Question

The rectangular surface area \((12\ \text{cm}\times 6\ \text{cm})\) of the black body, at a temperature \(127^{\circ }\text{C}\), emits heat energy at the rate of Q units per second. If the length and breadth of the surface area are each reduced to half of its initial value and the temperature is increased to \(327^{\circ }\text{C}\), then the emission of heat energy will be

• A. \(\frac{9Q}{4}\)
• B. \(\frac{81Q}{16}\)
• C. \(\frac{81Q}{32}\)
• D. \(\frac{81Q}{64}\)

Hint:

Use Stefan's law: the rate of emission is proportional to area times the fourth power of absolute temperature.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Stefan's law: rate \(\propto A\,T^{4}\), with T in kelvin.

Step 2: Area and temperature
Halving length and breadth makes the area \(\dfrac{1}{4}\) of the original.
\(T_{1}=127+273=400\ \text{K}\), \(T_{2}=327+273=600\ \text{K}\), so the ratio is \(\dfrac{3}{2}\).

Step 3: Calculation
New rate \(=Q\times\dfrac{1}{4}\times\left(\dfrac{3}{2}\right)^{4}=Q\times\dfrac{1}{4}\times\dfrac{81}{16}=\dfrac{81Q}{64}\)

Final Answer:
The new rate of emission is \(\frac{81Q}{64}\). \[ \boxed{\text{(D) }\dfrac{81Q}{64}} \]