Suppose the skin temperature of a naked person is 35.8 °C when the person is standing inside a room whose temperature is 25.0 °C. The skin area of the individual is 1.73 m². (a) Assuming the emissivity is 0.604, find the net loss of radiant power from the body. (b) Determine the number of food Calories of energy (1 food Calorie= 4186 J) that is lost in 14.5 hours due to the net loss rate obtained in part (a). Metabolic conversion of food into energy replaces this loss.

Suppose the skin temperature of a naked person is 35.8 °C when the person is standing inside a room whose temperature is 25.0 °C. The skin area of the individual is 1.73 m². (a) Assuming the emissivity is 0.604, find the net loss of radiant power from the body. (b) Determine the number of food Calories of energy (1 food Calorie= 4186 J) that is lost in 14.5 hours due to the net loss rate obtained in part (a). Metabolic conversion of food into energy replaces this loss.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Calorimetry

Measuring the heat transferred during chemical reactions or other physical processes is known as calorimetry. It comes under the branch of thermochemistry.

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Suppose the skin temperature of a naked person is 35.8 °C when the person is standing inside a room whose temperature is 25.0 °C.
The skin area of the individual is 1.73 m². (a) Assuming the emissivity is 0.604, find the net loss of radiant power from the body. (b)
Determine the number of food Calories of energy (1 food Calorie = 4186 J) that is lost in 14.5 hours due to the net loss rate obtained in
part (a). Metabolic conversion of food into energy replaces this loss.

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