Problem 62 - College Physics

Question

A spherical pot of hot coffee contains \(0.75 \mathrm{~L}\) of liquid (essentially water) at an initial temperature of \(95^{\circ} \mathrm{C}\). The pot has an emissivity of \(0.60,\) and the surroundings are at a temperature of \(20.0^{\circ} \mathrm{C}\). Calculate the coffee's rate of heat loss by radiation.

Step-by-Step Solution

Answer

The rate of heat loss by radiation is calculated using the Stefan-Boltzmann Law with given emissivity, temperatures in Kelvin, and surface area derived from volume.

Step 1: Understand the Problem

We need to calculate the rate of heat loss from the coffee, which cools by radiation. We will use the Stefan-Boltzmann Law, which relates temperature and heat loss by radiation.

Step 2: Note Given Values

The coffee volume is 0.75 L, the initial temperature of the coffee is 95°C, the emissivity of the pot is 0.60, and the surrounding temperature is 20°C. 1 L = 0.001 m³, so the volume of the coffee is 0.00075 m³.

Step 3: Convert Temperatures to Kelvin

Convert the temperatures from Celsius to Kelvin using the formula: \(T(\text{K}) = T(\text{C}) + 273.15\). Thus, the temperature of the coffee, \(T_c = 95 + 273.15 = 368.15\, K\), and the surrounding temperature, \(T_s = 20 + 273.15 = 293.15\, K\).

Step 4: Calculate Surface Area of the Pot

Assuming the pot is a perfect sphere, use the volume to find the radius, then the surface area. The volume of a sphere is \(V = \frac{4}{3}\pi r^3\). Solve for \(r\) and use \(A = 4\pi r^2\) for surface area.\[\begin{align*} V &= \frac{4}{3} \pi r^3 = 0.00075 \ r &= \left(\frac{3 \times 0.00075}{4 \pi}\right)^{1/3} \ A &= 4\pi r^2. \end{align*}\]

Step 5: Apply the Stefan-Boltzmann Law

The rate of heat loss \(P\) is calculated using \(P = \varepsilon \sigma A (T_c^4 - T_s^4)\), where \(\varepsilon = 0.60\) is the emissivity, \(\sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4\) is the Stefan-Boltzmann constant. Calculate using values derived earlier for \(A\), \(T_c\), and \(T_s\).

Step 6: Compute Results

Using the calculated values from previous steps, plug these into the Stefan-Boltzmann Law formula to solve for \(P\). Ensure all units are consistent for accuracy.

Key Concepts

Heat Transfer

Heat transfer is the process by which thermal energy is exchanged between physical systems. It can occur via three primary mechanisms: conduction, convection, and radiation. In this particular problem, heat loss from the coffee occurs through radiation, which is the emission of energy as electromagnetic waves or as moving subatomic particles. Radiation does not require a medium, such as air or water, to transfer heat. This means that it can occur even in a vacuum.

In many real-world scenarios, different modes of heat transfer can happen simultaneously. However, in this exercise, the focus is on radiative heat transfer. Understanding how and why heat is transferred is essential in developing systems for heating, cooling, and managing energy efficiency across various applications.

Radiation

Radiation is one of the core processes for heat transfer, where energy is emitted by a body in the form of electromagnetic waves. Every body with a temperature above absolute zero emits some form of thermal radiation. The amount of energy radiated depends on the temperature and surface properties of the object.

The Stefan-Boltzmann Law is a fundamental principle used to calculate the heat transfer by radiation. The law states that the power radiated by a black body is directly proportional to the fourth power of its absolute temperature. For real objects, which are not perfect black bodies, this concept is refined by using a factor called emissivity. Hence, energy radiated can be calculated using the formula: \(
P = \varepsilon \sigma A (T_c^4 - T_s^4)\)
Where:

• \(P\) is the rate of heat loss (Watt),
• \(\varepsilon\) is emissivity (a unitless measure ranging from 0 to 1),
• \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \, \text{W/m}^2 \text{K}^4)\),
• \(A\) is the surface area (\(\text{m}^2\)),
• \(T_c\) and \(T_s\) are the temperatures of the coffee and surroundings in Kelvin, respectively.

Temperature Conversion

To perform calculations involving temperature, converting between different temperature scales is crucial in ensuring precision. The Celsius scale is commonly used in everyday applications, while Kelvin is the SI unit for thermodynamic temperature and is widely used in scientific calculations. Converting from Celsius to Kelvin is straightforward and is achieved with the formula:

\(T(\text{K}) = T(\text{°C}) + 273.15\)

In this problem, the temperatures of both the coffee and the surroundings need to be converted from Celsius to Kelvin. This conversion is necessary because the Stefan-Boltzmann Law requires absolute temperatures in Kelvin for accurate calculations of radiative heat transfer. Understanding how to convert temperatures is a fundamental skill to prevent common calculation errors in physics and engineering problems.

Emissivity

Emissivity is a measure of an object's ability to emit infrared energy compared to that of a perfect black body. It is a dimensionless value ranging between 0 and 1. A black body has an emissivity of 1, meaning it emits the maximum possible radiation at a given temperature. In contrast, an object with an emissivity of 0 does not radiate at all.

The pot in this exercise has an emissivity of 0.60, indicating that it emits 60% of the radiation a perfect black body would at the same temperature. Emissivity affects how efficiently an object can radiate heat energy and is an essential factor in calculations using the Stefan-Boltzmann Law. It is influenced by factors like the material’s surface structure, texture, and temperature. Without accounting for emissivity, heat transfer calculations would be inaccurate, leading to potential errors in understanding energy dynamics in various systems.