00:01
So for this problem, starting off with part a, we have qh is equal to w over e.
00:10
And so ph is equal to pw over e, which is equal to 100 megawatts divided by 0 .40.
00:26
And that gives you 2 .50 times 10 to the third megawatts.
00:30
For part b, the heat input in one day would be 2 .50 times 10 to the 9 watts times 8 .64 times 10 to the 4 seconds, and that gives you 2 .16 times 10 to the 14 joules.
00:51
And so the mass of coal used per day would be 2 .16 times 10 to the 14 joules, divided by 2 .65 times 10, 10 to the 7 joules per kilogram, and that gives you 8 .15 times 10 to the 6 kilograms.
01:13
For part c, we have the absolute value of qh is equal to the absolute value of w, plus the absolute value of qc, and so pc is equal to ph minus pw, which is 2 .50 times 10 to the 5 megawatts minus 1 ,000 megawatt, and that is 1 .50 times 10 to the third megawatts.
01:47
For part d, the heat input to the river is 1 .50 times 10 to the 9 joules per second.
02:01
We're using q equals mc, change in t, and we know change in t is 0 .5 degrees, sorry, 0 .4 degrees celsius...