Physics for Scientists and Engineers With Modern Physics
Physics for Scientists and Engineers With Modern Physics
9th Edition
ISBN: 9781133953982
Author: SERWAY, Raymond A./
Publisher: Cengage Learning
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Chapter 20, Problem 82CP

(a)

To determine

The reason for the same rate of energy transfer through spherical surface.

(a)

Expert Solution
Check Mark

Answer to Problem 82CP

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)

To determine

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

(b)

Expert Solution
Check Mark

Answer to Problem 82CP

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)

To determine

The rate of energy transfer through the shell.

(c)

Expert Solution
Check Mark

Answer to Problem 82CP

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)

To determine

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

(d)

Expert Solution
Check Mark

Answer to Problem 82CP

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)

To determine

The temperature within the cell as a function of radius.

(e)

Expert Solution
Check Mark

Answer to Problem 82CP

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)

To determine

The temperature at radius. 5.00cm .

(f)

Expert Solution
Check Mark

Answer to Problem 82CP

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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