What are (a) the heat QH extracted from the hot reservoir and (b) the efficiency of a heat engine described by the pV diagram in the figure?
Answer 1
(a) the work done by the cycle is the area bounded by the process on the PV diagram.
The area of the triangle is 0.5*(400×10^-6 m^3)*(200×10^3 Pa) = 40 J
According to the first law, the net heat in the system is equal to the work (for one cycle). QH – 100 – 180 = 40
QH = 320J
(b) Efficiency is work output divided by energy input.
efficiency = 40/320 = 0.125 or 12.5%