Posted by Nancy on .
A typical nuclear reactor produces 1mw of power per day. What is the minimum rate of mass loss required to produce this energy.
Don't know where to start

nuclear chemistry 
andrew,
E=mc^2
10^6 J/s are being produced (1Mw)
10^6 J/s = m (3.0e8 m/s)
solve for mass to find amount of mass lost in one second.
~ .0033kg 
nuclear chemistry 
Nancy,
wouldn't you divide by the speed of light squared?
nuclear chemistry  andrew, Saturday, November 10, 2012 at 8:26pm
E=mc^2
10^6 J/s are being produced (1Mw)
10^6 J/s = m (3.0e8 m/s)
solve for mass to find amount of mass lost in one second.
~ .0033kg 
nuclear chemistry 
Colin,
I know this is too late for you Nancy and you Andrew, but I thought it might be useful for other students out there.
So, to solve this problem we:
First convert 1.0 MW to W then to J/s
1.0 MW x 10^6 W/MW x ((1.00 J/S)/W) = 1.0 x 10^6 J/s
Next we use the binding energy equation to find how many Kg/s are required to power the reactor:
E=mc^2 where E= 1.0 x 10^6 J/s now it's important to remember that
1J= 1Kg x m^2/s^2 so if we replace the J in the numerator of our units we get E= (1.0 x 10^6 kg x m^2/s^3)
c= 2.998 x 10^8 m/s
we solve for m to get m= E/(c^2) now plug in you values and solve for m.
m= (1.0 x 10^6 kg x m^2/s^3)/((2.998 m/s)^2)
and find m= 1.1 x 10^11 kg/s
From here it's just simple stoichiometry to find the mass rate loss in g/day.
The final answer should be
9.6 x 10^4 g/day.
Hope this helps some folks.