Calculating Efficiency and Thermal Source Rate in a Steam-Electric Power Plant

[rlc]

Homework Statement

A steam-electric power plant delivers 900 MW of electric power. The surplus heat is exhausted into a river with a flow of 5.51×105 kg/s, causing a change in temperature of 1.35 oC.
What is the efficiency of the power plant?
What is the rate of the thermal source?

Homework Equations

Q=cm(delta T)
c=4.186 J/g*Celsius

The Attempt at a Solution

I know how to answer the first question already:
Q=(4186 J/kg C)(5.51E5 kg/s)(1.35 C)
Q=3113756100 J/s (J/s=Watt)
Convert from Watt to Megawatt --> 3113.76 MW
900+3113.76=4013.76=Total output
900/total output=0.2242=22.42 %

I don't know how to calculate the rate of the thermal source.
What am I missing?

 

[NTW]

Input power = 4013,76 MW = 900 MW + 3113,76 MW

That's the input, i.e., the power delivered by the high-temp source to the machine, and not the output.

The original, high-temp power (4013,76 MW) is transformed by the machine in 900 MW of electricity and 3113,76 of 'rejected', low-temp power.

The calculation of the efficiency is right: 900/Total input

 

[rlc]

That makes sense, but how do you calculate the rate of the thermal source?

 

[NTW]

That makes sense, but how do you calculate the rate of the thermal source?

I don't know what you mean by that. I suspect that it may be the rate of energy delivery of the thermal source. That's the power.

 

[rlc]

Yeah, the question was asked in a weird way.
I tried 4013 MW and it said that that is correct for the rate.

Thank you for helping me!