00:01
So this question belongs to the laws of thermodynamics in which we have a power plant that produces electrical energy at the rate w equals to 1 ,300 megawatt, that is equals to 1 ,300 multiplied by 10 to the power 6 watt with efficiency e equals to 0 .31.
00:19
The excess heat is dumped into a river that carries volume flow rate v.
00:23
That is equals to 1 ,500 meter per second.
00:27
So we have to determine how much does the temperature? temperature increases of the river delta t so efficiency for the power plant it is equals to work done divided by heat supplied and work done divided by heat supplied can be written as heat rejected heat rejected qr qr plus work done w okay so substituting value so we get efficiency is 0 .3 1 this is equals to work done which is 3 300 made 10 to the power 6 divided by q rejected plus 1 ,300 multiplied by 10 to the power 6.
01:03
So from here after solving, we get q rejected.
01:07
This is equal to this comes out to be 2893 .55 multiplied by 10 to the power 3 joule per second.
01:17
That is what.
01:18
Now, this heat will be supplied to the river water.
01:21
So we can write that this qr, this will be equals to ms delta t.
01:26
And here, this rate, this, this, a, cue is in the rate form.
01:30
That is per second...