Question

A heat engine takes thermal energy QH from a hot reservoir and uses part of this energy to perform work W. Assuming that QH cannot be changed, how can the efficiency of the engine be improved?

a) Increase the work W, the thermal energy QC rejected to the cold reservoir increasing as a result.
b) Decrease the work W, the rejected QC decreasing as a result.
c) Increase the work W, the rejected QC remaining unchanged.
d) Increase the work W, the rejected QC decreasing as a result.
e) Decrease the work W, the rejected QC remaining unchanged

Answer 1

d) Increase the work W, the rejected QC decreasing as a result.

Step-by-step explanation:

By the second law of thermodynamics the efficiency of a heat engine working between two reservoirs is:

`\eta=(W)/(Q_(H))` (1)

With W the work and
`Q_(H)` the heat of the hot reservoir, note in (1) that efficiency is directly proportional to the work and inversely proportional to the heat of the hot reservoir, so if we remain
`Q_(H)` constant we should increase the work to increase the efficiency.

Also, efficiency is:

`\eta=1-(Q_(C))/(Q_(H))` (2)

With
`Q_(C)` the heat released to the cold reservoir, it is important to note that because second law of thermodynamics the efficiency of a heat engine should be between 0 and 1 (
`0\leq\eta\leq1`), so the ratio
`(Q_(C))/(Q_(H))` always is positive and its maximum value is 1, that implies if
`Q_(H)` remains constant and efficiency increases,
`Q_(C)` will decrease and the ratio
`(Q_(C))/(Q_(H))` too.

So, the correct answer is d)