A nuclear power plant has an electrical power output of 1200 MW and operates with an efficiency of 31%. If excess energy is carried away from the plant by a river with a flow rate of 1.0 x 10^6 kg/s, what is the rise in temperature of the flowing water?

Answer 1

Energy converted to heat
= 1200*10^6W/0.31
=3.87*10^9W
=3.87&10^9 j/s
Now divide by 4.18 to convert to cal/s
and divide by the mass of water (in grams) per sec. to get the increase in temperature in °C.

Answer 2

Thanks for your help. For the answer I got it to be 0.925837321 by following your directions. When I submitted it to my online homework it said the answer was wrong and off by more than 10%.

Answer 3

Good Lord. How did you get .925 from 3.87E9/4.18 ?

Answer 4

By taking 3.87E9/4.18 then dividing it by the flow rate of the water (1.0E6 kg/s = 1.0E9 g/s)

Answer 5

I see the error.

Energy produced+heat=total

but it is given that energy produced/total is .31, which means heat produced is .69 of energy produced, so..

heat produced= .69*1200*10^6W/0.31

Now do the conversions.
check my thinking.

Answer 6

Thanks for your help! I got the correct answer

Answer 7

Sorry Stefan, and thank you Bob!