KCET · Physics · Thermal Properties of Matter
Two slabs are of the thicknesses \(d_{1}\) and \(d_{2}\). Their thermal conductivities are \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) respectively. They are in series. The free ends of the combination of these two slabs are kept at temperatures \(\theta_{1}\) and \(\theta_{2}\). Assume \(\theta_{1}>\theta_{2}\). The temperature \(\theta\) of their common junction is
- A \(\frac{\mathrm{K}_{1} \theta_{1}+\mathrm{K}_{2} \theta_{2}}{\theta_{1}+\theta_{2}}\)
- B \(\frac{\mathrm{K}_{1} \theta_{1} \mathrm{~d}_{1}+\mathrm{K}_{2} \theta_{2} \mathrm{~d}_{2}}{\mathrm{~K}_{1} \mathrm{~d}_{2}+\mathrm{K}_{2} \mathrm{~d}_{1}}\)
- C \(\frac{\mathrm{K}_{1} \theta_{1} \mathrm{~d}_{2}+\mathrm{K}_{2} \theta_{2} \mathrm{~d}_{1}}{\mathrm{~K}_{1} \mathrm{~d}_{2}+\mathrm{K}_{2} \mathrm{~d}_{1}}\)
- D \(\frac{\mathrm{K}_{1} \theta_{1}+\mathrm{K}_{2} \theta_{2}}{\mathrm{~K}_{1}+\mathrm{K}_{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{K}_{1} \theta_{1} \mathrm{~d}_{2}+\mathrm{K}_{2} \theta_{2} \mathrm{~d}_{1}}{\mathrm{~K}_{1} \mathrm{~d}_{2}+\mathrm{K}_{2} \mathrm{~d}_{1}}\)
Step-by-step Solution
Detailed explanation
For first slab

Heat current, \(\mathrm{H}_{1}=\frac{\mathrm{K}_{1}\left(\theta_{1}-\theta\right) \mathrm{A}}{\mathrm{d}_{1}}\)
For second slab,
Heat current, \(\mathrm{H}_{2}=\frac{\mathrm{K}_{2}\left(\theta-\theta_{2}\right) \mathrm{A}}{\mathrm{d}_{2}}\)
As slabs are in series
\(\mathrm{H}_{1} =\mathrm{H}_{2} \)
\( \therefore \frac{\mathrm{K}_{1}\left(\theta_{1}-\theta\right) \mathrm{A}}{\mathrm{d}_{1}} =\frac{\mathrm{K}_{2}\left(\theta-\theta_{2}\right) \mathrm{A}}{\mathrm{d}_{2}} \)
\( \Rightarrow \theta =\frac{\mathrm{K}_{1} \theta_{1} \mathrm{~d}_{2}+\mathrm{K}_{2} \theta_{2} \mathrm{~d}_{1}}{\mathrm{~K}_{2} \mathrm{~d}_{1}+\mathrm{K}_{1} \mathrm{~d}_{2}}\)

Heat current, \(\mathrm{H}_{1}=\frac{\mathrm{K}_{1}\left(\theta_{1}-\theta\right) \mathrm{A}}{\mathrm{d}_{1}}\)
For second slab,
Heat current, \(\mathrm{H}_{2}=\frac{\mathrm{K}_{2}\left(\theta-\theta_{2}\right) \mathrm{A}}{\mathrm{d}_{2}}\)
As slabs are in series
\(\mathrm{H}_{1} =\mathrm{H}_{2} \)
\( \therefore \frac{\mathrm{K}_{1}\left(\theta_{1}-\theta\right) \mathrm{A}}{\mathrm{d}_{1}} =\frac{\mathrm{K}_{2}\left(\theta-\theta_{2}\right) \mathrm{A}}{\mathrm{d}_{2}} \)
\( \Rightarrow \theta =\frac{\mathrm{K}_{1} \theta_{1} \mathrm{~d}_{2}+\mathrm{K}_{2} \theta_{2} \mathrm{~d}_{1}}{\mathrm{~K}_{2} \mathrm{~d}_{1}+\mathrm{K}_{1} \mathrm{~d}_{2}}\)
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