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Question

https://edugen. wileyplus.com/edugen/student/mainfr.uni FULL SCREEN PRINTER VERSION BACK NEXT Chapter 13, Problem 29 ningWare 13.2 reviews the concepts that are involved in this problem. Suppose the skin temperature of a naked person is 34.7 °C when the person is standing inside a room whose temperature is 21.2 °C. The skin area of the individual is 1.75 m2. (a) Assuming the emissivity is 0.573, 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 10.4 hours due to the net loss rate obtained in part (a). Metabolic conversion of food into energy replaces this loss. (a) Number Units (b) Number Units LINK TO TEXT Question Attempts: O of 6 used SAVE POR LATER SUBMIT A SUBMIT ANSWER 3 .-.pdf · Show All

Explanation / Answer

a) Q / t = e * sigma * A T4

= 0.573 * 5.67 * 10-8 * 1.75 * (273 + 34.7)4 = 509.67 J/s

[Q / t]' = e * sigma * A T4

= 0.573 * 5.67 * 10-8 * 1.75 * (273 + 21.2)4 = 425.9 J/s

net heat loss = 509.67 - 425.9

net loss = 83.7 J/s

b) In 10.4 hours

Total heat loss = 83.7 * 10.4 * 60 * 60 = 3.13 * 106 J/hr / 4186

= 748 cal/h

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