Question

You become lost while hiking outside wearing only a bathing suit. a) Calculate the power radiated from your body, assuming that your body's surface area is about \(2.00 \mathrm{~m}^{2}\) and your skin temperature is about \(33.0^{\circ} \mathrm{C} .\) Also, assume that your body has an emissivity of \(1.00 .\) b) Calculate the net radiated power from your body when you are inside a shelter at \(20.0^{\circ} \mathrm{C} .\) c) Calculate the net radiated power from your body when your skin temperature dropped to \(27.0^{\circ} \mathrm{C}\)

Short answer

Question: Calculate the power radiated from a person's body in three scenarios: when the skin temperature is \(33.0^{\circ} \mathrm{C}\), when they're inside a shelter with an ambient temperature of \(20.0^{\circ} \mathrm{C}\), and when their skin temperature drops to \(27.0^{\circ} \mathrm{C}\). Answer: The power radiated from the body is approximately 801.07 W when the skin temperature is \(33.0^{\circ} \mathrm{C}\). Inside the shelter, the net radiated power is approximately 163.75 W. And when the skin temperature drops to \(27.0^{\circ} \mathrm{C}\), the net radiated power is approximately 45.15 W.

Step-by-step solution

Convert the temperature to Kelvin

We need to convert the skin temperature from Celsius to Kelvin by adding 273.15. T = 33.0 + 273.15 = 306.15 K

Apply the Stefan-Boltzmann law

The Stefan-Boltzmann law states that the power radiated by a body is: \(P = e \cdot A \cdot \sigma \cdot T^4\) Substitute the given values into the formula. e = 1.00 (emissivity) A = 2.00 m² (surface area) \(\sigma = 5.67 \times 10^{-8} \mathrm{W/m^2 K^4}\) (Stefan-Boltzmann constant)

Calculate the power radiated

Using the formula from Step 2, we can find the power radiated: \(P = 1.00 \cdot 2.00 \cdot 5.67 \times 10^{-8} \cdot (306.15)^4 = 801.07 \mathrm{W}\) The power radiated from the body is approximately 801.07 W. #b) Net radiated power inside a shelter#

Convert the shelter temperature to Kelvin

Convert the shelter temperature from Celsius to Kelvin: T_shelter = 20.0 + 273.15 = 293.15 K

Calculate power absorbed inside shelter

Since the person is inside the shelter, they will also absorb radiation from the shelter. Calculate the power absorbed inside the shelter using the same Stefan-Boltzmann law, but replacing the person's temperature with the shelter's temperature. \(P_{absorbed} = e \cdot A \cdot \sigma \cdot T_{shelter}^4 = 1.00 \cdot 2.00 \cdot 5.67 \times 10^{-8} \cdot (293.15)^4 = 637.32 \mathrm{W}\)

Calculate the net radiated power

Now, we can find the net radiated power inside the shelter by taking the difference between the power radiated by the body and the power absorbed from the shelter. \(P_{net} = P - P_{absorbed} = 801.07 \mathrm{W} - 637.32 \mathrm{W} = 163.75 \mathrm{W}\) The net radiated power inside the shelter is approximately 163.75 W. #c) Net radiated power when skin temperature drops#

Convert the new skin temperature to Kelvin

Convert the new skin temperature from Celsius to Kelvin: T_new = 27.0 + 273.15 = 300.15 K

Calculate the new power radiated

Now, find the new power radiated by the body by plugging in the new skin temperature into the Stefan-Boltzmann law. \(P_{new} = e \cdot A \cdot \sigma \cdot T_{new}^4 = 1.00 \cdot 2.00 \cdot 5.67 \times 10^{-8} \cdot (300.15)^4 = 682.47 \mathrm{W}\)

Calculate the new net radiated power

Finally, calculate the new net radiated power by taking the difference between the new power radiated by the body and the power absorbed from the shelter. \(P_{net\_new} = P_{new} - P_{absorbed} = 682.47 \mathrm{W} - 637.32 \mathrm{W} = 45.15 \mathrm{W}\) The net radiated power when the skin temperature drops to \(27.0^{\circ} \mathrm{C}\) is approximately 45.15 W.

Key concepts

Radiated Power Calculation

Understanding the radiated power of an object is essential in the realms of physics and engineering, as it helps us understand how energy is transferred in the form of electromagnetic radiation. The Stefan-Boltzmann law provides a way to calculate this radiated power and it can be applied in practical situations such as understanding body heat loss.

The formula to calculate radiated power is:
\[P = e \times A \times \$sigma \times T^4\].Here,

• \(P\) is the radiated power in watts (W),
• \(e\) is the emissivity of the material (a dimensionless coefficient ranging from 0 to 1),
• \(A\) is the surface area of the radiating body in square meters (\(m^2\)),
• \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \mathrm{W/m^2 K^4}\), and
• \(T\) is the absolute temperature in Kelvin (K).

The emissivity represents how effectively a material radiates energy. An emissivity of 1.00, as assumed for the human skin in the exercise, means the body is a perfect blackbody radiator.

For effective learning, it is important to grasp that this formula tells us that the radiated power grows rapidly with increasing temperature—as indicated by the fourth power of temperature. This means that even small changes in temperature can lead to large changes in the radiated power.

Temperature Conversion to Kelvin

Temperature plays a pivotal role in calculations involving thermal radiation. To use the Stefan-Boltzmann law, temperatures must be converted to the Kelvin scale, which is an absolute temperature scale starting at absolute zero, the point where particles have the minimum thermal motion.

To convert Celsius to Kelvin, one must use the formula:
\[T(K) = T(^\circ C) + 273.15\].This conversion is crucial because the Stefan-Boltzmann law requires the absolute temperature for accurate calculations.

When students are performing these calculations, they need to be meticulous and ensure that they convert temperatures accurately to avoid errors in their final computed radiated power. Always remember that Kelvin is the SI unit for temperature and that no negative temperatures exist on the Kelvin scale.

Emissivity in Thermal Radiation

Emissivity is a key concept when dealing with thermal radiation. It quantifies the efficiency with which a surface emits thermal radiation compared to a perfect blackbody, which is assumed to have an emissivity of 1. Real-world objects have emissivities less than 1, implying they are not perfect radiators and will always emit less radiation than a blackbody at the same temperature.

Emissivity depends on factors like material properties, temperature, and surface roughness. It affects the radiated power calculation and hence the thermal management in various applications such as spacecraft design, building insulation, and even clothing materials.

In the example given, a human body with an emissivity of 1.00 is considered a perfect emitter, which simplifies the calculation. In reality, human skin has an emissivity less than 1; however, for the sake of exercise and understanding the concept, this assumption is used. It's crucial to understand that the concept of emissivity in thermal radiation is foundational for understanding how objects interact with their thermal environments.