Calculating Entropy Change in a Heat Engine Operating at Different Temperatures

[will_lansing]

Homework Statement

A heat engine operates between a high-temperature reservoir at 690 K and a low-temperature reservoir at 320 K. In one cycle, the engine absorbs 6700 J of heat from the high-temperature reservoir and does 2200 J of work. What is the magnitude of the net change in entropy as a result of this cycle?

Homework Equations

delta S =Q/T

The Attempt at a Solution

Qh= 6700
Th=690
Qc=Qh-W=6700-2200=4500
Tc=320

delta S= (-6700/690)+(4500/320)= 4.4 J/K

But i got the answer wrong, Where did i make a mistake, did i not round off correctly, or is the formula i used incorrect.

 

[Gokul43201]

Homework Statement

A heat engine operates between a high-temperature reservoir at 690 K and a low-temperature reservoir at 320 K. In one cycle, the engine absorbs 6700 J of heat from the high-temperature reservoir and does 2200 J of work. What is the magnitude of the net change in entropy as a result of this cycle?

Homework Equations

delta S =Q/T

The Attempt at a Solution

Qh= 6700
Th=690
Qc=Qh-W=6700-2200=4500

I believe I see a sign error there. Start over from the equation for the First Law and be careful with the signs.

 

[will_lansing]

I still don't see where i went wrong. can you please explain a bit more.

 

[Andrew Mason]

Homework Statement

A heat engine operates between a high-temperature reservoir at 690 K and a low-temperature reservoir at 320 K. In one cycle, the engine absorbs 6700 J of heat from the high-temperature reservoir and does 2200 J of work. What is the magnitude of the net change in entropy as a result of this cycle?

Homework Equations

delta S =Q/T

The Attempt at a Solution

Qh= 6700
Th=690
Qc=Qh-W=6700-2200=4500
Tc=320

delta S= (-6700/690)+(4500/320)= 4.4 J/K

But i got the answer wrong, Where did i make a mistake, did i not round off correctly, or is the formula i used incorrect.

Your method is correct. And you are correct to use two significant figures. It is just a problem with when to round off your figures. I think they want you to calculate each of the entropy changes of the hot reservoir and cold reservoir separately to two significant figures and then take the total.

The change in entropy of the engine is 0 since it returns to its initial state in one complete cycle. In one cycle, the hot reservoir has lost 6700 joules and the cold reservoir has gained 4500 joules. So the total entropy change is that of the reservoirs only, which is:

\Delta S = -\Delta Q_h/T_h + \Delta Q_c/T_c = -6700/690 + 4500/320 = -9.7 + 14 = 4.3 J/K

AM