Question

In a composite slab there are two materials having coefficients of thermal conductivity K and 2K, thickness x and 4x repectively. The temperature of the two outer surfaces of a composite slab are \(T_2\) and \(T_1\) \((T_2 > T_1)\). \(T_2\) is on side K and \(T_1\) is on side 2K. The rate of heat transfer through the slab in a steady state is \([\frac{A(T_2-T_1)K}{x}]\cdot f\), where f is equal to

• A. \(1\)
• B. \(\frac{1}{2}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{1}{3}\)

Hint:

Energy emitted per second is proportional to area and the fourth power of absolute temperature.

Answer 1

The Correct Option is D

Step 1: Ratio form:
$\frac{E'}{E} = \frac{A'}{A}\left(\frac{T'}{T}\right)^4$.

Step 2: Insert:
$\frac{E'}{E} = \frac14\left(\frac{800}{400}\right)^4 = \frac{16}{4} = 4$ (C).

Final Answer:
4E. \[ \boxed{4E} \]