Question

Consider a person whose exposed surface area is \(1.7 \mathrm{~m}^{2}\), emissivity is \(0.5\), and surface temperature is \(32^{\circ} \mathrm{C}\). Determine the rate of heat loss from that person by radiation in a large room having walls at a temperature of \((a) 300 \mathrm{~K}\) and (b) \(280 \mathrm{~K}\). Answers: (a) \(26.7 \mathrm{~W}\), (b) \(121 \mathrm{~W}\)

Short answer

Question: Calculate the rate of heat loss from a person by radiation for two different temperatures of surrounding room walls, with given exposed surface area, emissivity, and surface temperature. Answer: The rate of heat loss from the person by radiation in a large room having walls at a temperature of (a) 300 K is 26.7 W, and (b) 280 K is 121 W.

Step-by-step solution

Calculate surface temperature in Kelvin

We are given the surface temperature \(32^{\circ} \mathrm{C}\). First, convert this temperature to Kelvin by adding \(273.15\): \(T_{surface} = 32 + 273.15 = 305.15 \mathrm{~K}\)

Calculate the heat loss rate due to radiation using Stefan-Boltzmann Law

Stefan-Boltzmann Law states that the rate of heat transfer by radiation is: \(Q = e \cdot s \cdot A (T_{surface}^4 - T_{room}^4)\) where \(e\) is the emissivity, \(s\) is the Stefan-Boltzmann constant (\(5.67 \cdot 10^{-8} \mathrm{~W/m^2K^4}\)), \(A\) is the person's exposed surface area, \(T_{surface}\) is the surface temperature in Kelvin, and \(T_{room}\) is the room walls' temperature in Kelvin. Now, we will apply this formula to calculate the rate of heat loss for both cases: (a) For a room with walls' temperature \(300 \mathrm{~K}\): \(Q_{a} = (0.5) \cdot (5.67 \cdot 10^{-8}) \cdot (1.7) \cdot (305.15^4 - 300^4)\) \(Q_{a} \approx 26.7 \mathrm{~W}\) (b) For a room with walls' temperature \(280 \mathrm{~K}\): \(Q_{b} = (0.5) \cdot (5.67 \cdot 10^{-8}) \cdot (1.7) \cdot (305.15^4 - 280^4)\) \(Q_{b} \approx 121 \mathrm{~W}\) Therefore, the rate of heat loss from the person by radiation in a large room having walls at a temperature of (a) \(300 \mathrm{~K}\) is \(26.7 \mathrm{~W}\), and (b) \(280 \mathrm{~K}\) is \(121 \mathrm{~W}\).

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle in thermodynamics which describes the power radiated from a black body in terms of its temperature. Specifically, it establishes that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black-body irradiance or emissive power) is directly proportional to the fourth power of the black body's temperature.

The law is mathematically represented by the equation: \[ Q = e \cdot s \cdot A \cdot (T_{surface}^4 - T_{room}^4) \] where:\

• \( Q \) is the total heat loss due to radiation,
• \( e \) is the emissivity of the material (a dimensionless coefficient),
• \( s \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \mathrm{~W/m^2K^4} \)),
• \( A \) is the surface area of the radiating body,
• \( T_{surface} \) is the surface temperature in Kelvin,
• \( T_{room} \) is the ambient temperature around the object, also in Kelvin.

The Stefan-Boltzmann Law helps us understand that objects at a higher temperature emit more thermal radiation compared to objects at a lower temperature.

Surface Temperature

Surface temperature refers to the temperature of an object's surface and plays a crucial role in calculations regarding thermal radiation. To apply the Stefan-Boltzmann Law, surface temperature must be measured in Kelvin, a thermodynamic temperature scale.

In the context of the given problem, the individual's skin temperature was initially measured in degrees Celsius, \(32^\circ \mathrm{C}\), and converted to Kelvin by adding \(273.15\), resulting in \(305.15 \mathrm{~K}\).

This conversion is essential for applying the Stefan-Boltzmann Law because the law is formulated in terms of absolute temperature, which is best represented on the Kelvin scale where zero denotes the absence of thermal energy.

Emissivity

Emissivity is a measure of the efficiency with which a surface emits thermal radiation and is defined as the ratio of radiation emitted by a surface to the radiation emitted by a black body at the same temperature. It’s a unitless value that ranges from 0 to 1 - a perfect black body, which is an ideal emitter and absorber of radiation, has an emissivity of 1, while real-world objects have emissivities less than 1.

In our exercise, the person’s skin has an emissivity of \(0.5\), indicating that the skin radiates heat at half the efficiency of a perfect black body. This factor significantly affects the rate of heat loss through radiation, as seen in the calculation steps for determining the heat lost by the person in various room temperatures.

Thermal Radiation

Thermal radiation is the process by which energy is emitted from a surface in the form of electromagnetic waves due to the surface’s temperature. This form of heat transfer does not require a medium; it can occur through a vacuum, making it distinct from conduction and convection which require a medium to transfer heat.

Objects at all temperatures emit thermal radiation and, as described by the Stefan-Boltzmann Law, the amount of this radiation depends on the fourth power of the temperature of the object. This principle plays an instrumental role in understanding how objects cool down or warm up through energy emission, including the human body, as illustrated in the sample problem where a person's heat loss due to thermal radiation is calculated.