Professor I have a question here. Is Carnot cycle totally isothermal? I mean there are two isothermal processes, which are isothermal compression and isothermal expansion. Others are adiabatic compression and adiabatic expansion. These adiabatics are also isothermal, at the same time ?
Great question. The adiabatic compression and expansion are not also isothermal because there are temperature changes during those processes. Please see this simulation: learncheme.com/simulations/thermodynamics/thermo-1/carnot-cycle-on-ideal-gas/
First law should be W=Qh-Qc & not Qh+Qc+W=0 ? The value of Qh is positive because heat is added into the system & the value of Qc is negative because Qc is removed from the system. The value of W comes out as a result of calculation to be positive because the value of heat added is greater than the value of heat the system rejected. The way u have written the first law is confusing, +ve Sign for the variable W on the LHS confuses me. We should not change the sign of the variable only the value. It causes confusing. If I understand it clearly, your lectures would become the most enjoyable lectures in the world, because u talk with calmness & clarity. Please clarify this confusion to me. Thanks
No, the first law is correct as written because Qh is positive and Qc and W are both negative, so the sum of the three (Qh + Qc +W) is zero because the energy of the cycle does not change so all the energy added as heat is removed as either work or heat at low temperature.
Efficiency cannot be 100%. That’s all that thermodynamics teaches. It does not put an upper bound on the efficiency. If someone can make a machine for which losses tend to zero, you have a Perpetual Motion machine for all practical purposes.
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Professor I have a question here. Is Carnot cycle totally isothermal? I mean there are two isothermal processes, which are isothermal compression and isothermal expansion. Others are adiabatic compression and adiabatic expansion. These adiabatics are also isothermal, at the same time ?
Great question. The adiabatic compression and expansion are not also isothermal because there are temperature changes during those processes. Please see this simulation: learncheme.com/simulations/thermodynamics/thermo-1/carnot-cycle-on-ideal-gas/
if the Qh is not given how do we find find Qh?
First law should be W=Qh-Qc & not Qh+Qc+W=0
?
The value of Qh is positive because heat is added into the system & the value of Qc is negative because Qc is removed from the system.
The value of W comes out as a result of calculation to be positive because the value of heat added is greater than the value of heat the system rejected.
The way u have written the first law is confusing,
+ve Sign for the variable W on the LHS confuses me.
We should not change the sign of the variable only the value. It causes confusing.
If I understand it clearly, your lectures would become the most enjoyable lectures in the world, because u talk with calmness & clarity.
Please clarify this confusion to me.
Thanks
No, the first law is correct as written because Qh is positive and Qc and W are both negative, so the sum of the three (Qh + Qc +W) is zero because the energy of the cycle does not change so all the energy added as heat is removed as either work or heat at low temperature.
LearnChemE Now im confused. Doesnt QH=QL+W since energy is conserved? So then QH-QL-W=0 or QL+W-QH=0?
Thanks
Perpetual motion geek need to watch videos like this...haha
Efficiency cannot be 100%. That’s all that thermodynamics teaches. It does not put an upper bound on the efficiency.
If someone can make a machine for which losses tend to zero, you have a Perpetual Motion machine for all practical purposes.
Thank u so much.. very helpful
quiet interesting