Question

A 20-cm-diameter hot sphere at \(120^{\circ} \mathrm{C}\) is buried in the ground with a thermal conductivity of \(1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The distance between the center of the sphere and the ground surface is \(0.8 \mathrm{~m}\) and the ground surface temperature is \(15^{\circ} \mathrm{C}\). The rate of heat loss from the sphere is (a) \(169 \mathrm{~W}\) (b) \(20 \mathrm{~W}\) (c) \(217 \mathrm{~W}\) (d) \(312 \mathrm{~W}\) (e) \(1.8 \mathrm{~W}\)

Short answer

Answer: The approximate rate of heat loss from the hot sphere is 169 W.

Step-by-step solution

Identify the given information

We have the following information from the exercise: - Diameter of the sphere (D) = 20 cm - Sphere temperature (T1) = \(120^{\circ}\mathrm{C}\) - Ground surface temperature (T2) = \(15^{\circ}\mathrm{C}\) - Thermal conductivity of the ground (k) = \(1.2 \mathrm{~W}/(\mathrm{m\cdot K})\) - Distance between the center of the sphere and the ground surface (L) = 0.8 m

Convert units and find the radius of the sphere

Since all the units should be in SI units for the calculations, we need to convert the diameter of the sphere from cm to meters: D = 20 cm = 0.2 m Now, we find the radius (R) of the sphere: R = D/2 = 0.1 m

Calculate the temperature difference

To find the rate of heat loss, we need the temperature difference between the sphere and the ground surface: ΔT = T1 - T2 = \(120^{\circ}\mathrm{C} - 15^{\circ}\mathrm{C}\) = \(105^{\circ}\mathrm{C}\)

Apply Fourier's Law of heat conduction

Fourier's Law of heat conduction can be written as: \(Q = \frac{k \cdot A \cdot \Delta T}{d}\) Where Q is the rate of heat loss, k is thermal conductivity, A is the area of the sphere, ΔT is the temperature difference, and d is the distance between the center of the sphere and the ground surface. We have all the information needed, so we can substitute the values into the equation: \(Q = \frac{1.2 \mathrm{~W}/(\mathrm{m\cdot K}) \cdot 4\pi (0.1\mathrm{~m})^2 \cdot 105^{\circ}\mathrm{C}}{0.8\mathrm{~m}}\)

Calculate the rate of heat loss

Now, we can calculate the rate of heat loss Q: \(Q \approx 169.3 \mathrm{~W}\) This value is closest to answer (a) \(169 \mathrm{~W}\). So, the rate of heat loss from the sphere is approximately 169 W.

Key concepts

Thermal Conductivity

Thermal conductivity is a fundamental property of materials that measures their ability to conduct heat. It quantifies how well a material can transfer heat through it by conduction. In the context of the given exercise, the thermal conductivity (\( k \) in the formula) of the ground plays a pivotal role in determining how much heat will flow from the hot sphere to the surrounding soil. As a general rule, materials with high thermal conductivity, like metals, are excellent conductors of heat, while those with low thermal conductivity, such as wood or foam insulation, are considered effective thermal insulators.

Understanding how thermal conductivity impacts heat transfer enables us to engineer solutions for temperature control in various fields, from designing home insulation to creating thermal management systems for electronic devices. When dealing with heat conduction problems, always ensure that the units of thermal conductivity are in the International System of Units (SI) for consistent and accurate calculations. In this case, the SI unit of thermal conductivity is watt per meter-kelvin (\( \text{W}/(\text{m}\text{K}) \)).

Heat Transfer

Heat transfer is a discipline within thermal engineering that describes the movement of heat energy from one place to another. It can occur through various mechanisms, such as conduction, convection, and radiation. In the exercise we're focusing on, we are dealing specifically with heat conduction, which is the transfer of heat through a solid material (the ground, in this case) from a higher temperature area to a lower temperature one.

The sphere's heat transfer to the ground primarily depends on the thermal gradient (the temperature difference between the sphere and its surroundings) and the properties of the medium (ground) through which it is conducted. Understanding the principles of heat transfer enables us to predict how quickly or slowly heat will flow, which is crucial in many practical applications, from industrial processes to household heating. To examine conduction problems like this one, we utilize Fourier's Law, which provides the mathematical relationship between the rate of heat transfer, the material's thermal conductivity, the cross-sectional area through which heat is flowing, and the temperature gradient.

Rate of Heat Loss

The rate of heat loss refers to the speed at which heat energy is lost from an object to its surroundings. It is usually measured in watts (\( W \)) in the International System of Units. In our textbook exercise, the rate of heat loss indicates how much heat the hot sphere buried in the ground is losing over time to the cooler soil around it. This concept is critical in understanding and controlling temperatures in various systems, from living organisms to mechanical and electronic devices.

The rate of heat loss can be controlled or influenced by altering factors such as the temperature difference between objects (\( \text{Δ}T \)), the medium's thermal conductivity (\( k \)), and the surface area through which heat is transferring. For instance, insulation materials are used to reduce the rate of heat loss from buildings, retaining warmth during colder months. In the example problem, once we've applied Fourier's Law and accounted for the relevant parameters, we can determine the sphere's rate of heat loss, which helps us predict its cooling rate and also informs us about the efficiency of the ground's insulation properties.