00:01
Now in this question we have to find the heat and the work transfer.
00:05
Now for that we have the given value which is p1 is equals to 100 kpa.
00:11
Temperature is given as t1 which is 120 degree celsius and we have volume of the cylinder is 1 m3.
00:26
Sigma 1 is equals to 1 .79 m2 per kg.
00:32
Because the process is isothermal then the value for t2 here it will be equals to t1 which is equals to 120 degree c.
00:43
The value for x2 is equals to 0 .5.
00:49
Now because we have t2 as 120 degree celsius so uf will be equals to 504 ks upon kg.
01:02
Value for ug is equals to 2530 kg upon ks.
01:11
The value for vf is equals to 0 .00106 m3 per kg and the value for vg is equals to 0 .891 m3 per kg.
01:30
Thus from here we can write that mass of h2o that is mass of water is equals to v1 upon sigma 1 which will be giving us 1 upon 1 .79 that is ultimately equals to 0 .558 kg.
01:51
Thus finding the specific volume we will get the specific volume that is v2 is equals to vf plus x2 multiplied by vg minus vf.
02:03
Putting the value and solving it we get 0 .00106 plus 0 .5 multiplied by 0 .851 minus 0 .00106.
02:20
So solving it we get v2 is equals to 0 .446 m3 per kg.
02:30
This is the value for v2.
02:32
Now writing down the specific internal energy that will be given by u2 it will be equals to uf plus x2 multiplied by ug minus uf.
02:46
Thus it will be 504 plus 0 .5 multiplied by 2530 minus 504...