Nuclear Missles and destruction?

[PhysicsJoe101]

Now a power plant produces 15 million kilowatt-hours of electricity per day, enough to power a city like Salem. So this leaves me with three questions, let’s see if you guys can help me out with them.

What would the amount of energy be when converted to Joules?

How much does the above energy weigh?

The energy given off by nuclear weapons is measured in kilotons, where 1 kiloton equals 4 x 1012 Joules. How much mass is converted to energy in a 900 kiloton bomb?

Knowing that a lot of nuclear weapons exist, I want to see what the practical implications are for safety.
Thanks in advance.

PhysicsJoe

 

[Astronuc]

1 Joule = 1 W-s, so 1 kJ = 1 kW-s, where s = second. 1 Watt (W) = 1 Joule/sec.

Energy does not weigh anything. Matter in a gravity field has weight = mass * acceleration of gravity.

In nuclear fission, e.g. in U-235, approximately 200 MeV is released, of which about 175 MeV goes to the kinetic energy of the fission products and prompt gammas. The composite nucleus (of U-235 + n) has atomic mass of 236 amu, and 1 amu = 1.66053886 × 10⁻²⁷ kg. 1 MeV = 1.6022×10⁻¹³J

or the specific energy of fission is approximately 7.155 x 10⁷ J/kg. or 71.55 MJ/kg.

 

[russ_watters]

A typical nuclear power plant produces more like 50 million kWh per day.

 

[Morbius]

Now a power plant produces 15 million kilowatt-hours of electricity per day, enough to power a city like Salem. So this leaves me with three questions, let’s see if you guys can help me out with them.

What would the amount of energy be when converted to Joules?

15e+06 kilowatt-hours = 15e+06 kilowatt-hours * 1000 watts/kilowatt * 3600 seconds/hour
= 5.4e+13 watt-sec = 5.4e+13 Joules

How much does the above energy weigh?

Energy doesn't "weigh" anything! Weight is how much force the Earth exerts on a quantity of matter.

The energy given off by nuclear weapons is measured in kilotons, where 1 kiloton equals 4 x 1012 Joules. How much mass is converted to energy in a 900 kiloton bomb?

For this you use Einstein's famous equation E = mc^2

E = 900 kilotons = 900 kilotons * 4e+12 Joules/kiloton = 3.6e+15 Joules
= 3.6e+15 kg-m^2/s^2
= m c^2 = m ( 2.998e+08 m/s)^2 = m ( 8.988e+16 m^2/s^2)

Solving for m; m = 0.04 kg = 40 grams.

Knowing that a lot of nuclear weapons exist, I want to see what the practical implications are for safety.

Safety of what? Safety of nuclear reactors? The safety of a nuclear reactor has
absolutely NOTHING to do with how much mass is converted to energy in a weapon.

Nuclear reactors and nuclear weapons are VERY, VERY, DIFFERENT animals.

Dr. Gregory Greenman
Physicist

 

[Morbius]

A typical nuclear power plant produces more like 50 million kWh per day.

Russ,

Correct.

A typical nuclear power reactor produces about 1000 Megawatts = 1 Million Kilowatts.

In a day, there are 24 hours. So in a day a typical 1 Gw(e) = 1000 Mw(e) power plant
produces 24 Million kWh per day.

1000 Mw(e) is a round number. [Some larger reactors will do about 1300 Mw(e).]

Many, if not most; nuclear power plants have 2 reactors or 2 Units.

If each unit is 1000 Mw(e); then a 2 unit plant will produce 48 million kWh per day.

Dr. Gregory Greenman
Physicist

 

[Morbius]

or the specific energy of fission is approximately 7.155 x 10⁷ J/kg. or 71.55 MJ/kg.

Astronuc,

The specific energy of fission is how much energy you get from fissioning a given amount
of Uranium. However, there's still mass in the form of fission products on the product
side of the equation in addition ot the energy released.

I believe the question is not how much energy we can get from a given mass of uranium.

The question is how much mass was converted to energy.

For that; we use Einstein's famous equation E = mc^2

Dr. Gregory Greenman
Physicist

 

[theCandyman]

But if PhysicsJoe101 wants to look at other yield weapons, all he has to do is divide by that value Astronuc gave him to find the minimum amount of U-235 that was fissioned.

 

[PhysicsJoe101]

Thanks

Wow, thanks for the answers guys.