Posted by **Nancy** on Saturday, November 10, 2012 at 8:23pm.

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**, 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 -
**Nancy**, Sunday, November 11, 2012 at 9:08am
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**, Friday, August 26, 2016 at 3:31pm
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.

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