Question

11. With the window blinds closed, creepy Joey likes to stand naked in his living room for hours at a time. The surface of his body is 1.5m2, his skin temperature maintains a constant temperature of 34.0°C and his skin emissivity is 0.62. a. Calculate the rate of heat loss due to radiative heat transfer from Joey's skin surface. b. It turns out that, for comfort, Joey keeps his living room at a constant room temperature of 23.0°C. Calculate the rate of heat Joey absorbs in his living room through radiation? c. Determine Joey's net radiative heat loss. d. How much energy must Joey's body generate to maintain a skin surface temperature of 34.0°C in 24.0 hours?

Answer 1

Using Stefan's law we can calculate the amount of heat lost and heat absorbed by a body.

From stefanes low amount of rædvahon nemitted per conit time from energy an area A is - emissivity 2 0.62 6- Stefan's constant - 5.67x10 uz & AT4 = 0.62 x 5.67 x 108 x 15 x 307.1 A aza 1.5m2 469 watt is the rate T = absolute temperature 2737 34-0 = 307.1k of al which Joey tooses hoab dare to rodvatuon the snitting body cb amount of radu choon energy per unit time absorbed an CULO A is ما u' = е би точ 3 0.62*5.67*16 x 1.5 x 296.14 temperature of absorb emitting borroom. = 40.5 esatt is the rate at which Joey absorbs heat througe saduation to his room. Net radiative loss of heal per unit time Unet re-ul = es Атч - es AT, 4 4 69 - 405 64 watt. d ) to maintain skin temperature of 34.0°c, hear lenergs generated in Joey's body E = met rate of 10ss of heat X time 64 watt X a 4 hours ou JS-1 x 86, noos 5.53x108 joule