Question

We often cut a watermelon in half and put it into the freezer to cool it quickly. But usually we forget to check on it and end up having a watermelon with a frozen layer on the top. To avoid this potential problem, a person wants to set a timer so that it will go off when the temperature of the exposed surface of the watermelon drops to \(3^{\circ} \mathrm{C}\). Consider a \(25-\mathrm{cm}\)-diameter spherical watermelon that is cut into two equal parts and put into a freezer at \(-12^{\circ} \mathrm{C}\). Initially, the entire watermelon is at a uniform temperature of \(25^{\circ} \mathrm{C}\), and the heat transfer coefficient on the surfaces is $22 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming the watermelon to have the properties of water, determine how long it will take for the center of the exposed cut surfaces of the watermelon to drop to \(3^{\circ} \mathrm{C}\).

Short answer

Answer: It will take approximately 1056 seconds or approximately 17.6 minutes for the temperature at the center of the exposed cut surface to drop to 3°C.

Step-by-step solution

Calculate the surface area of the watermelon

First, we need to find the surface area of the exposed cut surface of the half-watermelon. With a diameter of \(25\,\mathrm{cm}\), the radius is \(12.5\,\mathrm{cm}\) or \(0.125\,\mathrm{m}\). The surface area of the half-sphere is half of the surface area of a sphere, which is: \(A = \dfrac{1}{2}( 4 \pi r^2) = \dfrac{1}{2}( 4 \pi (0.125)^2) = 0.098\,\mathrm{m}^2\)

Calculate the temperature difference

The temperature difference between the surface-to-center temperature and the freezer temperature is: \(\Delta T = 3^\circ\mathrm{C} - (-12^\circ\mathrm{C}) = 15\,\mathrm{K}\)

Compute the heat transfer by convection

Now that we have the temperature difference and surface area, we can calculate the heat transfer by convection using Newton's law of cooling: \(Q = hA \Delta T\) Where \(Q\) is the heat transfer, \(h\) is the heat transfer coefficient (\(22\,\mathrm{W/m^2\cdot K}\)), \(A\) is the surface area, and \(\Delta T\) is the temperature difference. Substituting the given values, we have: \(Q = (22\,\mathrm{W/m^2\cdot K})(0.098\,\mathrm{m^2})(15\,\mathrm{K}) = 32.34\,\mathrm{W}\)

Calculate the heat capacity of the half-watermelon

We will assume that the half-watermelon has the properties of water, with a specific heat capacity \(c_p = 4.18\,\mathrm{kJ/kg\cdot K}\). The mass of the watermelon can be estimated by: \(m = \dfrac{4}{3} \pi r^3 \rho\) Where \(\rho\) is the density of water, which is approximately \(1000\,\mathrm{kg/m^3}\). Substituting the radius (\(0.125\,\mathrm{m}\)) and density, we get: \(m = \dfrac{4}{3} \pi (0.125)^3 (1000\,\mathrm{kg/m^3}) = 8.18\,\mathrm{kg}\) The heat capacity of the half-watermelon is: \(C = mc_p = (8.18\,\mathrm{kg})(4.18\,\mathrm{kJ/kg\cdot K}) = 34.15\,\mathrm{kJ/K}\)

Calculate the time required for the temperature to drop

Finally, we can compute the time required for the temperature to drop by dividing the heat capacity by the heat transfer: \(t = \dfrac{C}{Q} = \dfrac{34.15\,\mathrm{kJ/K}}{32.34\,\mathrm{W}}\) Since \(1\,\mathrm{W} = 1\,\mathrm{J/s}\), we need to convert \(34.15\,\mathrm{kJ/K}\) into Joules: \(C = 34.15\,\mathrm{kJ/K} \times \dfrac{1000\,\mathrm{J}}{1\,\mathrm{kJ}} = 34150\,\mathrm{J/K}\) Now, compute for \(t\): \(t = \dfrac{34150\,\mathrm{J/K}}{32.34\,\mathrm{W}} = 1056\,\mathrm{s}\) Hence, it will take approximately \(1056\,\mathrm{s}\) or \(\approx 17.6\,\mathrm{minutes}\) for the temperature at the center of the exposed cut surface of the half-watermelon to drop to \(3^\circ\mathrm{C}\).