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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Law of Cooling

Newton's Law of Cooling is a fundamental principle in heat transfer and thermodynamics. This law helps us understand how temperature changes over time for an object when it is in a different temperature environment. It states that the rate of temperature change of an object is directly proportional to the difference between the object's current temperature and the surrounding temperature. In mathematical terms, this can be expressed as: \( \frac{dT}{dt} = hA \frac{T_a - T}{mc_p} \)
Here:

• \( dT/dt \) is the rate of change of temperature over time.
• \( h \) is the heat transfer coefficient, which represents how efficiently heat is transferred to or from the object's surface.
• \( A \) denotes the surface area through which heat is being transferred.
• \( T_a \) is the ambient temperature, the temperature of the surrounding environment.
• \( T \) is the current temperature of the object.
• \( m \) represents the mass of the object and \( c_p \) is the specific heat capacity.

By integrating this equation, we can determine how long it will take for the object to reach a certain temperature. This is particularly useful in our watermelon example, where we wish to avoid freezing by setting a timer based on this law.

Specific Heat Capacity

Specific heat capacity is a vital concept in understanding thermal properties of materials. It is the amount of heat required to change the temperature of one kilogram of a substance by one degree Celsius. In formulas, it is often represented by \( c_p \). For our example with the watermelon, we assume it has properties similar to water. The specific heat capacity of water is \( 4180 \, \text{J/kg} \cdot \text{K} \). This factor plays a crucial role in determining how quickly or slowly an object will heat up or cool down.
Here's how it fits into the equation:

• In Newton's Law of Cooling, \( c_p \) helps determine the rate of temperature change for an object.
• Higher specific heat means the object can absorb more heat before its temperature increases significantly.
• For the watermelon, a high specific heat means it will take a considerable amount of time to cool down to the desired temperature, since it can store more thermal energy.

Understanding specific heat capacity is crucial not only in theoretical calculations but also in practical scenarios like heating or cooling foods, such as our watermelon.

Heat Transfer Coefficient

The heat transfer coefficient \( h \) plays a crucial role in the process of heat exchange between a surface and the surrounding fluid, like air or water. It is a measure of the effectiveness with which heat is conducted through a surface. Typically, its units are watts per square meter per degree Kelvin \, \( \text{W/m}^2 \cdot \text{K} \).

Its significance in the cooling of the watermelon can be understood through the following points:

• It defines how efficiently heat can be transferred from the watermelon to the cold air in the freezer.
• A higher heat transfer coefficient means faster cooling, as heat is removed more efficiently.
• In our scenario, the heat transfer coefficient is given as \( 22 \, \text{W/m}^2 \cdot \text{K} \), indicating a moderate rate of cooling.

Knowing the heat transfer coefficient helps in predicting the rate at which the watermelon will reach the target temperature of \( 3^{\circ} \text{C} \). This insight is vital for managing freezing times effectively.