Question

A 6-cm-high rectangular ice block $(k=2.22 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( and \)\alpha=0.124 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}$ ) initially at \(-18^{\circ} \mathrm{C}\) is placed on a table on its square base \(4 \mathrm{~cm} \times 4 \mathrm{~cm}\) in size in a room at $18^{\circ} \mathrm{C}$. The heat transfer coefficient on the exposed surfaces of the ice block is \(12 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\). Disregarding any heat transfer from the base to the table, determine how long it will be before the ice block starts melting. Where on the ice block will the first liquid droplets appear? Solve this problem using the analytical one-term approximation method.

Short answer

U = 6.887 W/m².K #tag_title#Step 2: Calculate the time it takes for the ice block to start melting#tag_content#Now that we have the overall heat transfer coefficient, we can use the temperature distribution equation in a semi-infinite solid, which is given by: \(T(x, t) = T_s - (T_s - T_i) erf(\frac{x}{2\sqrt{\alpha t}})\) where: T(x, t) = temperature at a certain depth x and time t (°C) T_s = surface temperature (°C) T_i = initial temperature of the ice block (°C) erf = error function α = thermal diffusivity of ice (m²/s) To find the time it takes for the ice block to start melting, we need to determine when the surface temperature (T_s) reaches the melting point of ice, which is 0°C. Thus, T(x, t) will be 0°C. The given values are: T_i = -10°C α = 1.31 x 10⁻⁶ m²/s In the equation, we are solving for t, while x = 0 at the surface of the ice block: \(0 = 0 - (-10) erf(\frac{0}{2\sqrt{1.31x10^{-6}t}})\) Since erf(0) = 0, the equation results in 10 = 0, which is not possible. This means that the error function approximation cannot be used in this case. Instead, we use Fourier's law of heat conduction to find the heat flux (q) reaching the surface of the ice block: \(q = U(T_\infty - T_s)\) Given the heat transfer coefficient U = 6.887 W/m².K, and surrounding temperature T_∞ = 20°C: \(q = 6.887(20 - 0) = 137.74 W/m²\) Now, we use the heat balance equation to find the time taken for the ice block to start melting: \(q = \rho c_p L \frac{dm}{dt}\) Where: ρ = density of ice (kg/m³) c_p = specific heat of ice (J/kg.K) L = heat of fusion for ice (J/kg) dm/dt = mass rate of melting Given the values: ρ = 900 kg/m³ c_p = 2100 J/kg.K L = 333500 J/kg The surface area of the ice block, A = 0.04 m², and the volume V = 0.0012 m³. From these, we can find the mass of the ice block: Mass = ρ × V = 900 kg/m³ × 0.0012 m³ = 1.08 kg Now, we can find the mass rate of melting (dm/dt): \(137.74 W/m² = 2100 J/kg.K × 0.0012 m³ × \frac{dm}{dt}\) \(dm/dt = \frac{137.74}{2526}\) dm/dt = 0.05451 kg/s Finally, we find the time it takes for the ice block to start melting: Time = mass / (dm/dt) = 1.08 kg / 0.05451 kg/s = 19.8 s #tag_title#Step 3: Determine the location where the first liquid droplets will appear#tag_content#To find the location where the first liquid droplets will appear, we will consider the temperature gradient along the ice block. Since the heat flux is not uniform across the surface, droplets will first form in regions with the highest heat flux. Therefore, the first liquid droplets will appear at the corners of the ice block, where the heat flux from the surrounding is highest. The non-uniform heat flux causes the ice to melt more rapidly at the sharper corners, which is why the first droplets will form there.

Step-by-step solution

Determine the overall heat transfer coefficient on the exposed surfaces

To determine the overall heat transfer coefficient (U), we need to use the formula that combines conduction and convection heat transfer: \(U = \frac{1}{\frac{1}{h}+\frac{L}{k}}\) where: U = overall heat transfer coefficient (W/m².K) h = heat transfer coefficient on the exposed surface (W/m².K) L = thickness of the ice block (m) k = thermal conductivity of ice (W/m.K) Using the given values: h = 12 W/m².K, k = 2.22 W/m.K, L = 0.06 m (6 cm) \(U = \frac{1}{\frac{1}{12}+\frac{0.06}{2.22}}\)