Thermodynamics contradiction?

A closed, uninsulated vessel having a volume of 28ft^3 contains 50lbmass of r-12 initially at 100 degrees ferenhiet. The vessel is placed outside where the temperature is 50 degrees ferenheit and, over a period of time, the temperature of the r-12 is 60 ferenheit.


Now, common sense tells you that if the contents of the tanks, whatever it may be (in this case r-12), is initially at 100 degrees and is surrounded by a lower, outside temp, then heat transfer will occur and eventually the tank will reach the outside temp (50 degrees) and during the process pass through 60 degrees. but when i work out the change in entropy I get that the process is impossible.


v=28ft^3/50lb=0.56ft^3/lb


@v and 100 degrees, u1=81.854 Btu/lb and s2=.17590 Btu/lb


for second temp its in liq-vap region. after finding those values i used q=u2-u1.


then, r-12 change in entropy= s2-s1


then s2-s1=Q/T to find change in entropy in the outside air


(s2-s1)r-12 + (s2-s1)air %26lt; 0. so process is impossible? huh?|||You are going to reach thermal equilibrium if the outside is a sink like your problem statement assumes. You do not really believe that you've found an exception to the laws of thermodynamics, correct?





That would lead the simpler explanation of messing something up computationally, which is very, very easy to do with thermo. Check a table for r12 to see if you pass through the saturated liquid portion of the phase diagram. This may require "lever" analysis, part of the mixutre is liquid, part gas so you've got to average the values.





This may be the whole of your problem. There are several ways to do these problems, but I'd draw out the state change first to see where you're going.|||Simple enough. You made a mistake in your math. I am too lazy to check where the "contradiction" occurs and what you did wrong because I know that the vessel will eventually be in thermal equilibrium. I also know that the process is not impossible. But I do not care much to correct you. You are smart enough to do that on your own.