Question

A rectangular black body of temperature \(27^{\circ }C\) has surface area \(6\) cm \(\times 3\) cm and rate of radiation \(E\). If its temperature is increased by \(300^{\circ }C\) and its linear dimensions are reduced to \(\left(\frac{1}{3}\right)^{rd}\) of their initial value then the rate of radiation is

• A. \(\frac{8}{9}E\)
• B. \(\frac{16}{9}E\)
• C. \(\frac{4}{9}E\)
• D. \(\frac{32}{9}E\)

Hint:

Radiation rate is proportional to area times the fourth power of the absolute temperature; use kelvin temperatures.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Stefan-Boltzmann law: rate of radiation \(E\propto AT^{4}\), with \(T\) in kelvin.

Step 2: Temperatures
\(T_{1}=27+273=300\) K. A rise of \(300^{\circ}\)C gives \(T_{2}=600\) K.
So \(\frac{T_{2}}{T_{1}}=2\).

Step 3: Area
Each linear dimension becomes \(\frac{1}{3}\), so the area becomes \(\frac{1}{9}\) of the original.

Step 4: Calculation
\(E'=E\times\frac{1}{9}\times2^{4}=\frac{16}{9}E\), option (B).

Final Answer:
The new rate of radiation is 16/9 times E. \[ \boxed{\text{(B) }\dfrac{16}{9}E} \]