Efficiency of a Carnot Engine

[vetgirl1990]

The efficiency of a Carnot Engine is described by the relationship: Tc/Th = Qc/Qh, so that e(Carnot) = 1 - Tc/Th

For heat engines, can their efficiency also be related to temperature as well?
Or is the description of their efficiency just: e(heat engine) = W / Qh = 1 - Qc/Qh

I am inclined to say that the only reason that a Carnot Engine's efficiency can be related to temperature like that, is because of the cyclic nature of the Carnot Cycle... But I'm not entirely sure.

The reason I am asking this question, is because I am trying to understand how to solve the following problem: "A heat engine operating between 200C and 80C achieves 20% of the maximum possible efficiency. What energy input will enable the engine to perform 10kJ of work?"
Tc/Th = Qc/Qh, W = Qh-Qc
Therefore, Tc/Th = (Qh - W) / Qh --> Tc/Th = 1 - W/Qh
So plugging in the above values, Qh = 16.7kJ

My solution is only valid if I made the correct assumption that Carnot efficiency can be applied to a heat engine efficiency.

 

[stockzahn]

The efficiency of a Carnot Engine is described by the relationship: Tc/Th = Qc/Qh, so that e(Carnot) = 1 - Tc/Th

I'm not sure how you derive the Carnot efficiency from T_c/T_h = Q_c/Q_h, but the final formula is correct.

For heat engines, can their efficiency also be related to temperature as well?
Or is the description of their efficiency just: e(heat engine) = W / Qh = 1 - Qc/Qh

η_C = W / Q_h = (Q_h - Q_c) / Q_h = 1 - Q_c / Q_h = 1 - (T_c ⋅ Δs) / (T_h ⋅ Δs) = 1 - T_c / T_h

I am inclined to say that the only reason that a Carnot Engine's efficiency can be related to temperature like that, is because of the cyclic nature of the Carnot Cycle... But I'm not entirely sure.

The Carnot efficiency is the theoretical maximal efficiency for thermodynamic cycles (and only for thermodynamic cycles) - so yes.

The reason I am asking this question, is because I am trying to understand how to solve the following problem: "A heat engine operating between 200C and 80C achieves 20% of the maximum possible efficiency. What energy input will enable the engine to perform 10kJ of work?"
Tc/Th = Qc/Qh, W = Qh-Qc
Therefore, Tc/Th = (Qh - W) / Qh --> Tc/Th = 1 - W/Qh
So plugging in the above values, Qh = 16.7kJ

My solution is only valid if I made the correct assumption that Carnot efficiency can be applied to a heat engine efficiency.

1) Your calculations are not correct. I recommend to take the Carnot efficiency as you stated it at the beginning of your post (η_C = 1 - T_c / T_h).
2) The Carnot efficiency is the maximum possible efficiency - what if the cycle only achieves 20 % of it?
3) Don't forget, that the temperatures in the statement are in °C.

 

[DrDu]

The reason why the efficiency of a Carnot process can be expressed in terms of two temperatures is the fact that it is the only process where heat exchange only takes place at two well defined temperatures. In other processes, temperature changes continuously during heat exchange (e.g. along isochores).

 

[Chandra Prayaga]

The reason why the efficiency of a Carnot process can be expressed in terms of two temperatures is the fact that it is the only process where heat exchange only takes place at two well defined temperatures. In other processes, temperature changes continuously during heat exchange (e.g. along isochores).

As DrDu says, the Carnot cycle is a reversible cycle between two temperatures, and you can prove that the efficiency of the cycle is as stated by you. I have just a couple of points to clarify.
The Carnot cycle is a heat engine. It is the most efficient possible one.
efficiency e = W/Qh is not just a description, it follows from the definition of efficiency as (useful work output/heat input to engine)