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

Using the Stefan-Boltzmann Law: \[ P = \sigma A T^4 \] where \( P = 30 \, \text{kW} = 30 \times 10^3 \, \text{W} \), \( A \) is the surface area of the sphere, and \( T \) is the temperature. The surface area of a sphere is given by: \[ A = 4 \pi r^2 \] Substitute \( r = 0.03 \, \text{m} \): \[ A = 4 \pi (0.03)^2 \approx 0.0113 \, \text{m}^2 \] Now use the Stefan-Boltzmann equation to solve for \( T \): \[ 30 \times 10^3 = (5.67 \times 10^{-8}) \times 0.0113 \times T^4 \] Solving for \( T \): \[ T^4 \approx \frac{30 \times 10^3}{(5.67 \times 10^{-8}) \times 0.0113} \approx 4.6 \times 10^3 \] Thus, \( T \approx 4600 \, \text{K} \). 

Answer 2

This question uses the Stefan-Boltzmann law for a small sphere acting as a perfect absorber and emitter. Rather than solving for temperature directly, each option can be checked by plugging it back into the law and comparing to the given 30 kW.

1. 5600 K: Raising 5600 to the fourth power gives a huge number, around \( 9.8 \times 10^{14} \). Multiplying by \( \sigma A \approx 6.41 \times 10^{-10} \) (using the sphere's actual surface area) gives a power far above 30 kW, so this temperature is much too high.
2. 4600 K: Its fourth power is close to \( 4.48 \times 10^{14} \). Multiplying by the same \( \sigma A \) still gives a power close to 287,000 W, roughly ten times more than the 30 kW stated in the question, so this is also too high.
3. 3600 K: Its fourth power is around \( 1.68 \times 10^{14} \), giving a power near 108,000 W once multiplied by \( \sigma A \), still well above the 30 kW target.
4. 2600 K: Its fourth power is close to \( 4.57 \times 10^{13} \). Multiplying by \( \sigma A \approx 6.41 \times 10^{-10} \) gives a power close to 29,300 W, very near the 30 kW absorbed by the sphere.

Testing each temperature against the actual formula shows the power radiated grows extremely fast with temperature, since it depends on the fourth power. Only 2600 K reproduces something close to the sphere's actual 30 kW.

So the correct answer is 2600 K.