0:00
Hi there.
00:01
So for this problem, we are given the carnal cycle on a d -a's diagram with a scaled.
00:11
So we are given that the value of x -x, it's equal to 0 .6 joules per kelvin.
00:23
And what we need to find for part a of this problem is the net heat transfer.
00:31
So we need to find the net heat.
00:39
Now, using the expression that we know for the heat, that is the temperature times the change in the entropy, we will have that we note that the heat enters the cycle along the top path, this top path in here, and leaves move along the bottom path that is in here, at a position that is 250 kelvin, then we will find that the heat is inward is equal to the value for the temperature that in this case is at the top one, which is 400 kelvin, times the change in entropy.
01:52
As you can see, since this value is 0 .6, then each value of this is 0 .1, 0 .2, 0 .3, 0 .4 .4.
02:10
0 .5.
02:13
So with that said, we have that the final value, it's 0 .6, and the initial value, it's 0 .1 joules per kelvin.
02:28
So we will have in here that that is the difference.
02:32
So 0 .6 joules per kelvin, oh, sorry, and minus 0 .1 joules per kelvin.
02:40
So from this we obtain that the heat that is end is equal to 200 joules...