Chapter 22, Problem 81P

(a)

To determine

The rate of energy absorbed by the water.

Answer to Problem 81P

The rate of energy absorbed by the water is $0.92\text{kW}$.

Explanation of Solution

Write the expression for rate of energy absorption.

  $\frac{Q}{\Delta t}=\frac{m{c}_{\text{w}}\Delta T}{\Delta t}$                                                                            (I)

Here, $Q/\Delta t$ is the rate of energy absorption, $m$ is the mass of heat, $c_{\text{w}}$ is the specific heat capacity of water, and $\Delta T$ is the change in temperature.

Conclusion:

Substitute $350\text{g}$ for $m$, $4186\text{J/kg}\cdot \text{K}$ for $c_{\text{w}}$, $100.0°\text{C}-25.0°\text{C}$ for $\Delta T$, and $2.00\text{min}$ for $\Delta t$ in equation (I).

  $\begin{array}{c}\frac{Q}{\Delta t}=\frac{\left(350\text{g}\right)\left(\frac{0.001\text{kg}}{1\text{g}}\right)\left(4186\text{J/kg}\cdot \text{K}\right)\left(100.0°\text{C}-25.0°\text{C}\right)}{\left(2.00\text{min}\right)\left(\frac{60.0\text{s}}{1\text{min}}\right)}\\ =\frac{\left(0.350\text{kg}\right)\left(4186\text{J/kg}\cdot \text{K}\right)\left(75\text{K}\right)}{\left(120.0\text{s}\right)}\\ =915.7\text{W}\\ =0.92\times {10}^3\text{W}\end{array}$

Convert the rate of energy absorption into kilowatt.

  $\begin{array}{c}\frac{Q}{\Delta t}=0.92\times {10}^3\text{W}\left(\frac{1\text{kW}}{{10}^3\text{W}}\right)\\ =0.92\text{kW}\end{array}$

Therefore, the rate of energy absorbed by the water is $0.92\text{kW}$.

(b)

To determine

The average intensity of the microwaves.

Answer to Problem 81P

The average intensity of the microwaves is $1.0\times {10}^5{\text{W/m}}^2$.

Explanation of Solution

Write the expression for average intensity of the microwaves.

  $I=\frac{P}{A}$                                                                                         (II)

Here, $I$ is the average intensity of the microwaves, $P$ is the power, and $A$ is the area.

Write the expression for power.

  $P=\frac{Q}{\Delta t}$                                                                                      (III)

Substitute equation (III) in the equation (II).

  $I=\frac{Q}{A\Delta t}$                                                                                     (IV)

Substitute equation (I) in the equation (IV).

  $I=\frac{m{c}_{\text{w}}\Delta T}{A\Delta t}$                                                                                 (V)

Conclusion:

Substitute $350\text{g}$ for , $4186\text{J/kg}\cdot \text{K}$ for , $100.0°\text{C}-25.0°\text{C}$ for , $88.0{\text{cm}}^2$ for $A$, and $2.00\text{min}$ for $\Delta t$ in above equation to find $I$.

  $\begin{array}{c}I=\frac{\left(350\text{g}\right)\left(\frac{0.001\text{kg}}{1\text{g}}\right)\left(4186\text{J/kg}\cdot \text{K}\right)\left(100.0°\text{C}-25.0°\text{C}\right)}{\left(88.0{\text{cm}}^2\right)\left(\frac{{10}^{-4}{\text{m}}^2}{1{\text{cm}}^2}\right)\left(2.00\text{min}\right)\left(\frac{60.0\text{s}}{1\text{min}}\right)}\\ =\frac{\left(0.350\text{kg}\right)\left(4186\text{J/kg}\cdot \text{K}\right)\left(75\text{K}\right)}{\left(88.0\times {10}^{-4}{\text{m}}^2\right)\left(120.0\text{s}\right)}\\ =10.4\times {10}^4{\text{W/m}}^2\\ =1.0\times {10}^5{\text{W/m}}^2\end{array}$

Therefore, the average intensity of the microwaves is $1.0\times {10}^5{\text{W/m}}^2$.

(c)

To determine

The rms electric and magnetic fields.

Answer to Problem 81P

The rms electric field inside the waveguide is $6.3\text{kV/m}$ and the magnetic field is $2.1\times {10}^{-5}\text{T}$.

Explanation of Solution

Write the expression for rms electric field inside the waveguide.

  $I={\epsilon }_0{E}_{\text{rms}}^{2}c$                                                                                    (VI)

Here, ${\epsilon }_0$ is the permittivity of free space, $E_{\text{rms}}$ is the rms electric field, and $c$ is the velocity of light.

Compare the equation (V) and (VI) to solve for $E_{\text{rms}}$.

  $\begin{array}{c}{\epsilon }_0{E}_{\text{rms}}^{2}c=\frac{m{c}_{\text{w}}\Delta T}{A\Delta t}\\ E_{\text{rms}}^2=\frac{m{c}_{\text{w}}\Delta T}{{\epsilon }_{0}cA\Delta t}\\ E_{\text{rms}}=\sqrt{\frac{m{c}_{\text{w}}\Delta T}{{\epsilon }_{0}cA\Delta t}}\end{array}$                                                                       (VII)

Write the rms magnetic field inside the waveguide.

  $B_{\text{rms}}=\frac{E_{\text{rms}}}{c}$                                                                                     (VIII)

Here, $B_{\text{rms}}$ is the rms magnetic field.

Conclusion:

Substitute $350\text{g}$ for , $4186\text{J/kg}\cdot \text{K}$ for , $100.0°\text{C}-25.0°\text{C}$ for , $8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$, $3.00\times {10}^8\text{m/s}$ for $c$, $88.0{\text{cm}}^2$ for $A$, and $2.00\text{min}$ for  in above equation to find $E_{\text{rms}}$ in equation (VII).

  $\begin{array}{c}{E}_{\text{rms}}=\sqrt{\frac{\left(350\text{g}\right)\left(\frac{0.001\text{kg}}{1\text{g}}\right)\left(4186\text{J/kg}\cdot \text{K}\right)\left(100.0°\text{C}-25.0°\text{C}\right)}{\left(8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(3.00\times {10}^8\text{m/s}\right)\left(88.0{\text{cm}}^2\right)\left(\frac{{10}^{-4}{\text{m}}^2}{1{\text{cm}}^2}\right)\left(2.00\text{min}\right)\left(\frac{60.0\text{s}}{1\text{min}}\right)}}\\ =\sqrt{\frac{\left(0.350\text{kg}\right)\left(4186\text{J/kg}\cdot \text{K}\right)\left(75.0\text{K}\right)}{\left(8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(3.00\times {10}^8\text{m/s}\right)\left(88.0\times {10}^{-4}{\text{m}}^2\right)\left(120.0\text{s}\right)}}\\ =0.63\times {10}^4\text{V/m}\end{array}$

Convert rms electric field into kV/m.

  $\begin{array}{c}{E}_{\text{rms}}=6.3\times {10}^3\text{V/m}\left(\frac{{10}^{-3}\text{kV/m}}{1\text{V/m}}\right)\\ =6.3\text{kV/m}\end{array}$

Substitute $6.3\times {10}^3\text{V/m}$ for $E_{\text{rms}}$ and $3.00\times {10}^8\text{m/s}$ for $c$ in equation (VIII) to find $B_{\text{rms}}$.

  $\begin{array}{c}{B}_{\text{rms}}=\frac{6.3\times {10}^3\text{V/m}}{3.00\times {10}^8\text{m/s}}\\ =2.1\times {10}^{-5}\text{T}\end{array}$

Therefore, the rms electric field inside the waveguide is $6.3\text{kV/m}$ and the magnetic field is $2.1\times {10}^{-5}\text{T}$.