Question

A sphere of 3 cm radius acts like a black body. It is in equilibrium with its surrounding and absorbs 30 kW of power radiated to it from surroundings. The temperature of the sphere is \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \). What is the temperature of the sphere?

• A. 5600 K
• B. 4600 K
• C. 3600 K
• D. 2600 K

Hint:

The Stefan-Boltzmann law is useful for calculating the temperature of a body in thermal equilibrium with its surroundings.

Answer 1

The Correct Option is B

Step 1: Find the surface area of the sphere.
The radius is 3 cm = 0.03 m, so the surface area is:
\[ A = 4\pi r^2 = 4\pi (0.03)^2 \approx 0.01131\,\text{m}^2 \]

Step 2: Apply the Stefan-Boltzmann law.
Since the sphere is a perfect black body in equilibrium, the power it absorbs equals the power it radiates:
\[ P = \sigma A T^4 \]

Step 3: Solve for \(T^4\).
\[ T^4 = \frac{P}{\sigma A} = \frac{30 \times 10^3}{5.67 \times 10^{-8} \times 0.01131} \approx 4.68 \times 10^{13} \]

Step 4: Take the fourth root and state the Final Answer.
\[ T \approx \left(4.68 \times 10^{13}\right)^{1/4} \approx 2615\,K \]
\[ \boxed{T \approx 2600\,K} \]

Answer 2

A different route to the same answer starts by working out how much power is radiated per square metre of the sphere's surface, then solving for temperature from that flux value.

1. 5600 K: A flux this high would need vastly more absorbed power than 30 kW spread over such a small sphere, since flux climbs with the fourth power of temperature. This option would correspond to a much larger absorbed power than stated.
2. 4600 K: Working backward from this temperature gives a flux, and hence a total power, many times larger than the 30 kW the sphere actually absorbs.
3. 3600 K: Still corresponds to a flux, and a total radiated power, well above the sphere's actual 30 kW.
4. 2600 K: The sphere's surface area is about 0.0113 square metres, so the flux needed to radiate 30 kW from that area is \( \frac{30000}{0.0113} \approx 2.65 \times 10^{6} \) W per square metre. Solving \( \sigma T^4 = 2.65 \times 10^{6} \) for T gives \( T^4 \approx 4.68 \times 10^{13} \), and taking the fourth root gives a temperature close to 2600 K.

Working from flux first instead of jumping straight to the full power equation makes it clear how sensitive the fourth power is: even a small change in temperature swings the radiated power by a large factor, which is why only 2600 K matches the sphere's actual 30 kW output.

The correct answer is 2600 K.