Thermodynamics: Waste heat and total ice converted to steam

1. Oct 25, 2012

jolierouge

1. The problem statement, all variables and given/known data
A coal-fi red plant generates 600 MW of electric power. The plant uses
4:8106 kg of coal each day. The heat produced by the combustion of coal is 3:3107 J/kg.
The steam that drives the turbines is at a temperature of 300C, and the exhaust water is at
37C. (a) How much thermal energy is exhausted each day by this plant? (b) How much ice
at -10C can be turned into steam at 100C per year by the exhaust from this power plant?

2. Relevant equations
Q=mL
Q=mcΔT

3. The attempt at a solution
For question a) all I did was take the coal used each day and multiplied it by the heat produce by its combustion. I found the thermal energy exhausted each day to be 1.584 x 10^14J. (I think this is right)
Question B is where I am having troubles.
This is the equation I got:
Q=mCice(0-(-10))+mLice+mCwater(100-0)+mLvaporization
q/(Cice(10)+Lice+Cwater(100)+Lvice)=m
where:
Lice=334 J/g
Cice=2.1J/gK
Cwater=4.2 J/gk
Lvice=2.256J/g
Q=1.584 x 10^14
Is this right? I getting confused about what energy I should and shouldn't take when trying to find the mass.

2. Oct 25, 2012

haruspex

That's the energy released by burning the coal, but some went into producing electricity. The exhausted heat is whatever's left.
You switched Lvap to Lvice in there, and the value seems much too low.

3. Oct 25, 2012

jolierouge

So, for the first part I take the "Q" I calculated and subtract the energy converted to energy which is 600MW (which is 600*10^6J) and I should be left with the thermal energy exhausted? And I use this for the q in the second part?

Ah, I see where I switched Lvap. And I see why my Lvap is too low...conversion issues(my bad).

(By the way thanks!)