Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
Question
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Chapter 14, Problem 90P

(a)

To determine

The rate of heat conduction along the bar.

(a)

Expert Solution
Check Mark

Answer to Problem 90P

The rate of heat conduction along the bar is 0.32 W .

Explanation of Solution

Write the Fourier’s law of heat conduction.

=κAΔTd (I)

Here, is the rate of heat conduction, κ is the thermal conductivity of material of the bar, A is the cross sectional area, ΔT is the change in temperature and d is the length of the bar

Write the equation for ΔT .

ΔT=T1T2 (II)

Here, T1 is the temperature of the first end of the bar and T2 is the temperature of  the second end of the bar

Put equation (II) in equation (I).

=κAT1T2d (III)

Conclusion:

Given that the cross sectional area of the bar is 1.0×106 m2 and its length is 0.10 m , the thermal conductivity of copper is 401 W/mK , temperature of the first end of the bar is 104°C and the temperature of the other end is 24°C .

Substitute 401 W/mK for κ , 1.0×106 m2 for A , 0.10 m for d , 104°C for T1 and 24°C for T2 in equation (III) to find .

=(401 W/mK)(1.0×106 m2)104°C24°C0.10 m=0.32 W

Therefore, the rate of heat conduction along the bar is 0.32 W .

(b)

To determine

The temperature gradient in the bar.

(b)

Expert Solution
Check Mark

Answer to Problem 90P

The temperature gradient in the bar is 800 K/m .

Explanation of Solution

Write the equation for the temperature gradient.

temperature gradient=ΔTd

Put equation (II) in the above equation.

temperature gradient=T1T2d (IV)

Conclusion:

Substitute 104°C for T1 , 24°C for T2 and 0.10 m for d in equation (IV) to find the temperature gradient .

temperature gradient=104°C24°C0.10 m=80 K0.10 m=800 K/m

Therefore, the temperature gradient in the bar is 800 K/m .

(c)

To determine

The rate of heat conduction if two bars are placed in series between the same temperature baths.

(c)

Expert Solution
Check Mark

Answer to Problem 90P

The rate of heat conduction if two bars are placed in series between the same temperature baths is 0.16 W .

Explanation of Solution

When two similar bars are placed in series, the effective length will be doubled. Equation (III) can be used to find the rate of heat conduction with the value of length doubled.

Conclusion:

The new length is 0.20 m .

Substitute 401 W/mK for κ , 1.0×106 m2 for A , 0.20 m for d , 104°C for T1 and 24°C for T2 in equation (III) to find .

=(401 W/mK)(1.0×106 m2)104°C24°C0.20 m=0.16 W

Therefore, the rate of heat conduction if two bars are placed in series between the same temperature baths is 0.16 W .

(d)

To determine

The rate of heat conduction if two bars are placed in parallel between the same temperature baths.

(d)

Expert Solution
Check Mark

Answer to Problem 90P

The rate of heat conduction if two bars are placed in parallel between the same temperature baths is 0.64 W .

Explanation of Solution

When the two identical bars are placed parallel, the effective area gets doubled. Equation (III) can be used to find the rate of heat conduction with the value of area doubled.

Conclusion:

The new area is 2.0×106 m2 .

Substitute 401 W/mK for κ , 2.0×106 m2 for A , 0.10 m for d , 104°C for T1 and 24°C for T2 in equation (III) to find .

=(401 W/mK)(2.0×106 m2)104°C24°C0.10 m=0.64 W

Therefore, the rate of heat conduction if two bars are placed in parallel between the same temperature baths is 0.64 W .

(e)

To determine

The temperature at the junction, where the bars meet, in series case.

(e)

Expert Solution
Check Mark

Answer to Problem 90P

The temperature at the junction, where the bars meet, in series case is 64°C .

Explanation of Solution

The bars are identical. This implies the temperature at the junction will be midway between the temperatures of the constant-temperature paths.

Write the equation for the temperature at the junction.

T=T1+T22 (V

Here, T is the temperature of the junction

Conclusion:

Substitute 104°C for T1 and 24°C for T2 in equation (V) to find T .

temperature gradient=104°C+24°C2=128°C2=64°C

Therefore, the temperature at the junction, where the bars meet, in series case is 64°C .

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Chapter 14 Solutions

Physics

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