Question

What is the maximum efficiency of a heat engine that operates between 440°C and 120°C?
____________________%
If this engine generates 2000 J of mechanical energy, how many calories does it absorb from the hot reservoir, and how many calories does it transfer into the cold reservoir?
___________________cal (hot reservoir)
________________________cal (cold reservoir)

Answer 1

The maximum efficiency of a heat engine is given by -

?_max = 1 - T_c/T_h = 1 - (120+273)K/(440+273)K = 1 - (393 / 713) = 1 - 0.55 = 0.45

So, maximum efficiency in percentage = 0.45 x 100 = 45%

Now, given that the engine generates 2000 J of mechanical energy.

Suppose heat absorbed from the hot reservoir = Q.

So -

2000 / Q = 0.45

=> Q = 2000 / 0.45 = 4444.4 J = 1062 Cal.

And the heat transferred into the cold reservoir = 4444.4 - 2000 = 2444.4 J = 584 Cal.