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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