Can you please explain why here you write -w = nR( Th-Tc) lnV2/V1.. and not (Tc-Th) as explained in the previous video for the Carnot cycle? Is it because you switched the signs?
Yes. In the previous video, we had w = n R (Tc - Th) ln V₂/V₁ In this video, the numerator is −w, with a negative sign. I could have written that as −w = −n R (Tc - Th) ln V₂/V₁ But instead of adding the negative sign out front, I just switched the order of the two temperatures, using this instead: −w = n R (Th - Tc) ln V₂/V₁
The adiabatic steps are insulated, but the isothermal steps are not. So heat does transfer to/from the environment in those steps. See the preceding video on the Carnot cycle for more details: ua-cam.com/video/szyFQuPiP9s/v-deo.html I love your username, btw!
My name is Ben Jamin, and I approve this message.
Great👍
Can you please explain why here you write -w = nR( Th-Tc) lnV2/V1.. and not (Tc-Th) as explained in the previous video for the Carnot cycle? Is it because you switched the signs?
Yes. In the previous video, we had
w = n R (Tc - Th) ln V₂/V₁
In this video, the numerator is −w, with a negative sign. I could have written that as
−w = −n R (Tc - Th) ln V₂/V₁
But instead of adding the negative sign out front, I just switched the order of the two temperatures, using this instead:
−w = n R (Th - Tc) ln V₂/V₁
The system is insulated ,so the Q_h given as heat will be used in doing work ,but how does it expel heat as Q_c,if all the heat becomes work ?
The adiabatic steps are insulated, but the isothermal steps are not. So heat does transfer to/from the environment in those steps.
See the preceding video on the Carnot cycle for more details: ua-cam.com/video/szyFQuPiP9s/v-deo.html
I love your username, btw!