Question

A sphere of surface area \(4\ \text{m}^2\) at temperature \(400\ \text{K}\) and having emissivity \(0.5\) is located in an environment of temperature \(200\ \text{K}\). The net rate of energy exchange of the sphere is
\[ \text{Stefan Boltzmann constant } \sigma=5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4} \]

• A. \(3260.8\ \text{W}\)
• B. \(1632.4\ \text{W}\)
• C. \(2721.6\ \text{W}\)
• D. \(4216.4\ \text{W}\)

Hint:

For radiation exchange with surroundings, use \(P=e\sigma A(T^4-T_0^4)\), not simply \(e\sigma AT^4\).

Answer 1

The Correct Option is C

Step 1: Use Stefan-Boltzmann law for net radiation.
The net rate of energy exchange is given by
\[ P=e\sigma A(T^4-T_0^4) \]
Here,
\[ e=0.5,\quad A=4\ \text{m}^2,\quad T=400\ \text{K},\quad T_0=200\ \text{K} \]

Step 2: Substitute the values.
\[ P=0.5\times 5.67\times10^{-8}\times4\left(400^4-200^4\right) \]
Now,
\[ 400^4=256\times10^8 \] and
\[ 200^4=16\times10^8 \]
So,
\[ 400^4-200^4=240\times10^8 \]

Step 3: Calculate the net power.
\[ P=0.5\times4\times5.67\times10^{-8}\times240\times10^8 \]
\[ P=2\times5.67\times240 \]
\[ P=2721.6\ \text{W} \]

Step 4: Final conclusion.
Hence, the net rate of energy exchange is
\[ \boxed{2721.6\ \text{W}} \]