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 the timer such that it will go off when the temperature of the exposed surface of the watermelon drops to \(3^{\circ} \mathrm{C}\). Consider a 25 -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 \(3^{\circ} \mathrm{C}\).

Step by step solution

01

Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between the object's temperature and the ambient temperature. The equation is: $$ \frac{dT}{dt} = hA \frac{T_a - T}{mc_p} $$ Where: \(T\) - Temperature of the object \(T_a\) - Ambient temperature \(A\) - Surface area of the object \(m\) - Mass of the object \(c_p\) - Specific heat of the object \(h\) - Heat transfer coefficient \(t\) - Time

02

Write down the given values

We have the following values provided in the problem: Diameter of the watermelon \(d = 0.25 \, \mathrm{m}\) Initial temperature \(T_i = 25^{\circ}\mathrm{C}\) Desired temperature \(T_f = 3^{\circ}\mathrm{C}\) Freezer temperature \(T_a = -12^{\circ}\mathrm{C}\) Heat transfer coefficient \(h = 22 \, \frac{\mathrm{W}}{\mathrm{m}^2 \cdot\mathrm{K}}\)

03

Calculate the surface area, mass and specific heat of the watermelon

First, we need to calculate the surface area of the exposed cut surface of the watermelon. Since the watermelon is cut into two equal parts, the exposed surface area of each part is a circle with a radius of \(r = \frac{d}{2} = 0.125 \, \mathrm{m}\). The surface area of one part is: $$ A = \pi r^2 = \pi (0.125)^2 \, \mathrm{m}^2 $$ Next, we need to calculate the mass of the watermelon. We can use the density of water, \(\rho = 1000 \, \frac{\mathrm{kg}}{\mathrm{m}^3}\), and the volume of a sphere \(V = \frac{4}{3} \pi r^3\) to find the mass. Since there are two equal parts, we will consider the mass of one part, \(m = \frac{1}{2} \rho V\): $$ m = \frac{1}{2}\rho\cdot \frac{4}{3}\pi r^3 $$ Finally, we need to find the specific heat of water, \(c_p = 4180 \, \frac{\mathrm{J}}{\mathrm{kg} \cdot\mathrm{K}}\).

04

Rearrange the Newton's Law of Cooling equation for time

In order to find the time it takes for the center of the exposed cut surface of the watermelon to cool down to \(3^{\circ} \mathrm{C}\), we need to rearrange the cooling equation for time, \(t\). First, let's rewrite the equation with the known values: $$ \frac{dT}{dt} = \frac{hA(T_a - T)}{mc_p} $$ Integrating both sides of the equation with respect to temperature and time: $$ \int_{T_i}^{T_f} \frac{m c_p}{h A (T_a - T)} dT = \int_0^t dt $$ Now we can find the time, \(t\), it takes for the center of the exposed cut surface to cool down to \(3^{\circ}\mathrm{C}\).

05

Solve for the time

Solving the integral equation for \(t\), we get: $$ t = \frac{mc_p}{hA} \int_{T_i}^{T_f} \frac{1}{T_a - T} dT $$ Plug in the given values and solve the integrals: $$ t = \frac{(0.5)(1000)(\frac{4}{3}\pi (0.125)^3)(4180)}{22\pi(0.125)^2}\int_{25}^{3} \frac{1}{-12 - T} dT $$ After simplifying and evaluating the integral, we obtain the final result for the time, \(t\).

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