Question

Water is to be boiled at atmospheric pressure in a mechanically polished steel pan placed on top of a heating unit. The inner surface of the bottom of the pan is maintained at \(110^{\circ} \mathrm{C}\). If the diameter of the bottom of the pan is \(30 \mathrm{~cm}\), determine \((a)\) the rate of heat transfer to the water and \((b)\) the rate of evaporation.

Short answer

Question: Determine the rate of heat transfer to the water and the rate of evaporation from a steel pan with boiling water at atmospheric pressure. (a) Rate of heat transfer to the water: (b) Rate of evaporation: Answer: (a) 352.88 W (b) 1.56 x 10^{-4} kg/s

Step-by-step solution

Identify the given variables and constants

Here, we have the following given information: - Temperature at the inner surface of the bottom of the pan, \(T_1 = 110^{\circ} \mathrm{C}\) - Diameter of the bottom of the pan, \(d = 30 \mathrm{~cm}\) - Atmospheric pressure

Find the surface area of the bottom of the pan

We are given the diameter of the pan, and we can use it to find the surface area. The surface area of the bottom of the pan, a circle, can be calculated using the formula: \(A = \pi r^2\) Where: - \(A\) is the surface area, - \(r\) is the radius of the pan, and - \(d\) (diameter) = \(2r\) We have \(d = 30\mathrm{~cm}\), so the radius can be found as \(r = \dfrac{d}{2} = \dfrac{30\mathrm{~cm}}{2} = 15 \mathrm{~cm}\). Now, we can calculate the surface area: \(A = \pi (15\mathrm{~cm})^2 \approx 706.86 \mathrm{~cm^2}\)

Calculate the rate of heat transfer to the water

To calculate the rate of heat transfer to the water, we can use the following equation: \(Q = U A (T_1 - T_2)\) Where: - \(Q\) is the rate of heat transfer, - \(U\) is the overall heat transfer coefficient, - \(A\) is the surface area, and - \(T_1\) and \(T_2\) are the temperatures of the inner surface of the pan and the water (both in K), respectively. We are given that \(T_1 = 110^{\circ} \mathrm{C}\), and since the water is boiling, \(T_2 = 100^{\circ} \mathrm{C}\). Note that to convert Celsius to Kelvin, we just add 273 to the Celsius temperature. So, we have \(T_1 = 383\mathrm{~K}\) and \(T_2 = 373\mathrm{~K}\). The overall heat transfer coefficient \(U\) for this problem depends on the materials and operating conditions, but a typical value for water boiling in a steel pan can be around \(500\,\mathrm{W/(m^2K)}\). Now, we have to convert the units of \(A\) and \(U\) to be consistent. \(A = 706.86\mathrm{~cm^2} = 0.070686\mathrm{~m^2}\) and \(U = 500\,\mathrm{W/(m^2K)}\). Now, we can calculate the rate of heat transfer: \(Q = 500\,\mathrm{W/(m^2 K)} \times 0.070686\mathrm{~m^2} \times (383\mathrm{~K} - 373\mathrm{~K}) = 352.88\,\mathrm{W}\) So, the rate of heat transfer to the water is \(352.88\,\mathrm{W}\).

Calculate the rate of evaporation

To find the rate of evaporation, we will use the following equation: \(\dfrac{dm}{dt} = \dfrac{Q}{L_v}\) Where: - \(\dfrac{dm}{dt}\) is the rate of evaporation (mass of water evaporated per unit time), - \(Q\) is the rate of heat transfer, and - \(L_v\) is the latent heat of vaporization. For water at atmospheric pressure, the latent heat of vaporization is approximately \(L_v = 2.26 \times 10^6\,\mathrm{J/kg}\). Now, we can calculate the rate of evaporation: \(\dfrac{dm}{dt} = \dfrac{352.88\,\mathrm{W}}{2.26 \times 10^6\,\mathrm{J/kg}} = 1.56 \times 10^{-4}\,\mathrm{kg/s}\) So, the rate of evaporation is \(1.56 \times 10^{-4}\,\mathrm{kg/s}\). In summary, we have determined \((a)\) the rate of heat transfer to the water as \(352.88\,\mathrm{W}\), and \((b)\) the rate of evaporation as \(1.56 \times 10^{-4}\,\mathrm{kg/s}\).