Question

The skin temperature of a nude person is $34.0^{\circ} \mathrm{C}$, and the surroundings are at $20.0^{\circ} \mathrm{C}$. The emissivity of skin is $0.900$, and the surface area of the person is $1.50 \mathrm{~m}^{2}$. (a)What is the rate at which energy radiates from the person? (b)What is the net energy loss from the body in 1 min by radiation?

   The skin temperature of a nude person is $34.0^{\circ} \mathrm{C}$, and the surroundings are at $20.0^{\circ} \mathrm{C}$. The emissivity of skin is $0.900$, and the surface area of the person is $1.50 \mathrm{~m}^{2}$. (a)What is the rate at which energy radiates from the person? (b)What is the net energy loss from the body in 1 min by radiation?

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College Physics

College Physics

Roger A. Freedman;… 2nd Edition

Chapter 14, Problem 108

Instant Answer

Solved by Expert Pronoy Sinha

07/04/2021

Step 1

0^{\circ} \mathrm{C}$, which we convert to Kelvin by adding 273. So, $T_1 = 34 + 273 = 307 \mathrm{K}$. We are also given the temperature of the surroundings $T_2 = 20.0^{\circ} \mathrm{C}$, which we also convert to Kelvin. So, $T_2 = 20 + 273 = 293 \mathrm{K}$. Show more…

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The skin temperature of a nude person is $34.0^{\circ} \mathrm{C}$, and the surroundings are at $20.0^{\circ} \mathrm{C}$. The emissivity of skin is $0.900$, and the surface area of the person is $1.50 \mathrm{~m}^{2}$. (a)What is the rate at which energy radiates from the person? (b)What is the net energy loss from the body in 1 min by radiation?

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Key Concepts

Time Integration of Power

The total energy transferred over a period is calculated by integrating power with respect to time. In problems addressing energy loss or gain over a discrete time interval, the net power output is multiplied by the duration of the period to determine the total energy exchanged.

Stefan?Boltzmann Law

This law relates the power radiated from a black body to its absolute temperature. It states that the energy emitted per unit area is proportional to the fourth power of the absolute temperature. This principle is crucial in calculating the radiative heat transfer from any object, including those that are not perfect black bodies when adjusted by the emissivity factor.

Emissivity

Emissivity is a dimensionless factor ranging from 0 to 1 that represents how efficiently a real surface radiates energy compared to an ideal black body. An object with emissivity 1 is a perfect emitter, while a lower value indicates less efficient radiation. It modifies the Stefan?Boltzmann equation to account for real materials.

Net Radiative Heat Transfer

Net radiative heat transfer is the difference between the energy radiated by an object and the energy absorbed from its surroundings. By applying the Stefan?Boltzmann law to both the object and its environment, the net rate of energy loss (or gain) can be determined as the difference in their respective T^4 terms, making it essential for understanding energy exchange between bodies.

Temperature Conversion

In thermodynamic calculations involving thermal radiation, temperatures must be expressed in Kelvin to ensure correct application of the laws of physics. Converting from Celsius to Kelvin is a necessary step, as the Stefan?Boltzmann law uses the absolute temperature scale.

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