Which of the following does not affect the rate of heat transfer through conduction?

Select a single answer.

Question

Which of the following does not affect the rate of heat transfer through conduction?

Select a single answer.

BlackTom AI · Accepted Answer

Fourier’s law of heat conduction: Q̇ = −kA (dT/dx), so the steady-state heat transfer rate through a uniform slab is Q̇ = k A (T1 − T2) / L.

1. Relevant knowledge points and concepts
- Fourier’s law relates heat transfer rate (Q̇) to thermal conductivity k, cross-sectional area A, and temperature gradient (ΔT/L).  
- For a single homogeneous layer the overall heat transfer coefficient U = k / L, so Q̇ = U A ΔT.  
- Mass (or density and specific heat) affects thermal inertia (transient response), not the steady-state conduction rate.

2. Step-by-step reasoning
- Start with Fourier’s law in one dimension for a slab: Q̇ = k A (T1 − T2) / L.  
- This expression shows the direct proportionality to k and A, inverse proportionality to thickness L, and dependence on ΔT.  
- Replace k/L by U to get Q̇ = U A (T1 − T2), making the role of U explicit.

3. Intermediate calculation steps
- Given Fourier’s law: Q̇ = k A (T1 − T2) / L.  
- Define U = k / L.  
- Substitute: Q̇ = U A (T1 − T2).  
- Note that mass m does not appear in any algebraic step.

4. Why the correct answer is correct
- The mass of the object (m) does not appear in the steady-state conduction formulas. Mass determines heat capacity C = m c, which controls how quickly temperature of an object changes with time (transient conduction), but it does not enter the steady-state flux equation. Therefore mass does not affect the steady-state rate of heat transfer by conduction.

5. Why the other options are incorrect
- Thermal conductivity of the object: appears directly (Q̇ ∝ k).  
- U-value of the object: U = k/L, directly used in Q̇ = U A ΔT.  
- Cross-sectional area: Q̇ ∝ A (more area, more heat flow).  
- Thickness of the conduction path: appears in denominator (Q̇ ∝ 1/L). All these affect the conduction rate.

6. Conclusion
Mass of the object is the variable that does not affect the steady-state rate of heat transfer through conduction. This matches the provided correct answer: Mass of the object.

Which of the following does not affect the rate of heat transfer through conduction? Select a single answer.单项选择题

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