Please help me with this basic thermpdynamics problem

[gharrington44]

A thermoelectric generator consists of a series of semiconductor elements heated on one side and cooled on the other. Electric current flow is produced as a result of energy transfer as hear. In a particular Experiment the current was measured to be .5A and the electrostatic potential difference was .8V. Energy transfer as heat to the hot side of the generator was taking place at a rate of 5.5W. Determine the rate of energy transfer as hear from the cold side and the energy conversion efficiency.


Now I thought I could just say that the current is carrying a power out of the cold side of the semiconductor.

So I thought that I could find Qc but just applying the power equation.

Qc = IV = (.5A)(.8V) = .4 W

But this is not the right answer. Could someone help me through this? The answer should be Qc = 5.1W

 

[member 553083]

and η = 0.92 The rate of energy transfer as heat from the cold side is Qc = P/V = 5.5W/0.8V = 6.875W. The energy conversion efficiency is η = Pout/Pin = 0.5A x 0.8V/5.5W = 0.92.