20 ! urgent answer physics ! : )

the output of a certain electric generating station is 2500 mw.

a) determine the energy output of the station each second, hour, day and year.

b. if the station is 41% efficient, how much energy input is required during a single day’s operation?

c. assume that 1.5 x 10 10 kg of water from a nearby lake circulates each day to cool the operating parts of the station. how much warmer is the return water than the intake water assuming that all the waste energy goes to heating the returned water?

2500 MW means

... 2500 megawatts
... 2.5 x 10⁹ watts
... 2.5 x 10⁹ joules per second          <== the most useful one.

a). 
second ... 2.5 x 10⁹ joules

hour          (2.5 x 10⁹ joule/sec) x (3,600 sec/hour) =  9 x 10¹² joule/hour  

day            (9 x 10¹² joule/hour) x (24 hour/day)  =  2.16 x 10¹⁴ joule/day

year          (2.16 x 10¹⁴ joule/day) x (365 day/year) =  7.884 x 10¹⁶ joule/yr

b).
                 2.16 x 10¹⁴ joule/day  = 41% of the input

Divide each side by 41%:

                   Input energy = (2.16 x 10¹⁴ joule/day)  /  0.41

                                         =   5.2683 x 10¹⁴ joule per day .

c).
The waste energy goes into the water. 
That's the other 59% of the daily input.
          0.59 x (5.2683 x 10¹⁴ joule per day) = 3.1083 x 10¹⁴ joule / day.

It takes 4179 joules to raise the temperature of 1 kg of water 1°C.

You have  3.1083 x 10¹⁴ joules, which you pump into  1.5 x 10¹⁰ kg
of water.  Every 4179 joules raises the temp of 1 kg by 1°C.

My mind is messed up by a lot of drugs for gout today, so I'll leave that
last calculation for you.  (And it wouldn't be such a bad idea for you to
check my earlier numbers too.)  

From negative to negative

F  you double the  force, you double the acceleration, but  if  you double the  mass, you velocity. all bodies accelerate at the  same  rate regardless of size or  mass.