Question

A 2-m-internal-diameter double-walled spherical tank is used to store iced water at \(0^{\circ} \mathrm{C}\). Each wall is \(0.5 \mathrm{~cm}\) thick, and the \(1.5-\mathrm{cm}\)-thick airspace between the two walls of the tank is evacuated in order to minimize heat transfer. The surfaces surrounding the evacuated space are polished so that each surface has an emissivity of \(0.15\). The temperature of the outer wall of the tank is measured to be $20^{\circ} \mathrm{C}\(. Assuming the inner wall of the steel tank to be at \)0^{\circ} \mathrm{C}\(, determine \)(a)$ the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -h period.

Short answer

Answer: Approximately 94.26 kg.

Step-by-step solution

Calculate the surface area of the inner wall

First, we need to find the surface area of the inner wall of the tank. The inner wall has a radius of 1 m (given that the diameter is 2 m). The formula for the surface area of a sphere is given by \(A = 4\pi r^2\). Therefore, the surface area of the inner wall of the tank is: \(A = 4\pi (1)^2 = 4\pi \mathrm{m^2}\).

Determine the heat transfer through radiation

Next, we need to find the rate of heat transfer through radiation between the inner wall and the outer wall of the tank. The formula for heat transfer through radiation is given by \(Q = \sigma A \epsilon (T_h^4 - T_c^4)\), where \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8}\mathrm{W/(m^2K^4)}\)), \(A\) is the surface area, \(\epsilon\) is the emissivity, \(T_h\) is the temperature of the hotter body, and \(T_c\) is the temperature of the colder body. We are given the temperature of the outer wall (\(20^{\circ} \mathrm{C}\)) and emissivity of the surfaces (\(0.15\)). In order to get the temperature values in Kelvin, we need to add \(273.15\) (\(0^{\circ} \mathrm{C} = 273.15 \mathrm{K}\) and \(20^{\circ} \mathrm{C} = 293.15 \mathrm{K}\)). We can now plug in the values into the formula to find the rate of heat transfer: \(Q = (5.67 \times 10^{-8}\mathrm{W/(m^2K^4)})(4\pi \mathrm{m^2})(0.15)(293.15^4 - 273.15^4)\) Calculating the heat transfer rate, we get \(Q \approx 109.67 \mathrm{W}\).

Find the amount of ice that melts during a 24-hour period

To find the amount of ice that melts during a 24-hour period, we need to consider the heat required to change the phase of ice at \(0^{\circ} \mathrm{C}\) to water at \(0^{\circ} \mathrm{C}\). The formula for heat transfer is given by \(Q = mc\Delta T + mL\) where \(m\) is the mass of ice, \(c\) is the specific heat capacity of ice (\(2100 \mathrm{J/(kgK)}\)), \(\Delta T\) is the temperature difference (0 in this case as both ice and water are at \(0^{\circ} \mathrm{C}\)), and \(L\) is the latent heat of fusion for ice (\(333,000 \mathrm{J/kg}\)). We want to find the mass (\(m\)) of ice that melts. Since \(\Delta T = 0\), the heat transfer equation simplifies to \(Q = mL\). The total heat transfer during a 24-hour period is given by \(Q_t = Qt\), where \(t\) is the time period (24 hours) in seconds: \(Q_t = (109.67 \mathrm{W}) \times (24 \times 3600 \mathrm{s})\) Now we can solve for the mass of ice that melts: \(m = \frac{Q_t}{L} = \frac{(109.67 \mathrm{W} \times 24 \times 3600 \mathrm{s})}{333,000 \mathrm{J/kg}}\) Calculating the mass, we get \(m \approx 94.26 \mathrm{kg}\).

Final Answers

Therefore, the answers to the given problem are: \((a)\) The rate of heat transfer to the iced water in the tank is approximately \(109.67 \mathrm{W}\). \((b)\) The amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24-hour period is approximately \(94.26 \mathrm{kg}\).