Question

Thermography is a technique for measuring radiant heat and detecting variations in surface temperatures that may be medically, environmentally, or militarily meaningful.

(a) What is the percent increase in the rate of heat transfer by radiation from a given area at a temperature of 34.2°C compared with that at 32.9°C, such as on a person's skin?

(b) What is the percent increase in the rate of heat transfer by radiation from a given area at a temperature of 34.2°C compared with that at 23.0°C, such as for warm and cool automobile hoods?

Answer 1

Rate of heat transfer is given by:

H = Q/t = sigma*e*A*T^4

sigma = stephen-boltzmann constant

e = emissivity of the body

A = surface area

Since sigma, e and A are constant, So % increase will be

% increase = (H2 - H1)/H1]*100% = [H2/H1 - 1]*100%

% increase = [(T2/T1)^4 - 1]*100%

Part A.

when T2 = 34.2 C = 307.2 K

T1 = 32.9 C = 305.9 K

So,

% increase = [(307.2/305.9)^4 - 1]*100% = 1.71%

Part B.

T2 = 34.2 C = 307.2 K

T1 = 23 C = 296 K

So,

% increase = [(307.2/296)^4 - 1]*100% = 16.02%

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