A simple problem confused me:
suppose a constant amount of heat Q and a solid body with heat capacity C and initial temperature T1.
I want to calculate the final temperature of the body when heat Q transferred to the body at the following conditions:
1- irreversible heat transfer
2- reversible heat transfer
3- quasi static heat transfer
???
Dear Sir
Kind regards..
You can not transfer any amount of heat to a body unless there is a temperature difference between the body and its environment. This amount of heat is dictated by the temperature difference. large temperature difference transfers more heat than small temperature difference. With reversible processes the temperature difference is zero which is an ideal and non-realistic case. for quasi static heat transfer, the difference is so small that causes no change in equilibrium and this takes very long time. In all cases the final temperature of the body is calculated from the following equation:- T2=T1+Q/(M*C)
Isn't irreversibility affect the finial temperature?
But it seem that at the reversible process the final temperature increase more?!
Hallo,
as a matter of fact, not in the way you pose the problem. If a given amount Q of thermal energy is exchanged, the final temperature will be as the colleague said T2=T1+Q/(M*C); you can realize this if you apply the energy balance to your system and express the variation of internal energy as the product of the specific thermal capacity times mass times temperature difference. As a matter of fact, the first-pricniple (energy balance) does not discern between reversible and irreversible processes (and heat and work would seem indistinguishable in this regard), you need to apply second-law balance equation to find out that non-reversible thermal exchanges introduce a term called entropy production (rate) which is absent in reversible processes. If you need a rather broad but accurate insight into the matter, I suggest (I take it German is your native tongue) you have a look at chapter 3 (for 2nd law) and 2 (first law) of the excellent "Thermodynamik - Grundlagen und technische Anwendungen" by Baher and Kabelac, published by Springer.
Marco
Is there any difference between "reversible" and "quasi static" heat transfer in entropy generation?
In Quasi Static process it the Q must be very small and therefore Tfinal will be T1.
in reversible process again T1 must be equal to Tfinal. But this time it is exactly same where as in quasi-static process there might be slight change as the process may be irreversible.
and in any case the final temperature ideally should be T1+Q/MC but it will be less than that
Entropy: A concept that is not a physical quantity
https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity
Entropy: A concept that is not a physical quantity????
Dear Zhang
Why this ans for this question???
In my view, both quasi-static and reversible process are equally likely similar. When we compare reversible and irreversible process, basic literature dictates that an irreversible process is sudden, hence less efficient, more loss, lower work output, since system is not allowed to attain equilibrium. It quite frankly can be compared to cooking of fries in oil. Too higher temperature results in burnt fries, but a medium level temperature results in best quality of fries, because oil and fries have enough time to equillibriate and transfer heat among themselves.
Now, let's back to theoretical physics where sudden changes in pressure will not allow gases to equillibriate volume, hence less work i.e. PV.
A common misconception arrives when people try to analyze this concept using mathematics, probably because both utilize same formula :
Q = m*C*del(T).
Q = Heat Supplied, C = Specific Heat Capacity and del(T) = change in temperature.
With same value of Q, and initial temperature Ti, final temperature Tf must also be same, but nobody pays attention to specific heat capacity value. In case of irreversible process specific heat capacity value is way higher compared to irreversible, hence final temperature in reversible is higher.
It could reiterated in another way, that since there are higher loss due to gas in non-equilibrium phase (mentioned before), we need higher amount of heat achieve similar final temperature.
It could be confirmed from ploy-tropic equations as well as curve of process and corresponding heat capacity achieved.
- Badges
- Science topic
- Similar topics
- Engineering
- Thermal Engineering
- Thermofluid
- Heat Transfer