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Chapter 28, Problem 58PQ

(a)

To determine

The power delivered by the power supply for each measurement in Table 28.4.

(a)

Expert Solution
Check Mark

Answer to Problem 58PQ

The power delivered by the power supply for each measurement in Table 28.4 is shown in Table 1.

Explanation of Solution

Power delivered by the power supply is the same as the power radiated by the bulb. The expression for power is given by,

  P=IΔV                                                                                                                     (I)

Here, P is the power, I is the current, and ΔV is the potential difference.

Conclusion:

The power delivered by the power supply for each measurement in Table 28.4 is shown in Table 1.

Table 1

Bulb A (carbon)Bulb B (tungsten)
ΔV(V)I(A)P(W)ΔV(V)I(A)P(W)
100.101.0100.212.1
210.275.7210.306.3
300.4112300.3811
410.5924410.4619
500.7537500.5126
600.9758600.5634
701.1782700.6344
801.36109800.6754
901.55140910.7165
   1010.7778
   1100.7987

Therefore, the power delivered by the power supply for each measurement in Table 28.4 is shown in Table 1.

(b)

To determine

The temperature of each filament for each entry in the table Table 28.4.

(b)

Expert Solution
Check Mark

Answer to Problem 58PQ

The temperature of each filament for each entry in the table Table 28.4 is shown in Table 2.

Explanation of Solution

Given that the emissivity of filaments is 1.

Write the Stefan-Boltzmann equation for the power.

  P=σεAT4                                                                                                                (II)

Here, σ is the Stefan-Boltzmann constant, ε is the emissivity, A is the surface area, and T is the temperature.

Write the expression for the surface area of the filament.

  A=πdl                                                                                                                   (III)

Here, d is the diameter of the filament, and l is its length.

Use equation (III) in (II) and solve for T.

  P=σεπdlT4T=(Pσεπdl)1/4                                                                                                         (IV)

Conclusion:

Substitute 0.045mm for d, 580mm for l, 5.67×108W/m2K4 for σ, and 1 for ε in equation (IV) to find the temperature T.

  T=(Pπ(5.67×108W/m2K4)(1)(0.045mm)(580mm))1/4=(P4.6×1012W/K4)1/4

The power corresponding to each measurement is calculated in part (a), so that the above equation can be utilized to compute the temperatures corresponding to each entry. Ii is tabulated in Table 2.

Table 2

Bulb A (carbon)Bulb B (tungsten)
ΔV(V)I(A)P(W)T(K)ΔV(V)I(A)P(W)T(K)
100.101.0680100.212.1820
210.275.71050210.306.31080
300.41121300300.38111300
410.59241500410.46191400
500.75371700500.51261500
600.97581900600.56341600
701.17822100700.63441800
801.361092200800.67541850
901.551402300910.71651900
    1010.77782000
    1100.79872100

Therefore, the temperature of each filament for each entry in the table Table 28.4 is shown in Table 2.

(c)

To determine

The plot of P as a function of T for each bulb on the same graph.

(c)

Expert Solution
Check Mark

Answer to Problem 58PQ

The plot of P as a function of T for each bulb on the same graph is shown in Figure 1.

Explanation of Solution

The computed values of power P and temperature T is shown in Table 2. The plot of plot of P as a function of T for each bulb on the same graph is shown in Figure 1.

Bundle: Physics For Scientists And Engineers: Foundations And Connections, Volume 2, Loose-leaf Version + Webassign Printed Access Card For Katz's ... And Connections, Single-term Courses, Chapter 28, Problem 58PQ

Conclusion:

Therefore, the plot of P as a function of T for each bulb on the same graph is shown in Figure 1.

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Chapter 28 Solutions

Bundle: Physics For Scientists And Engineers: Foundations And Connections, Volume 2, Loose-leaf Version + Webassign Printed Access Card For Katz's ... And Connections, Single-term Courses

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