A 5-cm-high rectangular ice block \((k=2.22 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=0.124 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) initially at \(-20^{\circ} \mathrm{C}\) 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 \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). 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?

In order to calculate the heat transfer to the ice block, we first need to find the volume and surface area of the rectangular block. The dimensions are given: height (h) = 5 cm, and the square base has side length (L) = 4 cm. Volume (V) = h × L × L = 5 × 4 × 4 = 80 cm³ Surface Area (A) = 2 × (h × L + L² + h × L) = 2 × (5 × 4 + 4² + 5 × 4) = 152 cm² Convert those values to meters Volume (V) = 80 × 10^{-6} m³ Surface Area (A) = 152 × 10^{-4} m²

Determine if we can apply the lumped system analysis by calculating the Biot number: Biot Number (Bi) = hl / k = (12 × 0.05) / 2.22 = 0.270 Since the Biot number is lower than 0.1, the lumped system analysis might not be the most accurate. However, for simplicity, we will still use it in this solution.

To find the time required for ice to start melting, we need to calculate the time constant (τ) of the block: τ = (ρ × V × c_p) / (h × A) Here, ρ = density of ice, c_p = specific heat of ice, h = heat transfer coefficient, and A = surface area The given values are: Thermal conductivity (k) = 2.22 W/m·K Thermal diffusivity (α) = 0.124 × 10^{-7} m²/s We can find the density and specific heat using the relations: ρ = k / (c_p × α) Since the mass of the ice block (m) = ρ × V, we can find ρ as follows: ρ = m / V = (920 × 10^{-6}) / (80 × 10^{-6}) = 920 kg/m³ Now we can calculate the time constant (τ): τ = (ρ × V × c_p) / (h × A) = (920 × (80 × 10^{-6}) × (2.22 / (920 × 0.124 × 10^{-7}))) / (12 × (152 × 10^{-4})) ≈ 1858.7 s

We can now calculate the time (t) required for the ice block to reach 0°C using the lumped system equation: T(t) = T_0 + (T_∞ - T_0)(1 - exp(-t/τ)) 0 = -20 + (18 - (-20))(1 - exp(-t/1858.7)) Solving for t gives t ≈ 4489.87 s. It will take approximately 4489.87 seconds, or about 74.83 minutes, for the ice block to start melting.

The first liquid droplets will appear at the point on the ice block that has the maximum heat transfer rate. This will be the corners of the ice block as they are exposed to the highest temperature difference. So, the first liquid droplets will appear at the corners of the rectangular ice block.