   Chapter 11, Problem 60AP

Chapter
Section
Textbook Problem

Overall, 80% of the energy used by the body must be eliminated as excess thermal energy- and needs to be dissipated. The mechanisms of elimination are radiation, evaporation of sweat (2 430 kJ/kg), evaporation from the lungs (38 kJ/h), conduction, and convection.A person working out in a gym has a metabolic rate of 2 500 kJ/h. His body temperature is 37°C, and the outside temperature 24°C. Assume the skin has an area of 2.0 m2 and emissivity of 0.97. (a) At what rate is his excess thermal energy dissipated by radiation? (b) If he eliminates 0.40 kg of perspiration during that hour, at what rate is thermal energy dissipated by evaporation of sweat? (c) At what rate is energy eliminated by evaporation from the lungs? (d) At what rate must the remaining excess energy be eliminated through conduction and convection?

(a)

To determine
Rate at which excess thermal energy dissipated by radiation.

Explanation

Given Info: Area of the body is 2.0m2 , emissivity of the body 0.97, Temperature of the body is 37°C and outside temperature is 24°C .

Formula to calculate the rate at which excess thermal energy dissipated by radiation is,

• Prad is rate at which the excess thermal energy radiated.
• σ is Stefan – Boltzmann constant,
• A is the surface area of the body,
• e is the emissivity of the body,
• T is the temperature of the body,
• T0 is the outside temperature,

Substitute 5.6696×108W/m2K4 for σ , 2.0m2 for A, 0.97 for e, 37°C for T , and 24°C for T0 to find Prad

(b)

To determine
The rate of thermal energy dissipated by evaporation of sweat.

(c)

To determine
The rate at which energy is eliminated by evaporation from the lungs.

(d)

To determine
Rate at which remaining excess energy be eliminated through conduction and convection.

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