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.Single choice

Similar Questions

This question relates to Case Study 1. (Download the Data Sheet and Case Studies from the link in the quiz instructions.) What is the magnitude of the conductive heat flow (in W) through the wall of the dome when it is 24°C inside and 6°C outside? Select a single answer.

Which statement is the best example of heat energy transfer by conduction?

Which statement is the best example of heat energy transfer by conduction?

A cylindrical flask is being used as an insulator to store tea and keep it hot on a cold day. How will increasing the thickness by 4 times of all walls and halving the contact surface area (without changing the type of material the walls are made of) of the insulator affect the rate of heat loss?

The collision of particles within a substance results in a transfer of energy by:

Which of the following will result in the smallest rate of heat transfer?

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler