Thanks!...I found the Qh by using the Qh=mh equation...
I found the Work by saying that there is 600MW of Power a day...so, 6 x 10^8 Watts of Power a day, 3600 times 24 will equal 8.64 x10^4 seconds in a day...
P=W/T 6 x 10^8W (8.64 x 10^4s) = Work
And then I have to solve for Qc...
Seriously Stuck..
A coal fired plant generates 600MW of electric power. The plant uses 4.8 x 10^6 kg of coal a day. The heat of combustion of coal is 3.3 x 10^7 J/kg. The steam that drives the turbines is at a temperature of 573K, and the exhaust water is 310K.
1.) What is the overall...