Some Carnot engine has an efficiency 0.700. The hot reservoir temperature is 900K. This means that the temperature of the cold reservoir is 270K. If the cold reservoir is dropped to 260K, the new efficiency is 0.701.

Question

If the output of the power is kept constant by increasing the amount of time it takes to complete a cycle, what's the ratio of the wasted transfer of heat after the change to before the change?

Answer 1

In your first paragraph, the efficiency of a Carnot engine would be 0.711 with a 260 cold reservoir, not 0.701.

Your second paragraph does not make sense to me. Increasing the time it takes to make a cycle will not increase the power.

The ratio of wasted heat with 260 cold reservoir to that with 270 cold reservoir is (1-.711)/(1-0.700) = 289/300 = 96%

Answer 2

That didn't work. I think you're supposed to find work in each case but I'm not sure how to do that.

Answer 3

Oh wait never mind. Seems like there's some rounding issues going on with the online system. Thanks for the help drwls.