Need help finding the efficiency of a heat engine.

Question

If a heat engine takes in 957 watts of heat from a source at a temperature 1,000 kelvin and dumps off 126 watts of heat to a heat sink at 300 kelvins. What is the efficiency of the heat engine?

Answer 1

Equate Efficiency to [Q_hot− Q_cold] ÷ Q_hot equal to [T_hot− T_cold] ÷ T_hot.

Place values to write [957 − Q_cold] Watts ÷ 957 Watts

= [1000.15 − 300.15] Kelvin ÷ 1000.15 Kelvin.

Reduction will isolate Q_cold as 287.20047 Watts.

Then [957 − 287.20047] / 957 and [1000.15 − 300.15] / 1000.15 will both calculate

Efficiency to 0.6998950157 or roughly 69.99%.

287.20047 Watts (or Joules Per Second) is the Power (Work Done Divided By Time Taken)

of the heat delivered to the heat sink by the heat engine.

Furthermore, Q_hot− Q_cold or [957 − 287.20047] or 669.79953 Watts gives the Power with

which the heat engine does the work.