00:01
Okay, so in this problem, we are working with a power plant, and we are given the power output and the efficiency and ask to estimate the heat discharged per second.
00:17
So let's go ahead and write down the quantities we know.
00:23
So work equals 550.
00:28
Let me make that near.
00:32
Milly, the prefix milly is 10 to the 6th, so that's 10 to the 6th, joules per second.
00:45
And we're also given that efficiency, e equals 38%, or 0 .38.
00:54
So, and then what we're being asked for is the heat discharge per second, which is ql.
01:03
So we know that our equation relating efficiency and work and heat is e equals w worked on over qh.
01:16
We also know that by conservation of energy, qh equals w plus ql.
01:25
So we can go ahead and plug that in right there.
01:28
Let me make that.
01:28
Let me clean that up a little bit.
01:32
There we go.
01:34
And with some rearranging of that formula, we can solve for ql saying that ql equals work times 1 over efficiency minus 1...