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Difficulty: HardHeat Transfer: Conduction

A technician replaces a standard soldering iron tip with a modified tip made of the exact same copper material. To reach into a tighter space, the new tip is designed to be twice as long and has half the cross-sectional area of the original tip. Assuming the temperature difference between the internal heating element and the wire joint remains constant, how will this modification affect the rate of conductive heat transfer to the joint?

  1. The transfer rate will decrease to one-fourth of its original value.Answer
  2. B
    The transfer rate will increase, as the longer surface promotes stronger convective heat flow to the joint.
  3. C
    The transfer rate will double, because the smaller tip has a lower heat capacity to overcome.
  4. D
    The transfer rate will increase to four times its original value, due to the inverse relationship between area and heat flow.

Answer

The transfer rate will decrease to one-fourth of its original value.
The rate of conductive heat transfer through a solid object depends directly on its cross-sectional area and inversely on its length. Halving the area cuts the flow rate by a factor of 12\frac{1}{2}, and doubling the length cuts the rate by another factor of 12\frac{1}{2}. Combining these constraints yields an overall transfer rate that is 14\frac{1}{4} of the original.

Step-by-Step Solution

1
Identify the factors affecting the rate of conductive heat transfer.
The rate is directly proportional to the cross-sectional area (AA) and inversely proportional to the length (LL).
Fourier's law of heat conduction defines how the physical dimensions of an object affect the flow of thermal energy.
2
Analyze the effect of halving the cross-sectional area.
Reducing the area to 12A\frac{1}{2}A reduces the heat transfer rate by a factor of 12\frac{1}{2}.
There is less physical material (fewer molecules in a given cross-section) available to pass the thermal energy forward.
3
Analyze the effect of doubling the length.
Increasing the length to 2L2L further reduces the heat transfer rate by a factor of 12\frac{1}{2}.
The thermal energy must travel a longer distance, which decreases the temperature gradient (the temperature drop per unit of length).
4
Combine the effects of both geometric changes.
Multiplying the two effects (12×12\frac{1}{2} \times \frac{1}{2}) results in a final heat transfer rate that is 14\frac{1}{4} of the original rate.
Both dimensional modifications act concurrently to restrict the flow of heat.

Key Concept

Geometric factors in thermal conduction
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Topics

Heat Transfer: Conduction

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