Key relationship
Q̇=(Th−Tc)/(L₁/(k₁A)+L₂/(k₂A)).
Physics · Class 11
Conduct heat through two layers
Thermal resistances in series add. The poorly conducting or thicker layer takes the larger temperature drop. In a steady state the same heat rate passes through both layers.
Open the interactive experiment →Conduct heat through two layers
தொடராக உள்ள அடுக்குகளின் வெப்பத் தடைகள் கூட்டப்படுகின்றன. குறைந்த கடத்துத்திறன் கொண்ட அடுக்கில் அதிக வெப்பநிலை வீழ்ச்சி இருக்கும்.
Key notes, relationships and model limits
Remember
Thermal resistances in series add. The poorly conducting or thicker layer takes the larger temperature drop. In a steady state the same heat rate passes through both layers.
Common trap
Temperature drops are not equal when the layers have different thermal resistances.
Model limits
One-dimensional steady conduction through layers in series; no contact resistance or side loss; conductivities are constant.
What you will learn
Combine layer resistances and find the interface temperature.
Try the experiment
Try both scenarios, change one control, and predict the result before checking the graph and readouts.
Think of it this way
Change one input while keeping the others fixed, as you would in a controlled experiment.
Worked example
With A=0.002 m², L₁=L₂=0.05 m, k₁=10 and k₂=1, resistances are 2.5 and 25 K/W. An 80 K difference drives 2.909 W.
Why this matters
The model connects conduct heat through two layers with an observable reading. Check the equation and assumptions before applying it.
Before you start
SI units · algebra · graph reading
Check your understanding
If one layer becomes thicker, what happens to steady heat rate for the same end temperatures?
- It increases
- It decreases
- It stays fixed
Read the explanation
Its resistance L/(kA) increases, so the same temperature difference drives a smaller rate.
Course connection: Class XI Chapter 10. Consult the roadmap for CBSE, NEET, JEE Main and Advanced scope.