A reversible heat engine operates between thermal reservoirs at 1000 K and 290 K. If 8610 J is extracted from the hot reservoir, how much work is done by the heat engine?

8610 (1000-290)/1000= what

ç = 1 - 290K/1000K = 0.29 = 29.0%

The thermal efficiency is defined as the amount of work done per amount of heat absorbed from hot reservoir or...
ç = W/Qh

so then you would just use
W = e*Qh

and plug in what you know

Hannah, reread the question. My response is correct. It does not ask for thermalefficiency.

To find the work done by the heat engine, we can use the Carnot efficiency formula:

Efficiency = 1 - (Tc/Th)

where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. The Carnot efficiency represents the maximum efficiency that can be achieved by any heat engine operating between the two given reservoirs.

In this case, Tc = 290 K and Th = 1000 K. Therefore, the Carnot efficiency is:

Efficiency = 1 - (290/1000) = 1 - 0.29 = 0.71

The efficiency of the heat engine is 0.71, which means that 71% of the energy extracted from the hot reservoir is converted into work.

Given that 8610 J is extracted from the hot reservoir, we can calculate the work done by the heat engine:

Work = Efficiency * Energy extracted from the hot reservoir = 0.71 * 8610 J ≈ 6113 J

Therefore, approximately 6113 J of work is done by the heat engine.