Physics Homework

Question

The temperature of an oven is kept constant at 905.0 K. A hole with a diameter of 27.0 mm is drilled in the wall of the oven. How much power is emitted by this hole? Hint: consider the hole as a black body.

Answer 1

To determine the power emitted by the hole, we can treat the hole as a black body radiator. The power emitted by a black body is given by the Stefan-Boltzmann law:

P = σ A T^4

where:

1. P is the power emitted (in watts),
2. σ is the Stefan-Boltzmann constant (5.670 × 10^-8 W/m²K⁴),
3. A is the area of the emitting surface (in m²),
4. T is the temperature of the black body (in Kelvin).

Step by step calculation:

1. Determine the area of the hole:
2. The area A of the hole is the area of a circle:

A = π r²

1. where r is the radius of the hole. Given the diameter of the hole is 27.0 mm = 0.0270 m, the radius is:

r = diameter / 2 = 0.0270 / 2 = 0.0135 m

1. Substituting into the area formula:

A = π (0.0135)² = 5.724 × 10⁻⁴ m²

1. Substitute the values into the Stefan-Boltzmann law:

P = σ A T^4

1. Substituting σ = 5.670 × 10^-8 W/m²K⁴, A = 5.724 × 10⁻⁴ m², and T = 905.0 K:

P = (5.670 × 10⁻⁸) (5.724 × 10⁻⁴) (905.0)^4

1. Calculate T^4

T^4 = (905.0)^4 = 6.697 × 10¹¹

1. Calculate the power:

P = (5.670 × 10⁻⁸) (5.724 × 10⁻⁴) (6.697 × 10¹¹)

P = (5.670 × 10⁻⁸) (3.831 × 10⁸)

P = 21.72 W

Final Answer:

The power emitted by the hole is approximately:

P = 21.7 W

Thanks, and hopefully this helps a student out!