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: Understanding the Concept:
Stefan's law: \(E = \sigma AT^4\), with T in kelvin.

Step 2: Change in area:
Length and breadth are both halved, so the area becomes \(\frac14\) of the original.

Step 3: Change in temperature:
\(T_1 = 127 + 273 = 400\) K and \(T_2 = 527 + 273 = 800\) K, so \(T\) doubles and \(T^4\) becomes 16 times.

Step 4: New energy:
\[ E' = E\times\frac14\times16 = 4E \]

Final Answer:
The new emitted energy is 4E, option (C). \[ \boxed{4E} \]