How High Can a Nuclear Bomb Energy Lift a Mass of Water?

[MarcL]

Homework Statement

A nuclear bomb containing 6.5kg of plutonium explodes. The sum of the rest masses of the products of the explosion is less than the original rest mass by one part in 10^4 .

c) What mass of water could the released energy lift to a height of 3.1 km?

Homework Equations

Well this is where I am stuck, I don't know what to use to plug in the height :/ ( I was given E=mc^2, E= mc^2 + k , etc...)

The Attempt at a Solution

There was an a and a b before hand ( the energy released from the explosion which was 5.85 * 10^13 J and the average power over 3.7 microseconds which was 1.58*10^19 Watts) but I really cannot seem to figure out what to do because I have no equations ( or none that I know of) that include the height.

 

[SammyS]

Homework Statement

A nuclear bomb containing 6.5kg of plutonium explodes. The sum of the rest masses of the products of the explosion is less than the original rest mass by one part in 10^4 .

c) What mass of water could the released energy lift to a height of 3.1 km?

Homework Equations

Well this is where I am stuck, I don't know what to use to plug in the height :/ ( I was given E=mc^2, E= mc^2 + k , etc...)

The Attempt at a Solution

There was an a and a b before hand ( the energy released from the explosion which was 5.85 * 10^13 J and the average power over 3.7 microseconds which was 1.58*10^19 Watts) but I really cannot seem to figure out what to do because I have no equations ( or none that I know of) that include the height.

Do you know the gravitational potential energy of a object with mass, m, raised to a height h ?


In my opinion, physics should be more than searching for equations/formulas which we can plug some random data into.

 

[MarcL]

I tried the traditional Pe=mgh... it didn't work.

 

[MarcL]

I don't think it's about formula and plugging it in. I understand what I am doing. I think I didn't make my statement clear. First of all I hate doing something that I don't understand ( i.e: plug in) I feel like a robot. But that is beside the point... I just do not know how to approach that part of the problem..

 

[SammyS]

I tried the traditional Pe=mgh... it didn't work.

I'm pretty sure that it is the traditional P.E. = mgh .

Show your attempt at working with this.

 

[MarcL]

I used the initial amount of Joules that I found ( so E=mc^2 which gave me 5.85*10^13J). I then plugged it in the equation knowing that no energy is loss or created ( it still applies in relativity I believe?). So (5.85*10^13) / ((9.82m/s^2)(3100m)) = 1.9*10^9 kg

 

[SammyS]

I used the initial amount of Joules that I found ( so E=mc^2 which gave me 5.85*10^13J). I then plugged it in the equation knowing that no energy is loss or created ( it still applies in relativity I believe?). So (5.85*10^13) / ((9.82m/s^2)(3100m)) = 1.9*10^9 kg

That all looks fine to me.