to the system (GTX295), and it is this big big cooler that will cause an increase in entropy to the environment (the surrounding air). The GTX295 will cause a greater increase in entropy to the environment than the HD5450. If we're talking about just 1 system, and if every thing stays constant, then an increase in power consumption will cause an increase in internal energy which also necessarily means an increase in temperature (according to the equation above). However, if we ask questions about 2 different systems (GTX295 and HD5450), then we must take entropy into account in order for the questions to make sense.
There, I hope I've cleared up some of your confusion. I've only had 1 year of physics and general chemistry in college a while back. So my explanation could be missing a lot of details.. any person with a firm grasp of the 2nd law of thermodynamics please feel free to correct me.