Question

A car engine burns gas at 495 K, and exhausts to the air at 293 K. If it ran at the highest possible efficiency, how much input heat would it take to do 10,000 J of work? (Unit=J)

Answer 1

  The heat taken to do the 10,000 J of work is 4081 J

Explanation:

The ratio of the temperature difference between the reservoirs and the temperature of the hot reservoir is termed as maximum thermal efficiency.

Maximum thermal efficiency is

                                                   1 - (Tc / Th).

where Tc represents the  absolute temperature of the cold reservoir

           Th represents the absolute temperature of the hot reservoir.

                                                    1 - (293/495) = 0.408  

If the input is 10,000 J, then 4081 J is used to do work.  The remaining 5919 J is rejected as heat into the air.