Chapter 24, Problem 67P

(a)

To determine

The intensity of light on the absorbing plate.

Answer to Problem 67P

The intensity of light on the absorbing plate is $625\text{kW}/{\text{m}}^2$.

Explanation of Solution

Given info: The diameter of the circular mirror is $1.00\text{m}$, the radius of the absorbing plate is $2.00\text{cm}$, the amount of water in the plate is $1.00\text{L}$, the initial temperature of the water is $20.0°\text{C}$ and the solar intensity is $1.00\text{kW}/{\text{m}}^2$ .

The formula to calculate the power is,

$P=I\left(\frac{\pi d^2}{4}\right)$

Here,

$I$ is the intensity of the sunlight on the mirror.

$d$ is the diameter of the circular mirror.

Substitute $1.00\text{m}$ for $d$ and $1.00\text{kW}/{\text{m}}^2$ for $I$  in the above equation to find the value of $P$.

$\begin{array}{c}P=\left(1.00\text{kW}/{\text{m}}^2\times \frac{{10}^3\text{W}/{\text{m}}^2}{1\text{kW}/{\text{m}}^2}\right)\left(\frac{\pi {\left(1.00\text{m}\right)}^2}{4}\right)\\ =785.5\text{W}\end{array}$

The formula to calculate the intensity on plate is,

$I_1=\frac{P}{\pi r^2}$ (1)

Here,

$P$ is the power of the radiations.

$r$ is the radius of the plate.

Substitute $2.00\text{cm}$ for $r$ and $785.5\text{W}$ for $P$ in the above equation to find the value of $P$.

$\begin{array}{c}{I}_1=\frac{\left(785.5\text{W}\right)}{\pi {\left(2.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2}\\ =625\times {10}^3\text{W}/{\text{m}}^2\times \frac{{10}^{-3}\text{kW}/{\text{m}}^2}{1\text{W}/{\text{m}}^2}\\ =625\text{kW}/{\text{m}}^2\end{array}$

Conclusion:

Therefore, the intensity of light on the absorbing plate is $625\text{kW}/{\text{m}}^2$.

(b)

To determine

The maximum magnitude of the electric field.

Answer to Problem 67P

The maximum magnitude of the electric field is $21.7\text{kV}/\text{m}$ .

Explanation of Solution

Given info: The diameter of the circular mirror is $1.00\text{m}$, the radius of the absorbing plate is $2.00\text{cm}$, the amount of water in the plate is $1.00\text{L}$, the initial temperature of the water is $20.0°\text{C}$ and the solar intensity is $1.00\text{kW}/{\text{m}}^2$.

The formula to calculate the intensity is,

$I_1=\frac{1}{2}c{\epsilon }_0{E}^2{}_{\text{max}}$ (2)

Here,

${\epsilon }_{\circ }$ is the permittivity of the free space.

$E_{\max }$ is the maximum magnitude of electric field.

Rewrite the equation (2) to find the value of $E_{\max }$.

$\begin{array}{l}{E}^2{}_{\text{max}}=\frac{2I_1}{c{\epsilon }_0}\\ E_{\text{max}}=\sqrt{\frac{2I_1}{c{\epsilon }_0}}\end{array}$

Substitute $625\text{kW}/{\text{m}}^2$ for $I_1$, $3\times {10}^8\text{m}/\text{s}$ for $c$ and $8.85\times {10}^{-12} \text{F}/\text{m}$ for ${\epsilon }_0$ in the above equation to find the value of $E_{\max }$.

$\begin{array}{c}{E}_{\text{max}}=\sqrt{\frac{2\left(625\text{kW}/{\text{m}}^2\right)}{\left(3\times {10}^8\text{m}/\text{s}\right)\left( 8.9\times {10}^{-12} \text{F}/\text{m}\right)}}\\ =21.7\text{kV}/\text{m}\end{array}$

Thus, the maximum magnitude of the electric field is $21.7\text{kV}/\text{m}$.

Conclusion:

Therefore, the maximum magnitude of the electric field is $21.7\text{kV}/\text{m}$.

(c)

To determine

The maximum magnitude of the magnetic field.

Answer to Problem 67P

The maximum magnitude of the electric field is $72.4\text{μT}$.

Explanation of Solution

Given info: The diameter of the circular mirror is $1.00\text{m}$, the radius of the absorbing plate is $2.00\text{cm}$, the amount of water in the plate is $1.00\text{L}$, the initial temperature of the water is $20.0°\text{C}$ and the solar intensity is $1.00\text{kW}/{\text{m}}^2$.

The expression for the magnetic field is,

$B_{\text{max}}=\frac{E_{\text{max}}}{c}$

Here,

$E_{\text{max}}$ is the maximum value of the magnitude of electric field.

$c$ is the speed of the light.

Substitute $21.7\text{kV}/\text{m}$ for $E_{\text{max}}$ and $3\times {10}^8\text{m}/\text{s}$ for $c$  in the above equation to find the value of $B_{\max }$.

$\begin{array}{c}{B}_{\text{max}}=\frac{\left(21.7\text{kV}/\text{m}\times \frac{{10}^3\text{V}/\text{m}}{1\text{kV}/\text{m}}\right)}{\left(3\times {10}^8\text{m}/\text{s}\right)}\\ =72.4\times {10}^{-6}\text{T}\times \frac{{10}^6\text{μT}}{1\text{T}}\\ =72.4\text{μT}\end{array}$

Conclusion:

Therefore, the maximum magnitude of the electric field is $72.4\text{μT}$.

(d)

To determine

The time interval to bring the water to its boiling point.

Answer to Problem 67P

The time interval to bring the water to its boiling point is $17.8\min$.

Explanation of Solution

Given info: The diameter of the circular mirror is $1.00\text{m}$, the radius of the absorbing plate is $2.00\text{cm}$, the amount of water in the plate is $1.00\text{L}$, the initial temperature of the water is $20.0°\text{C}$ and the solar intensity is $1.00\text{kW}/{\text{m}}^2$.

The formula to calculate the power consumed in phase change is,

$P_1=mC\left(\frac{T_1-T_2}{t}\right)$ (3)

Here,

$m$ is the mass of the water.

$C$ is the specific heat capacity of water.

$T_1$ is the boiling point temperature of the water.

$T_2$ is the initial temperature of the water in plate.

$t$ is the time consumed in transition.

The formula to calculate the mass is,

$m=V\rho$

Here,

$V$ is the volume of the plate.

$\rho$ is the density of water.

Substitute $1\text{L}$ for $V$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in the above equation to find the value of $m$.

$\begin{array}{c}m=\left(1\text{L}\times \frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\\ =1\text{kg}\end{array}$

The value of the power consumed in heating is,

$P_1=0.4P$

Substitute $785.5\text{W}$ for $P$ in above equation to find the value of $P_1$.

$\begin{array}{c}{P}_1=0.4\left(785.5\text{W}\right)\\ =314.2\text{W}\end{array}$

Substitute $1\text{kg}$ for $m$, $4186\text{J}/\text{kg}\cdot \text{K}$ for $C$, $20°C$ for $T_2$, $100°C$ for $T_1$ and $314.2\text{W}$ for $P_1$ in the equation (3) to find the value of $t$.

$\begin{array}{c}314.2\text{W}=\left(1\text{kg}\right)\left(4186\text{J}/\text{kg}\cdot \text{K}\right)\left(\frac{100°C-100°C}{t}\right)\\ t=\frac{4186\times 80}{314.2}\\ =1066\sec \times \frac{1\min }{60\sec }\\ =17.8\min \end{array}$

Thus, the time interval to bring the water to its boiling point is $17.8\min$.

Conclusion:

Therefore, the time interval to bring the water to its boiling point is $17.8\min$.