Question

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

Short answer

The coffee's rate of heat loss by radiation is approximately 81.8 W.

Step-by-step solution

Identify the Known Values

Firstly, identify the known values needed for calculations. The volume of the coffee, which we can approximate as a sphere, is given as 0.75 L, which is equivalent to 0.75 x 10^-3 m^3. The initial temperature of the coffee is 95°C and the surroundings are at 20°C. The emissivity of the pot is 0.60.

Calculate the Surface Area of the Sphere

The volume of a sphere is given by the formula \( V = \frac{4}{3}\pi r^3 \). We solve this for \( r \) and calculate the radius. Since \( r = \left(\frac{3V}{4\pi}\right)^{1/3} \) and \( V = 0.75 \times 10^{-3} \) m³, substituting the values will yield \( r \approx 0.057 \) m. The surface area \( A \) is then calculated using \( A = 4\pi r^2 \), which results in \( A \approx 0.041\text{ m}^2 \).

Use the Stefan-Boltzmann Law

The rate of heat loss by radiation is given by the Stefan-Boltzmann Law: \( P = \varepsilon \sigma A (T^4 - T_s^4) \). Here, \( \varepsilon = 0.60 \) is the emissivity, \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \) is the Stefan-Boltzmann constant, \( A \approx 0.041 \text{ m}^2 \) is the surface area, \( T = 95 + 273.15 = 368.15 \text{ K} \) and \( T_s = 20 + 273.15 = 293.15 \text{ K} \).

Substitute and Calculate the Rate of Heat Loss

Substitute all the known values into the Stefan-Boltzmann equation: \( P = 0.60 \times 5.67 \times 10^{-8} \times 0.041 \times (368.15^4 - 293.15^4) \). Calculating this gives the power \( P \approx 81.8 \text{ W} \).

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is essential in understanding thermal radiation and heat transfer. It states that the total energy radiated per unit surface area of a black body in unit time is directly proportional to the fourth power of the black body's temperature. Mathematically, this is expressed as:\[ P = \varepsilon \sigma A (T^4 - T_s^4) \] where:

• \( P \) is the power radiated, representing the rate of heat loss.
• \( \varepsilon \) is the emissivity, a measure of how effectively a real body radiates energy compared to a perfect black body.
• \( \sigma \) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\), a universal constant for black body radiation.
• \( A \) is the surface area of the radiating body.
• \( T \) and \( T_s \) represent the temperatures of the hot and surrounding environment in Kelvin, respectively.

This law helps us calculate the energy radiated by an object based on its temperature and surface characteristics.

Emissivity

Emissivity is a key concept in the study of thermal radiation, describing a material's effectiveness in emitting energy as thermal radiation. Ranging from 0 to 1, emissivity indicates how much radiation is emitted compared to a black body, which is the ideal emitter with an emissivity of 1. A material with an emissivity close to 1 radiates heat efficiently.
When considering a real-world object, such as a coffee pot in this exercise, emissivity affects how it radiates heat away. A higher emissivity indicates that the material will lose heat faster since it emits more thermal radiation. In the given problem, an emissivity of 0.60 suggests that the coffee pot is not a perfect black body, but it is relatively efficient at emitting heat.

Thermal Radiation

Thermal radiation is the energy emitted by objects due to their temperature. All bodies emit radiation spontaneously, including visible and invisible electromagnetic waves. The nature and intensity of the thermal radiation are influenced by the body's temperature and emissivity.
Unlike conduction and convection, thermal radiation does not require a medium to travel through and can occur in a vacuum. In the context of the spherical pot problem, the coffee emits thermal radiation to its cooler surroundings, leading to heat loss. The Stefan-Boltzmann Law helps quantify this energy loss, linking it to the temperature differential and the surface's emissivity.

Surface Area of a Sphere

In geometry, the sphere is a three-dimensional shape with all points on its surface equidistant from its center. For heat transfer calculations, the surface area of the sphere plays a crucial role in mathematically determining radiation losses. The formula for calculating the surface area \( A \) of a sphere based on its radius \( r \) is:\[ A = 4\pi r^2 \]
When given the volume, like in this problem, you can first find the radius using the formula for the volume \( V \) of a sphere \( V = \frac{4}{3}\pi r^3 \). From the known volume, you solve for \( r \) and subsequently find \( A \).
The specific area is vital, as it directly determines how much heat an object can radiate—it influences the rate of thermal radiation by setting the available surface through which energy escapes.