   Chapter 11, Problem 49P

Chapter
Section
Textbook Problem

A copper rod and an aluminum rod of equal diameter are joined end to end in good thermal contact. The temperature of the free end of the copper rod is held constant at 100.°C and that of the far end of the aluminum rod is held at 0°C. If the copper rod is 0.15 m long, what must be the length of the aluminum rod so that the temperature at the junction is 50.°C?

To determine
The length of the aluminum rod.

Explanation

Given info: Two rod have equal diameter, the free end of aluminum rod is at 0°C , free end of copper rod is at 100°C , the length copper rod is 0.15 m, and the temperature at the junction is 50.0°C .

Formula to calculate the rate at which energy transferred for aluminum rod is,

PAl=kAlA(TTAl)LAl

• PAl is the rate of energy transfer across the aluminum rod,
• kAl is the thermal conductivity of aluminum rod,
• A is the area of the rod,
• TAl is temperature of the free end of aluminum rod,
• T is temperature at the interface between two rods,
• LAl is the length of the rod of the aluminum rod,

Formula to calculate the rate at which energy transferred for iron rod is,

PCu=kCuA(TCuT)LCu

• PCu is the rate of energy transfer across the copper rod,
• kCu is the thermal conductivity of copper rod,
• TCu is temperature of the free end of copper rod,
• LCu is the length of the copper rod,

Energy transfer for each of the rods is same in steady state.

Equate both the rate of energy transfer equations and rewrite it in terms of T.

kAlA(TTAl)LAl=kCuA(TCuT)LCuLAlLCu=kAl

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