Calculate how many Kgs of water flows through the plant

[gigglin_horse]

Homework Statement

"A power station with an efficiency of 0.4 generates 10^8W of electric power and dissipates 1.5*10^8W of thermal energy to the cooling water that flows through it. Knowing that the specific heat of water in SI units is 4184J/KgºC, calculate how many Kgs of water flows through the plant each second if the water is heated through 3 Celsius degrees"

Homework Equations

The Attempt at a Solution

I have no idea. This is way above me.
Someone please explain this to me!

 

[CompuChip]

Let's look at any given second. Then how much energy is put into the water? The specific heat of water is given, so you know that it takes 4184 * 3 = 12552 J to heat one kilogram of water by three degrees, so you should then be able to calculate how many kilograms you can heat with the available energy.

 

[gigglin_horse]

I understand so far, but I don't understand the available energy.
Efficiency = T(hot)-T(cold)/T(hot) ...right?

What is the available energy?I tried
(1.5x10^8)-(10^8)/(1.5x10^8)=0.333333

Am I headed in the right direction?

 

[LowlyPion]

They already tell you how much is going to thermal dissipation in the water. The efficiency isn't really needed.

Figure the total number of joules dissipated and thence how many kilograms are needed to do it given the number of joules that 1 kg heated 3° carries.