00:01
Let us begin with subart a.
00:03
The initial temperature t1 is given as 40 degrees celsius.
00:07
The volume of a water b is given as 50 liters.
00:11
This is also equals to 50 into 10 power minus 3 meter cube.
00:18
From the steam table, we can get the specific volume for temperature p equals 40 degrees celsius.
00:25
That is, specific volume b1 equals 0 .001080.
00:32
Meter cube per kilogram.
00:34
Now the mass of the water m is calculated using the expression v by v, sorry, v by v1.
00:43
Substead the given values, we have the volume of the water as 50 into 10 power minus 3 meter cure divided by v1.
00:54
This specific volume is 0 .0010m meter 2 per kilogram this is equals to a value of 49 .61 kilograms.
01:10
Therefore the mass of the water m is calculated as 49 .61 kilograms.
01:17
This is the final answer for subpart a.
01:22
Let us move on to part b.
01:25
The heat is transferred to the water.
01:28
So the final state of the water is a saturated vapor.
01:32
It is given that it has a constant pressure p of 200kcal that is the final temperature p2 is equal to the saturation temperature saturation temperature so from steam table at pressure 200 kilo pascal the corresponding saturation temperature is given as 120 .218 degrees celsius so thus we can say that the final temperature p2 is equal to to 120 .218 degrees celsius.
02:07
This is the final answer for part b.
02:12
Moving on to subpart c, the total enthalpy change is given by the expression delta h equals h2 minus h1 into m kilojoules...