Chapter 16, Problem 9Q

(a)

To determine

The time for which the sun must shine in order to release energy produced by the mass-energy conversion of a carbon atom with mass $2\times {10}^{-26}\text{ kg}$. The sun’s luminosity is $3.90\times {10}^{26}\text{}\text{ W}$.

Answer to Problem 9Q

Solution:

$4.6\times {10}^{-36}\text{ s}$

Explanation of Solution

Given data:

The sun’s luminosity is $3.90\times {10}^{26}\text{}\text{ W}$, and the mass of carbon atom is $2\times {10}^{-26}\text{ kg}$.

Formula used:

Time taken in seconds is given as:

$t=\frac{E}{L_{\odot }}$

Here, $E$ is the energy released due to mass-energy equivalence in joule and $L_{\odot }$ is the luminosity in joule per second.

Einstein’s mass-energy conversion:

$E=m{c}^2$

Where, E is the energy release from the mass of m and c is speed of light.

Explanation:

Calculating the energy released mass-energy conversion of a carbon atom:

$E=m{c}^2$

Replace $2\times {10}^{-26}\text{ kg}$ for m and $3\times {10}^8{\text{m/s}}^{\text{2}}$ for c in the above equation:

$\begin{array}{c}E=(2\times {10}^{-26}\text{kg}){(}^3\\ =1.8\times {10}^{-9}\text{J}\end{array}$

Recalling the expression for time:

$t=\frac{E}{L_{\odot }}$

Substituting $1.8\times {10}^{-9}\text{ J}$ for $E$ and $3.90\times {10}^{26}\text{}\text{ W}$ for $L_{\odot }$ in the above expression:

$\begin{array}{c}t=\frac{1.8\times {10}^{-9}\text{ J}}{3.90\times {10}^{26}\text{}\frac{\text{J}}{\text{second}}}\\ =4.6\times {10}^{-36}\text{ seconds}\end{array}$

Conclusion:

The sun must shine $4.6\times {10}^{-36}\text{ s}$ to produce the energy produced by a carbon atom with mass $2\times {10}^{-26}\text{ kg}$.

(b)

To determine

The time for which the sun must shine in order to release energy produced by the mass-energy conversion of 1 kg substance. The sun’s luminosity is $3.90\times {10}^{26}\text{}\text{ W}$.

Answer to Problem 9Q

Solution:

$2.3\times {10}^{-10}\text{ s}$

Explanation of Solution

Given data:

The sun’s luminosity is $3.90\times {10}^{26}\text{}\text{ W}$, and the mass of the substance is 1 kg.

Formula used:

Time taken in seconds is given as:

$t=\frac{E}{L_{\odot }}$

Here, $E$ is the energy released due to mass-energy equivalence in joule and $L_{\odot }$ is the luminosity in joule per second.

Einstein’s mass-energy conversion:

$E=m{c}^2$

Where, E is the energy release from the mass of m and c is speed of light.

Explanation:

Calculating the energy released mass-energy conversion of a carbon atom:

$E=m{c}^2$

Replace $\text{1 kg}$ for m and $3\times {10}^8{\text{m/s}}^{\text{2}}$ for c in the above equation:

$\begin{array}{c}E=(1\text{kg}){(}^3\\ =9\times {10}^{16}\text{J}\end{array}$

Recalling the expression for the time:

$t=\frac{E}{L_{\odot }}$

Substituting $9.1\times {10}^{16}\text{ J}$ for $E$ and $3.90\times {10}^{26}\text{}\text{ W}$ for $L_{\odot }$ in the above expression:

$\begin{array}{c}t=\frac{9.1\times {10}^{16}\text{ J }}{3.90\times {10}^{26}\text{}\frac{\text{J}}{\text{second}}}\\ =2.3\times {10}^{-10}\text{ seconds}\end{array}$

Conclusion:

The sun must shine for $2.3\times {10}^{-10}\text{ seconds}$ to produce the energy produced by a substance with mass $\text{1 kg}$.

(c)

To determine

The time for which the sun must shine in order to release energy, produced by the mass-energy conversion of Earth with mass $6\times {10}^{24}\text{ kg}$. The sun’s luminosity is $3.90\times {10}^{26}\text{}\text{ W}$.

Answer to Problem 9Q

Solution:

$\text{1}\text{.4}\times {\text{10}}^{15}\text{ seconds}$

Explanation of Solution

Given data:

The sun’s luminosity is $3.90\times {10}^{26}\text{}\text{ W}$, and the mass of Earth is $6\times {10}^{24}\text{ kg}$.

Formula used:

Time taken in seconds is given as:

$t=\frac{E}{L_{\odot }}$

Here, $E$ is the energy released due to mass-energy equivalence in joule and $L_{\odot }$ is the luminosity in joule per second.

Einstein’s mass-energy conversion:

$E=m{c}^2$

Where, E is the energy release from the mass of m and c is speed of light.

Explanation:

Calculating the energy released mass-energy conversion of a carbon atom:

$E=m{c}^2$

Replace $6\times {10}^{24}\text{ kg}$ for m and $3\times {10}^8{\text{m/s}}^{\text{2}}$ for c in the above equation:

$\begin{array}{c}E=(6\times {10}^{24}\text{ kg}){(}^3\\ =5.4\times {10}^{41}\text{J}\end{array}$

Recalling the expression for time:

$t=\frac{E}{L_{\odot }}$

Substituting $5.4\times {10}^{41}\text{ J}$ for $E$ and $3.90\times {10}^{26}\text{}\text{ W}$ for $L_{\odot }$ in the above expression:

$\begin{array}{c}t=\frac{5.4\times {10}^{41}\text{ J}}{3.90\times {10}^{26}\text{}\frac{\text{J}}{\text{second}}}\\ \text{=1}\text{.4}\times {\text{10}}^{15}\text{ seconds}\end{array}$

Conclusion:

The sun must shine for $\text{1}\text{.4}\times {\text{10}}^{15}\text{ seconds}$ to produce the energy produced by a substance with mass $6\times {10}^{24}\text{ kg}$.