# Thermal energy power station problem

1. Feb 19, 2010

### dymand68

1. The problem statement, all variables and given/known data:
A power station with an efficiency of 0.3 generates 10^8 W of electric power and dissipates 2.33 multiplied by 10^8 W of thermal energy to the cooling water that flows through it. Knowing the specific heat of water in SI units is 4184 J/kg°C, calculate how many kilograms of water flow through the plant each second if the water is heated through 3 degrees Celsius.

2. Relevant equation:
Q=cm* change in temperature

3. The attempt at a solution:
2.33*10^8=4184*m*3
m=(2.33*10^8)/(4.184*3)

2. Feb 19, 2010

### Andrew Mason

Why are you using 4.184 and not 4184 in the denominator?

AM

3. Feb 19, 2010

### Sirsh

Q = mcΔt

Q = Energy in Joules
m = Mass in kilograms
c = Specific heat capacity Jkg-1K-1
Δt = Temperature in degrees

4184Jkg-1K-1 is how much energy that's needed to heat 1kg of water from 0degrees to 1degrees (quite alot of energy). You putting 4.148Jkg-1K-1 says that it only requires 4.148joules of energy to change in 1 degree.