Question

A human subject with mass 96 kg, body specific heat 3500 Jkg^-1K^-1 skin temperature 34.5 degrees Celsius, surface area 1.5m^2 and emissivity 0.7 enters a room whose walls are at temperature 21 degrees Celsius. At what rate does body temperature decrease? Answer in degrees per hour, as a positive number

Answer 1

rate of heat radiation by the body is given by

`(dQ)/(dt) = \sigma e A(T^4 - T_s^4)`

here we know that

`\sigma = 5.67 * 10^(-8)`

e = 0.7

`A = 1.5 m^2`

`T = 34.5 + 273 = 307.5 k`

`T_s = 21 + 273 = 294 k`

now from above formula rate of heat dissipation

`(dQ)/(dt) = (5.67 * 10^(-8)}(0.7)(1.5)(307.5^4 - 294^4)`

`(dQ)/(dt) = 87.5`

now we know that

`Q = ms \Delta T`

from above equation

`(dQ)/(dt) = ms(dT)/(dt)`

now we have

`m s (dT)/(dt) = 87.5`

here we have

m = 96 kg

s = 3500

`96 * 3500 * (dT)/(dt) = 87.5`

`(dT)/(dt) = (2.6 * 10^(-4)) \: ^0C/s`

`(dT)/(dt) = 0.94 ^0 C/hour`