COMEDK · Physics · 12. Thermal Properties of Matter
Two slabs are of the thicknesses \(d_1\) and \(d_2\). Their thermal conductivities are \(K_1\) and \(K_2\), 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 \(\frac{K_1 \theta_1+K_2 \theta_2}{\theta_1+\theta_2}\)
- B \(\frac{K_1 \theta_1 d_1+K_2 \theta_2 d_2}{K_1 d_2+K_2 d_1}\)
- C \(\frac{K_1 \theta_1 d_2+K_2 \theta_2 d_1}{K_1 d_2+K_2 d_1}\)
- D \(\frac{K_1 \theta_1+K_2 \theta_2}{K_1+K_2}\)
Answer & Solution
Correct Answer
(C) \(\frac{K_1 \theta_1 d_2+K_2 \theta_2 d_1}{K_1 d_2+K_2 d_1}\)
Step-by-step Solution
Detailed explanation
For first slab, Heat current, \(H_1=\frac{K_1\left(\theta_1-\theta\right) A}{d_1}\) For second slab, Heat current, \(H_2=\frac{K_2\left(\theta-\theta_2\right) A}{d_2}\) As slabs are in series, \(H_1=H_2\)…
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