# How long will it take for 1/2 of this reactor's cooling water to boil off?

runningphysics
Homework Statement:
If 4.5×10^5kg of emergency cooling water at 10 ∘C are dumped into a malfunctioning nuclear reactor whose core is producing energy at the rate of 200 MW , and if no circulation or cooling of the water is provided, how long will it be before half the water has boiled away?
Relevant Equations:
Q1=Lm
Q2=mc T
m=4.5*10^5
L=2257 kj/kg
c=4.184 kj/kg*K
I tried using the equation Q'*t= Q1+Q2. Where Q' is the energy of the reactor aka 200,000 kJ and t is the time. Take Q1 to be (1/2m*2257) and Q2 to be (1/2m*4.184*90). The 90 is the change in temperature for the phase change to occur from liquid water to gas, or boiling. Plugged everything in and got 2962.755 seconds. Convert to minutes by dividing by 60 seconds to get 49.379 minutes. This answer was wrong. Can anyone tell me what is wrong with this setup?

runningphysics
I figured it out. What needs to happen the mass for Q1 needs to be the total mass divided by 2. The Q2 mass needs to be the total mass. This is because 1/2 of the water is being boiled off, but the total amount of water needs to reach the 100 degrees Celsius.

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