As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4 . 0 cm and 2 . 5 cm respectively and the area

Question

As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4 . 0 cm and 2 . 5 cm respectively and the area of cross-section is 120 cm 2 for each plate. The equivalent thermal conductivity of the compound plate is 1 + 5 α K , then the value of α will be _____ .

TestHub · Accepted Answer

The thermal resistance of rod 1 will be $R_{1} = \frac{l_{1}}{K A} = \frac{4}{K A}$ and rod 2 will be $R_{2} = \frac{l_{2}}{2 K A} = \frac{2.5}{2 K A}$ .

Since they are connected in series, $R_{e q} = R_{1} + R_{2} = \frac{10.5}{2 K A}$ .

If equivalent thermal conductivity is $K_{e q}$ , then $R_{e q} = \frac{l_{1} + l_{2}}{K_{e q} A} = \frac{6.5}{K_{e q} A}$ . Therefore,

$\frac{10.5}{2 K A} = \frac{6.5}{K_{e q} A} \Rightarrow K_{e q} = \frac{26}{21} K = (1 + \frac{5}{21}) K$

Hence, $α = 21$ .

Physics - Heat & Thermo Question with Solution | TestHub

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