A person puts a few apples into the freezer at \(-15^{\circ} \mathrm{C}\) to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of \(20^{\circ} \mathrm{C}\), and the heat transfer coefficient on the surfaces is \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Treating the apples as 9 -cm-diameter spheres and taking their properties to be \(\rho=840 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.81 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.418 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\), determine the center and surface temperatures of the apples in \(1 \mathrm{~h}\). Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Short Answer

Expert verified
Based on the provided data and calculations, after being placed in a freezer for 1 hour, the center temperature of the apple will be approximately 7.14°C, and the surface temperature will be -15°C. During this time, each apple transfers about 23.63 kJ of heat to the freezer.

Step by step solution

01

Calculate the Biot number, Fo, and hD/k ratio

To start, let's calculate the Biot number, Fourier number (Fo) and the ratio hD/k. Biot Number (Bi) is given by the formula \(Bi=\frac{hD}{k}\), where h is the heat transfer coefficient, D is the diameter of the apple, and k is the thermal conductivity. Fourier number (Fo) is given by the formula \(Fo=\frac{\alpha t}{r^{2}}\), where α is the thermal diffusivity, t is the time, and r is the radius of the apple. Finally, the hD/k ratio will be used later to calculate heat transfer. $$Bi = \frac{hD}{k} = \frac{8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot 0.09 \mathrm{~m}}{0.418 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} = 1.7235$$ $$Fo = \frac{\alpha t}{r^{2}} = \frac{1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s} \cdot 3600 \mathrm{~s}}{0.045^2 \mathrm{~m^{2}}} = 0.3511$$
02

Calculate the temperature distribution

We'll use the transient heat conduction equation with a one-term approximation method. Considering a spherical object (apple), the temperature distribution can be found using the following equation: $$\frac{T(r, t) - T_{\infty}}{T_{i} - T_{\infty}} = \frac{4}{\sqrt{\pi}}\frac{1}{\sqrt{Fo}}e^{-(1.128Bi)^{2}}$$ where \(T(r, t)\) is the temperature at any radial position r and time t, \(T_{\infty}\) is the freezer temperature (-15\(^{\circ}\mathrm{C}\)), and \(T_{i}\) is the initial temperature of the apple (20\(^{\circ}\mathrm{C}\)). We'll find the center and surface temperatures, which are at \(r=0\) and \(r=0.045\)m, respectively. $$T_{center} = T(0, 1h) = 20 - \frac{4}{\sqrt{\pi}}\frac{1}{\sqrt{Fo}}e^{-(1.128Bi)^{2}}(20 - (-15))$$ $$T_{surface} = T(0.045, 1h) = 20 - (20 - (-15))$$ Now, plug in the values of Bi and Fo to find the temperatures: $$T_{center} \approx 7.14^{\circ}\mathrm{C}$$ $$T_{surface} \approx -15^{\circ}\mathrm{C}$$
03

Calculate the heat transfer

Finally, we'll calculate the amount of heat transfer from each apple. The heat transfer can be found using the lumped system analysis equation: $$Q = mc_{p}(T_{i} - T_{f})$$ where \(m\) is the mass of the apple, \(c_{p}\) is the specific heat capacity, \(T_{i}\) is the initial temperature, and \(T_{f}\) is the final temperature. The apple's mass can be found using its density and volume: $$m = \rho V = \rho \frac{4}{3} \pi r^{3} = 840 \mathrm{~kg/m^{3}}\cdot\frac{4}{3}\pi(0.045)^{3} \approx 0.245 \mathrm{~kg}$$ Since the internal and external temperatures are almost uniform, we can consider the average temperature of the apple at the end of one hour \(T_f\): $$T_f = \frac{T_{center} + T_{surface}}{2} \approx -3.93^{\circ}\mathrm{C}$$ Now, plug in the values to find the heat transfer: $$Q \approx 0.245 \mathrm{~kg} \cdot 3.81 \mathrm{~kJ/kg \cdot K} \cdot (20 - (-3.93))^{\circ}\mathrm{C} \approx 23.63 \mathrm{~kJ}$$ Each apple transfers about 23.63 kJ of heat to the freezer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Biot Number
The Biot number (Bi) is a dimensionless quantity in heat transfer analysis that helps in determining whether the heat conduction within a body is faster than the heat transfer across the boundary. It is calculated using the formula:
\[\begin{equation}Bi = \frac{hD}{k}\end{equation}\]
where \(h\) is the heat transfer coefficient, \(D\) is the characteristic length (typically the diameter for spheres), and \(k\) is the thermal conductivity. If Bi < 0.1, the temperature within the object can be assumed uniform due to rapid conduction compared to convection (lumped system analysis). However, if Bi is significantly larger than 0.1, temperature gradients within the object need to be considered for accurate analysis.
Fourier Number
The Fourier number (Fo) is another dimensionless number that characterizes transient heat conduction. It is a ratio of conductive heat transfer to the energy stored and is found using the formula:
\[\begin{equation}Fo = \frac{\alpha t}{r^{2}}\end{equation}\]
where \(\alpha\) is the thermal diffusivity, \(t\) is the time, and \(r\) is the radius (or half the characteristic length) of the body. The Fourier number provides insight into how much time is required for a material to reach a certain level of thermal equilibrium, with higher values indicating a closer approach to uniform temperature distribution.
Analytical One-Term Approximation Method
The analytical one-term approximation method simplifies the solution of the complex transient heat conduction equation by using a single term from the infinite series. This method is particularly useful when dealing with spherical, cylindrical, or planar geometries and gives a reasonable approximation when the first term dominates the series, which is often the case for short times or objects with high thermal diffusivity. By applying this method, estimates for temperature distribution, as shown in the exercise above, can be determined effectively without detailed numerical analysis.
Heat Transfer Coefficient
The heat transfer coefficient (\(h\)) quantifies the convective heat transfer between a solid surface and the surrounding fluid. It is measured in watts per square meter-kelvin \(\left(\frac{W}{m^2 \cdot K}\right)\). This value depends on the properties of the fluid, the flow conditions, and the surface geometry. A higher \(h\) means that heat will be transferred more efficiently from the surface into the fluid or vice versa.
Thermal Diffusivity
Thermal diffusivity (\(\alpha\)) is a measure of how quickly a material can conduct heat relative to how much heat it can store. It is given by:
\[\begin{equation}\alpha = \frac{k}{\rho c_{p}}\end{equation}\]
where \(k\) is thermal conductivity, \(\rho\) is density, and \(c_{p}\) is the specific heat capacity. The higher the thermal diffusivity, the faster the heat spreads through the material.
Specific Heat Capacity
Specific heat capacity (\(c_p\)) represents how much energy is needed to raise the temperature of a unit mass of a substance by one degree Celsius. It is a crucial property in heat transfer calculations because it indicates the ability of a material to store thermal energy. Specific heat capacity impacts how rapidly a material can change its temperature when exposed to heat or how well it can maintain temperature in the face of heat loss or gain.
Thermal Conductivity
Thermal conductivity (\(k\)) is a fundamental material property that measures the ability of a material to conduct heat. It is commonly found in the units \(\frac{W}{m \cdot K}\). Materials with high thermal conductivity can transfer heat quickly, such as metals, while materials with low thermal conductivity, such as plastics or wood, act as insulators. Thermal conductivity plays a vital role in determining the heat transfer rate through an object and is an essential parameter in calculating the Biot number and thermal diffusivity.
Lumped System Analysis
Lumped system analysis is an approach to simplify the otherwise complicated differential equations in transient heat conduction problems. By assuming that the temperature variation inside a body is negligible relative to the variation at the surface, we can treat the entire object at a single, uniform temperature. This is suitable when the Biot number is less than 0.1 and leads to an ordinary differential equation that's straightforward to solve. It helps in approximately calculating the total heat transfer over time as demonstrated in the given solution.
Spherical Heat Conduction
Spherical heat conduction refers to the transfer of heat within spherical objects. This process is governed by the spherical version of the heat equation, which is generally more complex than its Cartesian counterpart due to geometric considerations. In practice, formulas are often derived for specific boundary conditions and simplified scenarios, such as in this exercise, where the analytical one-term approximation method is used to approximate the temperature distribution within the apples, treated as spherical objects.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the physical significance of the Fourier number? Will the Fourier number for a specified heat transfer problem double when the time is doubled?

Conduct the following experiment at home to determine the combined convection and radiation heat transfer coefficient at the surface of an apple exposed to the room air. You will need two thermometers and a clock. First, weigh the apple and measure its diameter. You can measure its volume by placing it in a large measuring cup halfway filled with water, and measuring the change in volume when it is completely immersed in the water. Refrigerate the apple overnight so that it is at a uniform temperature in the morning and measure the air temperature in the kitchen. Then take the apple out and stick one of the thermometers to its middle and the other just under the skin. Record both temperatures every \(5 \mathrm{~min}\) for an hour. Using these two temperatures, calculate the heat transfer coefficient for each interval and take their average. The result is the combined convection and radiation heat transfer coefficient for this heat transfer process. Using your experimental data, also calculate the thermal conductivity and thermal diffusivity of the apple and compare them to the values given above.

Citrus trees are very susceptible to cold weather, and extended exposure to subfreezing temperatures can destroy the crop. In order to protect the trees from occasional cold fronts with subfreezing temperatures, tree growers in Florida usually install water sprinklers on the trees. When the temperature drops below a certain level, the sprinklers spray water on the trees and their fruits to protect them against the damage the subfreezing temperatures can cause. Explain the basic mechanism behind this protection measure and write an essay on how the system works in practice.

A stainless steel slab \((k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=\) \(\left.3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) and a copper slab \((k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=117 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) are placed under an array of laser diodes, which supply an energy pulse of \(5 \times 10^{7} \mathrm{~J} / \mathrm{m}^{2}\) instantaneously at \(t=0\) to both materials. The two slabs have a uniform initial temperature of \(20^{\circ} \mathrm{C}\). Using \(\mathrm{EES}\) (or other) software, investigate the effect of time on the temperatures of both materials at the depth of \(5 \mathrm{~cm}\) from the surface. By varying the time from 1 to \(80 \mathrm{~s}\) after the slabs have received the energy pulse, plot the temperatures at \(5 \mathrm{~cm}\) from the surface as a function of time.

When water, as in a pond or lake, is heated by warm air above it, it remains stable, does not move, and forms a warm layer of water on top of a cold layer. Consider a deep lake \(\left(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that is initially at a uniform temperature of \(2^{\circ} \mathrm{C}\) and has its surface temperature suddenly increased to \(20^{\circ} \mathrm{C}\) by a spring weather front. The temperature of the water \(1 \mathrm{~m}\) below the surface 400 hours after this change is (a) \(2.1^{\circ} \mathrm{C}\) (b) \(4.2^{\circ} \mathrm{C}\) (c) \(6.3^{\circ} \mathrm{C}\) (d) \(8.4^{\circ} \mathrm{C}\) (e) \(10.2^{\circ} \mathrm{C}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free