00:03
So in this question there is given a graph and from graph this is clear that we see that the temperature is constant at point d and c so where this is figure that means two reservoir in this figure above r 600 r1 reservoir of r is 600 k where r2 is 300 kelvin so that the high temperature reservoir is at 600k reservoir which is in the contact with the path ab.
00:41
So this is the path, this is in the contact with ab.
00:45
So we know that delta s for this path is q by t.
00:51
Now it is an isothermal process as temperature is not varying in this process.
01:03
So by this physical contact with the hot reservoir we can say the heat change is a very temperature is.
01:10
In the path of q ab is equal to temperature of a and entropy change of a to b.
01:20
So to find this heat exhaust along the process of a, b, we had to find the area under the t and s.
01:29
So the area under the t and s that will be as now this way the area like this, from here to this area so this is the area we have to find now qav is the area that we have defined here so qav will be equal to t a of entropy change at b point and entropy changed at a point will be again as 600 multiplied by 4 minus 2 so the answer of heat changed at the path q ab will be equal to 1 ,200 jude.
02:24
Now we know that from part a, the low temperature reservoir is 300 kelvin, the reservoir which is in physical contact with the path cd.
02:35
So this is the path cd and the reservoir is in contact of lower temperature.
02:41
So we know that delta s for hair is equal to q, by t.
02:47
Now q of cd, a heat change from path c to d will be equal to as temperature at c, delta s, c, d...