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, \( T \approx 4600 \, \text{K} \).