Question

A 30 -cm-diameter black ball at \(120^{\circ} \mathrm{C}\) is suspended in air. It is losing heat to the surrounding air at \(25^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of $12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ and by radiation to the surrounding surfaces at \(15^{\circ} \mathrm{C}\). The total rate of heat transfer from the black ball is (a) \(322 \mathrm{~W}\) (b) \(595 \mathrm{~W}\) (c) \(234 \mathrm{~W}\) (d) \(472 \mathrm{~W}\) (e) \(2100 \mathrm{~W}\)

Short answer

Answer: The approximate total rate of heat transfer is 918.79 W.

Step-by-step solution

Understand given information

We are given the diameter of the black ball (\(D = 30 cm\) or \(0.3 m\)), the temperature of the black ball (\(T_{ball} = 120^{\circ} \mathrm{C}\)), the surrounding air temperature (\(T_{air} = 25^{\circ} \mathrm{C}\)), and the heat transfer coefficient by convection (\(h = 12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)). The temperature of the surrounding surfaces is \(T_{surfaces} = 15^{\circ} \mathrm{C}\).

Calculate the ball's surface area

To find the surface area of the ball, we can use the formula \(A = 4 \pi r^{2}\), where \(r\) is the ball's radius. The radius can be found by dividing the diameter by 2: \(r = 0.3 m / 2 = 0.15 m\). Thus, the surface area is \(A = 4 \pi (0.15)^{2} = 0.2827 \mathrm{~m}^{2}\).

Calculate the rate of convective heat transfer

We can calculate the rate of convective heat transfer with the formula \(Q_{conv} = hA(T_{ball} - T_{air})\). Convert the temperatures to Kelvin: \(T_{ball} = 120 + 273.15 = 393.15 \mathrm{K}\) and \(T_{air} = 25 +273.15 =298.15 \mathrm{K}\). Now, \(Q_{conv} = 12 \times 0.2827 (393.15 - 298.15) = 321.61 \mathrm{~W}\).

Calculate the rate of radiative heat transfer

To calculate the rate of radiative heat transfer, we can use the formula \(Q_{rad} = \sigma \epsilon A (T_{ball}^{4} - T_{surfaces}^{4})\), where \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\)), \(\epsilon\) is the emissivity of the black ball (\(\epsilon = 1\) for a perfect black body), and \(T_{surfaces}\) is the temperature of the surrounding surfaces. Convert \(T_{surfaces}\) to Kelvin: \(T_{surfaces} = 15 + 273.15 = 288.15 \mathrm{K}\). Now, \(Q_{rad} = 5.67 \times 10^{-8} \times 1 \times 0.2827 (393.15^{4} - 288.15^{4}) = 597.18 \mathrm{~W}\).

Calculate the total rate of heat transfer

Now we can calculate the total rate of heat transfer by adding the convective rate and the radiative rate: \(Q_{total} = Q_{conv} + Q_{rad} = 321.61 \mathrm{~W} + 597.18 \mathrm{~W} = 918.79 \mathrm{~W}\). However, none of the given options match this result. It's possible that there is an error in the problem's given choices, or some assumptions were made along the way that led to a different answer. Based on the calculations, the total rate of heat transfer would be approximately 918.79 W.