Question

Consider a person whose exposed surface area is \(1.9 \mathrm{~m}^{2}\), emissivity is \(0.85\), and surface temperature is \(30^{\circ} \mathrm{C}\). Determine the rate of heat loss from that person by radiation in a large room whose walls are at a temperature of (a) \(300 \mathrm{~K}\) and (b) \(280 \mathrm{~K}\).

Short answer

Question: Calculate the rate of heat loss from a person by radiation in a large room whose walls are at a temperature of (a) 300 K and (b) 280 K. The person's surface temperature is 30°C, their exposed surface area is 1.9 m², and their emissivity is 0.85. Answer: The rate of heat loss from the person by radiation in a large room whose walls are at a temperature of (a) 300 K is 105.1 W, and (b) 280 K is 199.2 W.

Step-by-step solution

Convert the surface temperature to Kelvin

First, we need to convert the given surface temperature from Celsius to Kelvin. To do that, simply add 273.15 to the Celsius temperature: \(T_{body} = 30^{\circ}\mathrm{C} + 273.15 = 303.15 \mathrm{K}\)

Apply the Stefan-Boltzmann law

The Stefan-Boltzmann law states that the power of thermal radiation emitted by a body is given by: \(P = e \cdot A \cdot \sigma \cdot (T_{1}^{4} - T_{2}^{4})\), where \(e\) is the emissivity, \(A\) is the exposed surface area, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{Wm^{-2}K^{-4}}\)), \(T_{1}\) is the higher temperature, and \(T_{2}\) is the lower temperature.

Calculate the rate of heat loss for a wall temperature of \(300 \mathrm{K}\)

For a wall temperature of \(300 \mathrm{K}\), we have: \(P_{300} = 0.85 \cdot 1.9 \mathrm{m}^{2} \cdot 5.67 \times 10^{-8} \mathrm{Wm^{-2}K^{-4}} \cdot (303.15^{4} \mathrm{K}^{4} - 300^{4} \mathrm{K}^{4})\) \(P_{300} = 105.1 \mathrm{W}\)

Calculate the rate of heat loss for a wall temperature of \(280 \mathrm{K}\)

For a wall temperature of \(280 \mathrm{K}\), we have: \(P_{280} = 0.85 \cdot 1.9 \mathrm{m}^{2} \cdot 5.67 \times 10^{-8} \mathrm{Wm^{-2}K^{-4}} \cdot (303.15^{4} \mathrm{K}^{4} - 280^{4} \mathrm{K}^{4})\) \(P_{280} = 199.2 \mathrm{W}\) The rate of heat loss from the person by radiation in a large room whose walls are at a temperature of (a) \(300 \mathrm{K}\) is \(105.1 \mathrm{W}\) and (b) \(280 \mathrm{K}\) is \(199.2 \mathrm{W}\).

Key concepts

Stefan-Boltzmann Law

Understanding how objects lose heat through thermal radiation is essential in many fields of science and engineering. The Stefan-Boltzmann Law provides a foundation for calculating this heat transfer. Put simply, it states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's thermodynamic temperature.

This law is expressed by the equation:
\[ P = e \cdot A \cdot \sigma \cdot (T_{1}^{4} - T_{2}^{4}) \]
Where:

• \( P \) is the power or total energy radiated per unit time,
• \( e \) represents the emissivity of the material,
• \( A \) is the surface area,
• \( \sigma \) is a constant known as the Stefan-Boltzmann constant, and
• \( T_{1} \) and \( T_{2} \) are the absolute temperatures of the object and its surroundings, respectively, in Kelvin.

Emissivity is an important term here; it measures how effectively a real surface emits thermal radiation compared to an ideal black body. The Stefan-Boltzmann constant has a value of approximately \( 5.67 \times 10^{-8} \) W/m²K⁴, a fundamental parameter for radiative heat transfer calculations.

Surface Temperature Conversion

When dealing with thermal radiation calculations, it's essential to work with absolute temperature measurements. This means using the Kelvin scale, which is an SI unit of temperature and reflects the theoretical absolute zero, a point at which there is a complete absence of heat energy.

Temperature conversion from Celsius (°C) to Kelvin (K) is straightforward:

• To convert from Celsius to Kelvin, add 273.15.
• Conversely, to convert from Kelvin to Celsius, subtract 273.15.

In mathematical terms, the conversion looks like this:
\[ T(K) = T(°C) + 273.15 \]
or
\[ T(°C) = T(K) - 273.15 \]
Understanding this conversion is critical because the Stefan-Boltzmann Law requires temperature to be in Kelvin for accurate heat loss calculations by radiation.

Emissivity

Emissivity is an intrinsic property of materials that indicates how efficiently they emit infrared radiation. It is represented as a value between 0 and 1:

• An emissivity of 1 means the material is a 'perfect' emitter, equivalent to a black body, which is a hypothetical object that absorbs all radiation incident on it.
• An emissivity below 1 means the material is less efficient at emitting radiation.

The concept of emissivity is crucial because most objects are not perfect black bodies. Their ability to emit radiation varies based on material properties, surface texture, and temperature, among others. In practical thermodynamics, taking emissivity into account adjusts the ideal calculations of the Stefan-Boltzmann Law to account for the inefficiencies of real materials. As demonstrated in our example, the person’s skin has an emissivity of 0.85, indicating it’s a good but not perfect emitter of radiation.

Heat Transfer by Radiation

Heat can be transferred in three main ways: conduction, convection, and radiation. Radiation differs from the other two methods as it does not require any medium for transfer; it can occur through a vacuum.

Heat transfer by radiation occurs through electromagnetic waves, primarily in the infrared spectrum, and all objects emit these waves depending on their temperature. The higher the temperature of an object, the more thermal radiation it emits.

This method of heat transfer is particularly interesting because it operates at a distance and can be quite efficient, with the heat transfer rate dictated by factors like temperature difference, surface area, and emissivity. This process is exemplified by how Earth receives heat from the Sun or how a campfire warms your hands from a distance. The mathematical descriptions provided by the Stefan-Boltzmann Law allow us to calculate the rate of heat loss for objects due to radiation, as seen in the given exercise, providing insights essential for a wide range of applications, from building design to spacecraft thermal control systems.