Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 19, Problem 46CP

A spherical shell has inner radius 3.00 cm and outer radius 7.00 cm. It is made of material with thermal conductivity k = 0.800 W/m · °C. The interior is maintained at temperature 5°C and the exterior at 40°C. After an interval of time, the shell reaches a steady state with the temperature at each point within it remaining constant in time. (a) Explain why the rate of energy transfer P must be the same through each spherical surface, of radius r, within the shell and must satisfy

d T d r = P 4 π k r 2

(b) Next, prove that

5 d T = P 4 π k 0.03 0.07 r 2 d r

where T is in degrees Celsius and r is in meters. (c) Find the rate of energy transfer through the shell. (d) Prove that

5 T d T = 1.84 0.03 r r 2 d r

where T is in degrees Celsius and r is in meters. (e) Find the temperature within the shell as a function of radius. (f) Find the temperature at r = 5.00 cm, halfway through the shell.

(a)

Expert Solution
Check Mark
To determine
The reason for the same rate of energy transfer through spherical surface.

Answer to Problem 46CP

The rate of energy transfer through the spherical surface is same because the temperature gradient is constant.

Explanation of Solution

Let the rate of energy transfer is P , radius of the spherical surface is r , the elemental temperature is dT and elemental distance is dr .

The law of thermal conduction is,

P=kAdTdr

Here,

P is the power or rate of energy transfer.

k is the coefficient of thermal conductivity.

A is the surface area of surface.

T is the temperature in Kelvin.

r is the radial distance.

The expression for the surface area of the sphere ism,

A=4πr2

Substitute 4πr2 for a in above equation,

P=k4πr2dTdr

dTdr=P4πkr2 (1)

Since the value of the coefficient of thermal conductivity is constant and the radius of the spherical surface is also constant. The thermal gradient becomes constant.

The rate of energy transfer is directly proportional to the thermal gradient. Since the thermal gradient is constant so the rate of energy transfer through the spherical surface is same.

Conclusion:

Therefore, the rate of energy transfer through the spherical surface is same because the temperature gradient is constant

(b)

Expert Solution
Check Mark
To determine

To show: The given relation, 5dT=P4πk0.030.07r2dr .

Answer to Problem 46CP

The given relation, 5dT=P4πk0.030.07r2dr is valid.

Explanation of Solution

Let the temperature is T in degrees Celsius and the radius of the spherical surface is r .

Rearrange the equation (1) to prove the relation,

dT=P4πr2kdr

Integrate at both sides for temperature from 5°C to °C and for radial distances from 0.03m to 0.07m .

5dT=P4πk0.030.07drr25dT=P4πk0.030.07r2dr

Conclusion:

Therefore, the equation, 5dT=P4πk0.030.07r2dr is valid.

(c)

Expert Solution
Check Mark
To determine
The rate of energy transfer through the shell.

Answer to Problem 46CP

The rate of energy transfer through the shell is 18.47W .

Explanation of Solution

From the equation (1),

dTdr=P4πkr2dT=P4πr2kdr

Integrate at both sides for temperature from 5°C to 40°C and for radial distances from 0.03m to 0.07m .

540dT=P4πk0.030.07drr2[T]540=P4π×0.8[1r]0.030.07P=18.47W

Conclusion:

Therefore, the rate of energy transfer through the shell is 18.47W .

(d)

Expert Solution
Check Mark
To determine

To show: The given equation, 5TdT=P4πk0.03rr2dr .

Answer to Problem 46CP

The equation, 5TdT=1.840.03rr2dr is valid.

Explanation of Solution

Let the temperature is T in degrees Celsius and the radius of the spherical surface is r .

Rearrange the equation (1) to prove the relation,

dT=P4πr2kdr

Integrate at both sides for temperature from 5°C to T°C and for radial distances from rm to 0.07m ,

5TdT=P4πk0.03rdrr2

Substitute 40°C for T and 0.07m for r in above equation to calculate the value of power,

540dT=P4πk0.030.07r2dr[T]540=P4πk[1r]0.030.07P4πk=1.8375

P4πk=1.84 (3)

Put the value of the P4πk in equation (2),

5TdT=1.840.03rr2dr (4)

Conclusion:

Therefore, the equation, 5TdT=1.840.03rr2dr is valid.

(e)

Expert Solution
Check Mark
To determine
The temperature within the cell as a function of radius.

Answer to Problem 46CP

The temperature within the cell as a function of radius is T=56.33+1.84r

Explanation of Solution

From the equation (1),

5TdT=1.840.03rr2dr

Integrate the above equation,

[T5]=1.84[10.03+1r]T=56.33+1.84r (5)

Conclusion:

Therefore, the temperature within the cell as a function of radius is T=56.33+1.84r

(f)

Expert Solution
Check Mark
To determine
The temperature at radius. 5.00cm .

Answer to Problem 46CP

The temperature at radius. 5.00cm . is 19.53°C

Explanation of Solution

From the equation (5),

T=56.33+1.84r

Substitute 0.05cm for r in above equation,

T=56.33+1.845cmT=56.33+1.845cm×100cm1mT=19.53°C

Conclusion:

Therefore, the temperature in spherical shell at radius. 5.00cm is 19.53°C .

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Chapter 19 Solutions

Physics for Scientists and Engineers

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