Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 39, Problem 8P

(a)

To determine

The power radiated by the black body.

(a)

Expert Solution
Check Mark

Answer to Problem 8P

The power radiated by the black body is 70,870W.

Explanation of Solution

Write the expression for the power radiated by the black body.

    P=σAeT4                                                                                                         (I)

Here, P is the power radiated by the black body, A is the area, σ is the Stefan-Boltzmann constant, e is the emissivity and T is the temperature.

Conclusion:

Substitute 5.6696×108W/m2K4 for σ, 20.0cm2 for A, 1 for e and 5000K for T in equation (I) to find P.

    P=(5.6696×108W/m2K4)(20.0cm2)(1)(5000K)4=(5.6696×108W/m2K4)((20.0cm2)(104m21cm2))(1)(5000K)4=70,870W

Thus, the power radiated by the black body is 70,870W.

(b)

To determine

The wavelength at which the blackbody radiate most intensely.

(b)

Expert Solution
Check Mark

Answer to Problem 8P

The wavelength at which the blackbody radiate most intensely is 580nm.

Explanation of Solution

Write the equation for the wavelength at which the blackbody radiate most intensely.

    λmax=2.898×103mKT                                                                        (II)

Here, λmax is the maximum wavelength.

Conclusion:

Substitute 5000K for T in equation (II) to find λmax.

    λmax=2.898×103mK5000K=(5.796×107m)(109nm1m)580nm

Thus, the wavelength at which the blackbody radiate most intensely is 580nm.

(c)

To determine

The spectral power per wavelength interval at 580nm.

(c)

Expert Solution
Check Mark

Answer to Problem 8P

The spectral power per wavelength interval at 580nm is 8.0×1010W/m.

Explanation of Solution

Write the equation for C.

    C=hckBT                                                                                             (III)

Here, C is the constant, h is the Plank’s constant, c is the speed of light and kB is the Boltzmann constant.

Write the equation for D.

    D=2πhc2A                                                                                        (IV)

Here, D is a constant.

Write the equation for the power per wavelength interval.

    P(λ)=2πhc2Aλ5[exp(hc/λkBT)1]

Substitute C for hckBT and D for 2πhc2A.

    P(λ)=Dλ5[exp(C/λ)1]                                                                            (V)

Conclusion:

Substitute 6.626×1034m2kg/s for h, 3×108m/s for c, 1.38×1023J/K for kB and 5000K for T in equation (III) to find C.

    C=(6.626×1034m2kg/s)(3×108m/s)(1.38×1023J/K)(5000K)=0.00288×103m=2.88×106m

Substitute 6.626×1034m2kg/s for h, 3×108m/s for c and 20.0cm2 for T in equation (IV) to find C.

    D=2π(6.626×1034m2kg/s)(3×108m/s)2(20.0cm2)=2(3.14)(6.626×1034m2kg/s)(3×108m/s)2((20.0cm2)(104m21cm2))=7.50×1019Jm2/s

Substitute 2.88×106m for C, 7.50×1019Jm2/s for D and 580nm for λ in equation (V) to find P(λ).

    P(λ)=7.50×1019Jm2/s(580nm)5[exp(2.88×106m580nm)1]=7.50×1019Jm2/s((580nm)(109m1nm))5[exp(2.88×106m(580nm)(109m1nm))1]=7.99×1010W/m8.0×1010W/m

Thus, the spectral power per wavelength interval at 580nm is 8.0×1010W/m.

(d)

To determine

The spectral power per wavelength interval at 1.00nm.

(d)

Expert Solution
Check Mark

Answer to Problem 8P

The spectral power per wavelength interval at 1nm is 9.42×101226W/m.

Explanation of Solution

From the equation (V) in part (c), the spectral power per wavelength is.

    P(λ)=Dλ5[exp(C/λ)1]

Conclusion:

Substitute 2.88×106m for C, 7.50×1019Jm2/s for D and 1.00nm for λ in above equation to find P(λ).

P(λ)=7.50×1019Jm2/s(1nm)5[exp(2.88×106m1nm)1]=7.50×1019Jm2/s((1nm)(109m1nm))5[exp(2.88×106m(1nm)(109m1nm))1]=9.42×101226W/m

Thus, the spectral power per wavelength interval at 1nm is 9.42×101226W/m.

(e)

To determine

The spectral power per wavelength interval at 5.00nm.

(e)

Expert Solution
Check Mark

Answer to Problem 8P

The spectral power per wavelength interval at 5.00nm is 1.00×10227W/m.

Explanation of Solution

From the equation (V) in part (c), the spectral power per wavelength is.

    P(λ)=Dλ5[exp(C/λ)1]

Conclusion:

Substitute 2.88×106m for C, 7.50×1019Jm2/s for D and 5.00nm for λ in above equation to find P(λ).

    P(λ)=7.50×1019Jm2/s(5nm)5[exp(2.88×106m5nm)1]=7.50×1019Jm2/s((5nm)(109m1nm))5[exp(2.88×106m(5nm)(109m1nm))1]=1.00×10227W/m

Thus, the spectral power per wavelength interval at 5.00nm is 1.00×10227W/m.

(f)

To determine

The spectral power per wavelength interval at 400nm.

(f)

Expert Solution
Check Mark

Answer to Problem 8P

The spectral power per wavelength interval at 400nm is 5.44×1010W/m.

Explanation of Solution

From the equation (V) in part (c), the spectral power per wavelength is.

    P(λ)=Dλ5[exp(C/λ)1]

Conclusion:

Substitute 2.88×106m for C, 7.50×1019Jm2/s for D and 400nm for λ in above equation to find P(λ).

    P(λ)=7.50×1019Jm2/s(400nm)5[exp(2.88×106m400nm)1]=7.50×1019Jm2/s((400nm)(109m1nm))5[exp(2.88×106m(400nm)(109m1nm))1]=5.44×1010W/m

Thus, the spectral power per wavelength interval at 400nm is 5.44×1010W/m.

(g)

To determine

The spectral power per wavelength interval at 700nm.

(g)

Expert Solution
Check Mark

Answer to Problem 8P

The spectral power per wavelength interval at 700nm is 7.38×1010W/m.

Explanation of Solution

From the equation (V) in part (c), the spectral power per wavelength is.

    P(λ)=Dλ5[exp(C/λ)1]

Conclusion:

Substitute 2.88×106m for C, 7.50×1019Jm2/s for D and 700nm for λ in above equation to find P(λ).

    P(λ)=7.50×1019Jm2/s(700nm)5[exp(2.88×106m700nm)1]=7.50×1019Jm2/s((700nm)(109m1nm))5[exp(2.88×106m(700nm)(109m1nm))1]=7.38×1010W/m

Thus, the spectral power per wavelength interval at 700nm is 7.38×1010W/m.

(h)

To determine

The spectral power per wavelength interval at 1.00mm.

(h)

Expert Solution
Check Mark

Answer to Problem 8P

The spectral power per wavelength interval at 1.00mm is 0.260W/m.

Explanation of Solution

From the equation (V) in part (c), the spectral power per wavelength is.

    P(λ)=Dλ5[exp(C/λ)1]

Conclusion:

Substitute 2.88×106m for C, 7.50×1019Jm2/s for D and 1.00mm for λ in above equation to find P(λ).

    P(λ)=7.50×1019Jm2/s(1.00mm)5[exp(2.88×106m1.00mm)1]=7.50×1019Jm2/s((1.00mm)(103m1mm))5[exp(2.88×106m(1.00mm)(103m1mm))1]=0.260W/m

Thus, the spectral power per wavelength interval at 1.00mm is 0.260W/m.

(I)

To determine

The spectral power per wavelength interval at 10.0cm.

(I)

Expert Solution
Check Mark

Answer to Problem 8P

The spectral power per wavelength interval at 10.0cm is 2.60×109W/m.

Explanation of Solution

From the equation (V) in part (c), the spectral power per wavelength is.

    P(λ)=Dλ5[exp(C/λ)1]

Conclusion:

Substitute 2.88×106m for C, 7.50×1019Jm2/s for D and 10.0cm for λ in above equation to find P(λ).

    P(λ)=7.50×1019Jm2/s(10.0cm)5[exp(2.88×106m10.0cm)1]=7.50×1019Jm2/s((10.0cm)(102m1cm))5[exp(2.88×106m(10.0cm)(102m1cm))1]=2.60×109W/m

Thus, the spectral power per wavelength interval at 10.0cm is 2.60×109W/m.

(J)

To determine

The power radiated by the object as visible light.

(J)

Expert Solution
Check Mark

Answer to Problem 8P

The power radiated by the object as visible light is 19kW.

Explanation of Solution

The wavelengths 400nm and 700nm comes in the visible region. Thus, when taking power, the average power at these wavelengths are used.

Write the equation for the power radiated.

    P=P(λ¯)Δλ (VI)

Here, P is the radiated power in the visible region, P(λ¯) is the average power in the visible area and Δλ is the change in wavelength.

Conclusion:

The average power is calculated from the visible area.

  P(λ¯)=5.44×1010W/m+7.38×1010W/m2=6.41×1010W/m

Substitute 6.41×1010W/m for P(λ¯) and 700nm400nm for Δλ in equation (VI) to find P.

    P=(6.41×1010W/m)(700nm400nm)=(6.41×1010W/m)((300nm)(109m1nm))=(1.92×104W)(103kW1W)=19kW

Thus, the power radiated by the object as visible light is 19kW.

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