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 \(\dfrac{K_1 \theta_1 d_2+K_2 \theta_2 d_1}{K_1 d_2+K_2 d_1}\)
- B \(\dfrac{K_1 \theta_1+K_2 \theta_2}{K_1+K_2}\)
- C \(\dfrac{K_1 \theta_1+K_2 \theta_2}{\theta_1+\theta_2}\)
- D \(\dfrac{K_1 \theta_1 d_1+K_2 \theta_2 d_2}{K_1 d_2+K_2 d_1}\)
Answer & Solution
Correct Answer
(A) \(\dfrac{K_1 \theta_1 d_2+K_2 \theta_2 d_1}{K_1 d_2+K_2 d_1}\)
Step-by-step Solution
Detailed explanation
In a series combination of two slabs, the rate of heat flow \(H\) through each slab must be the same in the steady state. The rate of heat flow through the first slab is \(H = \dfrac{K_1 A (\theta_1 - \theta)}{d_1}\), where \(A\) is the cross-sectional area. The rate of heat…
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