00:01
Hi, we need to calculate the heat released during the conversion of 55 gm of water at 107°c to 50°c.
00:10
At 107°c, water exists as vapors.
00:16
As 100°c is the boiling point of vapor.
00:19
So first, this conversion will take place in three steps.
00:25
First, h2o at 107°c is cooled to h2o at 100°c.
00:35
Then this is in vapor form.
00:42
Then this is converted into liquid water at 100°c.
00:48
And then this is cooled down to h2o liquid at 50°c.
00:57
So i represent this by q1, q2 and q3.
01:03
So q1 is calculated as mc delta t, where m is the mass of water taken, c is the heat capacity of steam which is 1 .84 joules per gram degree celsius, and delta t is the change in temperature which is 100°c minus 107°c as this is final minus initial temperature.
01:27
This comes out to be 708 .4 joules.
01:32
Minus sign suggests that heat is released.
01:34
Now in q2 is related to the conversion of vapors into liquid.
01:41
For this, the formula is n delta h of vaporization, n is the moles of water which is equal to 55 gm multiplied by 18 gm per mole.
01:57
And the n delta h of vaporization is given as 40 .67 kj per mole...