Chapter 16, Problem 46A

(a)

To determine

The amount of energy released.

Answer to Problem 46A

The amount of energy released is $9\times {10}^{13}\text{ J}$ .

Explanation of Solution

Given:

The mass to energy in a fission reactor is $\frac{1}{1000}$ .

The mass of uranium is $m=1\text{ kg}$ .

Formula used:

The expression for the amount of energy is,

  $E=m{c}^2$

Here, $m$ is the mass and $c$ is the speed of the light.

Calculation:

Consider the speed of light as $c=3\times {10}^8\text{m}/\text{s}$ .

The amount of energy released by converting $0.1\%$ of mass is,

  $\begin{array}{l}E=m{c}^2\\ E=\frac{1}{1000}\times \left(1\text{ kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ E=9\times {10}^{13}\text{ J}\end{array}$

Conclusion:

Thus, the amount of energy released is $9\times {10}^{13}\text{ J}$ .

(b)

To determine

The cost of energy in dollars.

Answer to Problem 46A

The cost of energy in dollars is $2.7\times {10}^6\text{ dollars}$ .

Explanation of Solution

Given:

The mass to energy in a fission reactor is $\frac{1}{1000}$ .

The mass of uranium is $m=1\text{ kg}$ .

The cost of energy is 3 cents per MJ.

Calculation:

From part (a), the amount of energy released is $E=9\times {10}^{13}\text{ J}$ .

The cost of energy in dollars is,

  $\begin{array}{l}\text{Cost of energy in dollars}=E\times \text{Cost of energy in cents per MJ}\\ \text{Cost of energy in dollars}=\left(9\times {10}^{13}\text{ J}\right)\times \left(3\text{cents}/\text{MJ}\right)\times \frac{1\text{ dollar}}{100\text{ cents}}\times \frac{1\text{ MJ}}{{10}^6\text{J}}\\ \text{Cost of energy in dollars}=2.7\times {10}^6\text{ dollars}\end{array}$

Conclusion:

Thus, the cost of energy in dollars is $2.7\times {10}^6\text{ dollars}$ .