Chapter 34, Problem 34.73AP

(a)

To determine

The radius of the hemisphere.

Answer to Problem 34.73AP

The radius of the hemisphere is $0.161\text{m}$.

Explanation of Solution

Given info: The weight of the black cat is $5.50\text{kg}$, the weight of the kittens is $0.800\text{kg}$, the temperature of the hemisphere is $31.0°\text{C}$, the emissivity of the black cat is $0.970$ and the uniform density is $990\text{kg}/{\text{m}}^{\text{3}}$.

Explanation:

The formula to calculate the total mass is,

$m=m_{\text{c}}+4m_{\text{k}}$

Here,

$m_{\text{c}}$ is the mass of cat.

$m_{\text{k}}$ is the mass of the kitten.

Substitute $5.50\text{kg}$ for $m_{\text{c}}$ and $0.800\text{kg}$ for $m_{\text{k}}$ in the above equation to find the value of $m$.

$\begin{array}{c}m=5.50\text{kg}+4\left(0.800\text{kg}\right)\\ =8.7\text{kg}\end{array}$

The formula to calculate the mass of the hemisphere is,

$m=\left(\frac{2}{3}\pi r^3\right)\rho$ (1)

Here,

$r$ is the radius of the hemisphere.

$\rho$ is the density of the hemisphere.

Substitute $8.7\text{kg}$ for $m$ and $990\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in equation (1) to find the value of $r$ .

$\begin{array}{c}8.7\text{kg}=\left(\frac{2}{3}\pi r^3\right)\left(990\text{kg}/{\text{m}}^{\text{3}}\right)\\ r=\sqrt[3]{\frac{\left(8.7\text{kg}\right)\times 3}{\left(990\text{kg}/{\text{m}}^{\text{3}}\right)2\pi }}\\ =0.161\text{m}\end{array}$

Conclusion:

Therefore, the radius of the hemisphere is $0.161\text{m}$.

(b)

To determine

The area of the curved surface.

Answer to Problem 34.73AP

The area of the curved surface is $0.163{\text{m}}^2$ .

Explanation of Solution

Given info: The weight of the black cat is $5.50\text{kg}$, the weight of the kittens is $0.800\text{kg}$, the temperature of the hemisphere is $31.0°\text{C}$, the emissivity of the black cat is $0.970$ and the uniform density is $990\text{kg}/{\text{m}}^{\text{3}}$.

Explanation:

The formula to calculate the area is,

$A=2\pi r^2$

Substitute $0.161\text{m}$ for $r$ in the above equation to find the value of $A$ .

$\begin{array}{c}A=2\pi {\left(0.161\text{m}\right)}^2\\ =0.163{\text{m}}^2\end{array}$

Conclusion:

Therefore, the area of the curved surface is $0.163{\text{m}}^2$.

(c)

To determine

The power emitted by the cats.

Answer to Problem 34.73AP

The power emitted by the cats is $76.8\text{W}$.

Explanation of Solution

Given info: The weight of the black cat is $5.50\text{kg}$, the weight of the kittens is $0.800\text{kg}$, the temperature of the hemisphere is $31.0°\text{C}$, the emissivity of the black cat is $0.970$ and the uniform density is $990\text{kg}/{\text{m}}^{\text{3}}$.

Explanation:

The formula to calculate the power emitted is,

$P=\sigma A{T}^4$

Here,

$\sigma$ is the Stephan’s Boltzmann constant.

$T$ is the temperature of the surroundings.

Substitute $5.54\times 10{ }^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}$ for $\sigma$, $31.0°\text{C}$ for $T$ and $0.163{\text{m}}^2$ for $A$ in the above equation to find the value of $P$ .

$\begin{array}{c}P=\left(5.54\times 10{ }^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}} \right)\left(0.163{\text{m}}^2\right){\left(31.0°+273\text{K}\right)}^4\\ =76.8\text{W}\end{array}$

Conclusion:

Therefore, the power emitted by the cats is $76.8\text{W}$.

(d)

To determine

The intensity of radiation at the surface.

Answer to Problem 34.73AP

The intensity of radiation at the surface is $470\text{W}/{\text{m}}^2$ .

Explanation of Solution

Given info: The weight of the black cat is $5.50\text{kg}$, the weight of the kittens is $0.800\text{kg}$, the temperature of the hemisphere is $31.0°\text{C}$, the emissivity of the black cat is $0.970$ and the uniform density is $990\text{kg}/{\text{m}}^{\text{3}}$.

Explanation:

The formula to calculate the intensity of radiation is,

$I=\frac{P}{A}$

Substitute $76.8\text{W}$ for $P$ and $0.1634{\text{m}}^2$ for $A$ in the above equation to find the value of $I$.

$\begin{array}{l}I=\frac{76.8\text{W}}{0.1634{\text{m}}^2}\\ =470\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Therefore, the intensity of radiation at the surface is $470\text{W}/{\text{m}}^2$.

(e)

To determine

The amplitude of electric field in the electromagnetic wave.

Answer to Problem 34.73AP

The amplitude of electric field in the electromagnetic wave is $595\text{V}/\text{m}$ .

Explanation of Solution

Given info: The weight of the black cat is $5.50\text{kg}$, the weight of the kittens is $0.800\text{kg}$, the temperature of the hemisphere is $31.0°\text{C}$, the emissivity of the black cat is $0.970$ and the uniform density is $990\text{kg}/{\text{m}}^{\text{3}}$.

Explanation:

The formula to calculate the amplitude of the electric field is,

$E=\sqrt{2{\mu }_{\circ }cI}$

Here,

$I$ is the intensity of wave.

$c$ is the speed of the light.

${\mu }_{\circ }$ is the permeability of vacuum.

Substitute $470\text{W}/{\text{m}}^2$ for $I$, $4\pi \times {10}^{-7}$ for ${\mu }_{\circ }$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation to find the value of $E$ .

$\begin{array}{c}E=\sqrt{2\left(4\pi \times {10}^{-7}\right)\left(3\times {10}^8\text{m}/\text{s}\right)\left(470\text{W}/{\text{m}}^2\right)}\\ =595\text{V}/\text{m}\end{array}$

Thus, the amplitude of electric field in the electromagnetic wave is $595\text{V}/\text{m}$ .

Conclusion:

Therefore, the amplitude of electric field in the electromagnetic wave is $595\text{V}/\text{m}$ .

(f)

To determine

The amplitude of magnetic field in the electromagnetic wave.

Answer to Problem 34.73AP

The amplitude of magnetic field in the electromagnetic wave is $1.98\text{μT}$.

Explanation of Solution

Given info: The weight of the black cat is $5.50\text{kg}$, the weight of the kittens is $0.800\text{kg}$, the temperature of the hemisphere is $31.0°\text{C}$, the emissivity of the black cat is $0.970$ and the uniform density is $990\text{kg}/{\text{m}}^{\text{3}}$.

Explanation:

The formula to calculate the amplitude of the magnetic field is,

$B=\frac{E}{c}$

Substitute $595\text{V}/\text{m}$ for $E$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation to find the value of $B$ .

$\begin{array}{c}B=\frac{595\text{V}/\text{m}}{3\times {10}^8\text{m}/\text{s}}\\ =1.98\times {10}^{-6}\text{T}\times \frac{{10}^6\text{μT}}{1\text{T}}\\ =1.98\text{μT}\end{array}$

Conclusion:

Therefore, the amplitude of magnetic field in the electromagnetic wave is $1.98\text{μT}$.

(g)

To determine

The total power radiated by the family of cat.

Answer to Problem 34.73AP

The total power radiated by the family of cat is $119\text{W}$.

Explanation of Solution

Given info: The weight of the black cat is $5.50\text{kg}$, the weight of the kittens is $0.800\text{kg}$, the temperature of the hemisphere is $31.0°\text{C}$, the emissivity of the black cat is $0.970$ and the uniform density is $990\text{kg}/{\text{m}}^{\text{3}}$.

Explanation:

The formula to calculate the mass of the hemisphere is,

$m_{\text{k}}=\left(\frac{2}{3}\pi r_{\text{k}}{}^3\right)\rho$

Here,

$r_{\text{k}}$ is the radius of the hemisphere covered by kittens.

$\rho$ is the density of the hemisphere.

Substitute $0.800\text{kg}$ for $m_{\text{k}}$ and $990\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in equation (1) to find the value of $r_{\text{k}}$ .

$\begin{array}{c}0.800\text{kg}=\left(\frac{2}{3}\pi r_{\text{k}}{}^3\right)\left(990\text{kg}/{\text{m}}^{\text{3}}\right)\\ r_{\text{k}}=\sqrt[3]{\frac{\left(0.800\text{kg}\right)3}{\left(990\text{kg}/{\text{m}}^{\text{3}}\right)2\pi }}\\ =0.156\text{m}\end{array}$

The formula to calculate the power radiated by the kittens is,

$P_{\text{k}}=4\sigma \left(2\pi r_{\text{k}}{}^2\right)T^4$

Here,

$\sigma$ is the Stephan’s Boltzmann constant.

$T$ is the temperature of the surroundings.

Substitute $5.54\times 10{ }^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}$ for $\sigma$, $31.0°\text{C}$ for $T$ and $0.15\text{m}$ for $r_{\text{k}}$ in the above equation to find the value of $P_{\text{k}}$ .

$\begin{array}{c}{P}_{\text{k}}=4\times \left(5.54\times 10{ }^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}} \right)\left(2\pi {\left(0.15\text{m}\right)}^2\right){\left(31.0°+273\text{K}\right)}^4\\ =29.3\text{W}\end{array}$

The formula to calculate the mass of the hemisphere is,

$m_{\text{c}}=\left(\frac{2}{3}\pi r_{\text{c}}{}^3\right)\rho$

Here,

$r_{\text{c}}$ is the radius of the hemisphere covered by kittens.

$\rho$ is the density of the hemisphere.

Substitute $5.5\text{kg}$ for $m_{\text{c}}$ and $990\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in equation (1) to find the value of $r_{\text{c}}$ .

$\begin{array}{c}5.5\text{kg}=\left(\frac{2}{3}\pi r_{\text{c}}{}^3\right)\left(990\text{kg}/{\text{m}}^{\text{3}}\right)\\ r_{\text{c}}=\sqrt[3]{\frac{\left(5.5\text{kg}\right)3}{\left(990\text{kg}/{\text{m}}^{\text{3}}\right)2\pi }}\\ =0.29\text{m}\end{array}$

The formula to calculate the power radiated by the cat is,

$P_{\text{c}}=\sigma \left(2\pi r_{\text{c}}{}^2\right)T^4$

Here,

$\sigma$ is the Stephan’s Boltzmann constant.

$T$ is the temperature of the surroundings.

Substitute $5.54\times 10{ }^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}$ for $\sigma$, $31.0°\text{C}$ for $T$ and $0.29\text{m}$ for $r_{\text{c}}$ in the above equation to find the value of $P_{\text{c}}$ .

$\begin{array}{c}{P}_{\text{c}}=\left(5.54\times 10{ }^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}} \right)\left(2\pi {\left(0.31\text{m}\right)}^2\right){\left(31.0°+273\text{K}\right)}^4\\ =26.4\text{W}\end{array}$

The formula to calculate the total power radiated by the family of cat is,

$P_{\text{T}}=e\left(2\left(P_{\text{c}}+P_{\text{K}}\right)+\frac{\left(P_{\text{c}}+P_{\text{K}}\right)}{6}\right)$

Substitute $0.98$ for $e$, $28.638\text{W}$ for $P_{\text{c}}$ and $29.3\text{W}$ for $P_{\text{k}}$ in the above equation to find the value of $P_{\text{T}}$.

$\begin{array}{c}{P}_{\text{T}}=0.98\left(2\left(28.638\text{W}+29.3\text{W}\right)+\frac{\left(28.638\text{W}+29.3\text{W}\right)}{6}\right)\\ =119\text{W}\end{array}$

Conclusion:

Therefore, the total power radiated by the family of cat is $119\text{W}$.