Question

The Prairie Island nuclear power plant near Red Wing, Minnesota, produces about 2×10⁹watts. A watt is one Joule per second.

b. Calculate the Joules of energy produced by the Prairie Island plant in one day.

c. Recall that E = m c². Use E from part b. Calculate m, the minimum mass that was converted to energy in one day. (Because of inefficiency, the actual mass converted must be larger.)

d. During fission of U-235, 0.1% of uranium's mass is converted to energy.  What mass of U-235 is required for the mass loss you calculated in part c?

e.  Combustion of bituminous coal produces 30 kJ/g (Chemistry in Context Table 5.2). That is heat, not electrical power, but ignore inefficiency and suppose that all the heat is converted to electrical energy. What mass of coal will produce the daily energy calculated in part b?

Answer 1

Given,

Power= 2× 109 W

And 1 W= 1 J/s

We know that, Energy (E) = P× t

For a day, t= 24×60×60 sec= 86400 sec

Therefore, Energy produced in joules in one day= 2× 109× 86400 J

= 18835200 J

From Einstein equation: E= mc2

Where E is energy

m is the mass

c is the speed of light.

m= E/c× c

m= 18835200/( 3× 10⁸)²

m= 2.0926× 10^-9 g

Now, we have mass loss during fission of U-235

Percentage of mass loss= 0.1%

Initial mass(Mi) = 2.0926× 10^-8

Final mass (Mf) = x

We have mass loss formula:

Percentage loss = (Mi-Mf))Mi × 100

0.1= (2.0926× 10^-8 - x)/2.0926× 10^-8 × 100

x= 18.836× 10^-8 g

For the last part:

Heat = 30 KJ/g

This heat is converted into electrical energy=

Energy produced= mass× heat

= 2.0926× 10^-8× 30× 10³ . As H= 30 KJ/g

= 62.778× 10^-6 J

Using Energy mass relationship:

To calculate the mass of coal produced from above calculated energy value.

E= mc2

62.778× 10^-6= m(3× 10^-8)²

m=( 62.778/9)× 10^-22 g

= 6.9753 × 10^{-22 _g}