How Much Energy Input Is Needed for an Engine to Perform 10.2 kJ of Work?

[VelvetRebel]

Homework Statement

A heat engine operating between 78.5oC and 204oC achieves 20.6% of the maximum possible efficiency. What energy input will enable the engine to perform 10.2 kJ of work?

The only help I need with this is exactly what equation(s) to use. I would think thermal efficiency would be one equation(e=W/Q_h) but not sure what else to use.

 

[kjohnson]

You need to take a look at the carnot efficiency of the heat engine, this will tell you the maximum possible work for a heat engine operating between two temperature resivors.

 

[kerimek]

Dear student,

Thank you for your question. I can provide you with the necessary equations to solve this problem.

First, you are correct in using the thermal efficiency equation (e=W/Qh). This equation represents the ratio of the work output (W) to the heat input (Qh) of a heat engine.

In this case, we know the efficiency (e) is 20.6%, which can be written as 0.206. We also know the work output (W) is 10.2 kJ. Plugging these values into the equation, we get:

0.206 = 10.2 kJ/Qh

To solve for Qh, we can rearrange the equation to:

Qh = 10.2 kJ/0.206 = 49.51 kJ

This means that in order to produce 10.2 kJ of work, the engine requires 49.51 kJ of heat input.

Next, we can use the equation for the Carnot efficiency (eC = 1 - Tc/Th) to determine the maximum possible efficiency of the engine. In this equation, Tc represents the temperature of the cold reservoir (78.5oC) and Th represents the temperature of the hot reservoir (204oC).

Plugging in these values, we get:

eC = 1 - (78.5 + 273)/ (204 + 273) = 1 - 351.5/477 = 0.264 = 26.4%

This means that the maximum possible efficiency of the engine is 26.4%. Knowing this, we can use the equation for thermal efficiency again to determine the energy input needed to produce 10.2 kJ of work at this maximum efficiency:

0.264 = 10.2 kJ/Qh

Solving for Qh, we get:

Qh = 10.2 kJ/0.264 = 38.64 kJ

This means that in order to achieve 10.2 kJ of work at the maximum efficiency of 26.4%, the engine requires 38.64 kJ of heat input.

I hope this helps you solve the problem. Please let me know if you have any further questions.

Best,