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

$\frac{dQ}{dt} = \sigma e A(T^4 - T_s^4)$

here we know that

$\sigma = 5.67 \times 10^{-8}$

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

$\frac{dQ}{dt} = (5.67 \times 10^{-8}}(0.7)(1.5)(307.5^4 - 294^4)$

$\frac{dQ}{dt} = 87.5$

now we know that

$Q = ms \Delta T$

from above equation

$\frac{dQ}{dt} = ms\frac{dT}{dt}$

now we have

$m s \frac{dT}{dt} = 87.5$

here we have

m = 96 kg

s = 3500

$96 \times 3500 \times \frac{dT}{dt} = 87.5$

$\frac{dT}{dt} = (2.6 \times 10^{-4}) \: ^0C/s$

$\frac{dT}{dt} = 0.94 ^0 C/hour$