Water is boiled at atmospheric pressure by a horizontal polished copper heating element of diameter \(D=0.5\) in and emissivity \(\varepsilon=0.05\) immersed in water. If the surface temperature of the heating element is \(788^{\circ} \mathrm{F}\), determine the rate of heat transfer to the water per unit length of the heating element.

First, we need to convert the given surface temperature of the heating element from Fahrenheit to Kelvin using the following formula: \(T_K = \frac{5}{9} (T_F-32) + 273.15\) where \(T_F\) is the temperature in Fahrenheit and \(T_K\) is the temperature in Kelvin. \(T_K = \frac{5}{9} (788 - 32) + 273.15\) \(T_K \approx 694.15 \, K\) The surface temperature of the heating element in Kelvin is approximately 694.15 K.

The Stefan-Boltzmann law states that the heat transfer per unit area from a black body is proportional to the fourth power of the temperature, and it can be described by the equation: \(q = \varepsilon \sigma (T^4 - T_{\infty}^4)\) where \(q\) is the heat transfer per unit area, \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(\sigma = 5.67 \times 10^{-8} W/(m^2K^4)\)), and \(T\) and \(T_{\infty}\) are the temperatures of the heating element and the surrounding fluid (water) in Kelvin. However, since water is in contact with the heating element, we assume that the water temperature is at the boiling point. At atmospheric pressure, water boils at 100°C, or 373.15 K. So \(T_{\infty} = 373.15 \, K\). Now we can plug in the given values: \(q = 0.05 \times 5.67 \times 10^{-8} \times (694.15^4 - 373.15^4)\) \(q \approx 9.55 \, W/m^2\)

Next, we need to determine the heat transfer rate per unit length of the heating element. We can do this by multiplying the heat transfer rate per unit area (\(q\)) by the circumference of the heating element (\(2 \pi r\)): \(Q' = q (2 \pi r)\) where \(Q'\) is the heat transfer rate per unit length, \(q\) is the heat transfer rate per unit area, and \(r\) is the radius of the heating element. First, we will convert the diameter from inches to meters: \(D = 0.5\) in \(= 0.5 \times 0.0254\) m \(\approx 0.0127\) m Then, we can calculate the radius \(r = D/2 \approx 0.00635\) m. Now, we can plug in the values: \(Q' = 9.55 \times (2 \pi \times 0.00635)\) \(Q' \approx 0.3787 \, W/m\) The rate of heat transfer to the water per unit length of the heating element is approximately 0.3787 W/m.