A 5-in-diameter spherical ball is known to emit radiation at a rate of \(550 \mathrm{Btu} / \mathrm{h}\) when its surface temperature is \(950 \mathrm{R}\). Determine the average emissivity of the ball at this temperature.

Short Answer

Expert verified
Based on the given diameter of a spherical ball, the radiation it emits, and its surface temperature, determine the average emissivity of the ball at this temperature. The average emissivity of the ball at this temperature is approximately 0.544.

Step by step solution

01

Find the surface area of the ball

Firstly, we need to find the surface area of the ball. The formula for the surface area of a sphere is: \(A = 4 * \pi * r^2\) The diameter of the ball is 5 in, so the radius is half of the diameter: \(r = \frac{d}{2} = \frac{5 \ \text{in}}{2} = 2.5 \ \text{in}\) Now convert the radius to feet: \(r = 2.5 \ \text{in} * \frac{1 \ \text{ft}}{12 \ \text{in}} = \frac{5}{24} \ \text{ft}\) Now we can find the surface area of the ball: \(A = 4 * \pi * (\frac{5}{24})^2 \approx 1.653 \ \text{ft}^2\)
02

Apply the Stefan-Boltzmann law and solve for emissivity

Now we have the surface area of the ball (\(A\)). We can apply the Stefan-Boltzmann law and solve for emissivity (\(\epsilon\)): \(Q = \epsilon * A * \sigma * T^4\) Rearrange the equation to solve for emissivity: \(\epsilon = \frac{Q}{A * \sigma * T^4}\) Now plug in the given values: \(\epsilon = \frac{550 \ \frac{\mathrm{Btu}}{\mathrm{h}}}{1.653 \ \mathrm{ft}^2 * 1.714 * 10^{-9} \frac{\mathrm{Btu}}{\mathrm{h} \cdot \mathrm{ft}^2 \cdot \mathrm{R}^4} * (950 \ \mathrm R)^4}\) Solve for emissivity: \(\epsilon \approx 0.544\) The average emissivity of the ball at this temperature is approximately 0.544.

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