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: Write the net radiation law.
A body of emissivity $e$ and area $A$ at temperature $T$ in surroundings at $T_0$ exchanges net power $P=e\sigma A\left(T^4-T_0^4\right)$.
Step 2: List the data.
Here $e=0.5$, $A=4$ m$^2$, $T=400$ K, $T_0=200$ K and $\sigma=5.67\times10^{-8}$ W m$^{-2}$K$^{-4}$.
Step 3: Work out the fourth powers.
$400^4=256\times10^8$ and $200^4=16\times10^8$, so $T^4-T_0^4=240\times10^8$.
Step 4: Assemble the numbers.
\[ P=0.5\times 5.67\times10^{-8}\times 4\times 240\times10^8 \]
Step 5: Multiply step by step.
The powers of ten cancel: $10^{-8}\times10^8=1$. Then $0.5\times 4=2$, and $2\times 5.67\times 240=2721.6$.
Step 6: Conclude.
The net rate of energy exchange of the sphere is $2721.6$ W.
\[ \boxed{2721.6\ \text{W}} \]