Consider a system with delta U = 0. The system has higher temperature in the surrounding. If Work is done on the system, will the system
in this case wd= heat...it will be isothermal process
Essentially, when you say that work is done on the system, you are also saying that the system does work. The only difference would be of sign. If you, say, do 5 J of work on the system, then from the system's perspective, it has done -5J work. Now, from the first law, the heat evolved would be the work done on the system.
But I think by second law, heat cannot evolve as temperature is higher outside..Heat cannot flow from lower temperature to high..Is it correct?
Do we give preference to first law than the second?
What I do know is that both laws have to be satisfied, there is no question of preference.
If the temperature outside is higher, then heat flow should be from the surroundings to the system
May be the initial assumption that delta U is 0 is wrong..
Say all the heat that flows into the system will go into expansion work...
But by second law this cannot hold true.
Thus we can come to the conclusion that the conditions you said won't hold true, at leat not in natural conditions.
It may hold true if you artificially generate heat which flows from the system to the surroundings.
The second law is stated in VMC module as follows:
"It is impossible to construct a machine that is able to convey heat by a process from a colder to a hotter body unless work is done on the machine by some outside agency."
So the heat will be "forced" to go outside which is not in violation of the above statement..
I said that. The process has to be artificial, because otherwise the heat flowing in can't be all converted to work(I can't remember which statement of 2nd law said that)
Thus, what you are saying is that this process has to be non-spontaneous which will not occur in nature.
So the heat will keep on coming to the system and the work will be done on the system until equilibrium. (Btw, i have read that heat cannot be converted to do work).
Please correct if wrong, thanks for your help..