Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 20, Problem 36P

(a)

To determine

To Calculate:The thermal resistance of each cube.

(a)

Expert Solution
Check Mark

Answer to Problem 36P

Thermal resistance of copper cube is 0.0831 K/W.

Thermal resistance of aluminum cube is 0.141 K/W.

Explanation of Solution

Given: A copper cube and aluminum cube each having3.00cm long edges.

Formula used:

Thermal resistance R=|Δx|kA

  |Δx| is the thickness, k is the conductivity, and A is the area of the cube.

Calculation:

Thermal resistance of copper cube is calculated as follows:

Substitute 3 cm for, |Δx| 401 W/mK for K, and (3 cm)2 for A in the above equation.

  RCu=3cm( 401 W/mK) ( 3 cm )2( 1cm 1×10 -2 m)= 0.0831 K/W

Therefore, thermal resistance of copper cube is 0.0831 K/W.

Thermal resistance of aluminum cube is calculated as follows:

Substitute 3 cm for |Δx| , 237 W/mK for K, and (3 cm)2 for A in the above equation.

  RAl=3cm( 237 W/mK) ( 3 cm )2( 1cm 1×10 -2 m)= 0.141 K/W

Therefore, thermal resistance of aluminum cube is 0.141 K/W

Conclusion: Thermal resistance can be calculated by using heat conductivity.

(b)

To determine

To Calculate:The thermal resistance of two cube combination.

(b)

Expert Solution
Check Mark

Answer to Problem 36P

The thermal resistance of two-cube combination is 0.224 K/W.

Explanation of Solution

Given:

Cubes are connected in series.

A copper cube and aluminum cube each has 3.00cm long edges.

0.0831 K/W for RCu and 0.141 K/W for RAl

Calculation:

The cubes are in series, so the resultant thermal resistance is:

  Req = Rcu + RAI

Substitute 0.0831 K/W for RCu and 0.141 K/W for RAl , in the above equation.

  Req=0.0831 K/W+0.141 K/W= 0.224 K/W

Conclusion: Thus, the thermal resistance of two-cube combination is 0.224 K/W.

(c)

To determine

To Calculate:The thermal current I.

(c)

Expert Solution
Check Mark

Answer to Problem 36P

Thermal current I at the interface between the two cubes is 0.375 kW.

Explanation of Solution

Given: A copper cube and aluminum cube each has 3.00cm long edges.

  ΔT= (373 K - 293 K)

  Req=0.224 K/W

Calculation:

The expression for the thermal current I is

  I=ΔTReq

Here, ΔT is the difference in temperature and Req is the equivalent resistance.

Substitute the values

  I=(373K-293K)0.224K/W=357W( 1k 1× 10 3 )=0.357kW

Conclusion:

Therefore, thermal current I at the interface between the two cubes is 0.375 kW.

(d)

To determine

To Calculate:The temperature at the interface of two cubes.

(d)

Expert Solution
Check Mark

Answer to Problem 36P

  70.3°C

Explanation of Solution

Given:

I = 357 W

  RCu = 0.0831 K/W and RAl = 0.141 K/W

Calculation:

The temperature at the interface between the two cubes is

  Tinterface = 373 KΔTCu

Substitute IRCu for ΔTCu in the above equation.

  Tinterface = 373 KIRCu

  Tinterface = 373 K(( 357 W)( 0.0831 K/W))= 343.3 K= (343.3273)°C= 70.3°C

Conclusion: Therefore, the temperature at the interface between the two cubes is 70.3°C.

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