Question

For the human body, what is the rate of heat transfer by
conduction (in kJ/day) through the body's tissue with the following
conditions: the tissue thickness is 2.90 cm, the change in
temperature is 2.04°C, and the skin area is 1.35 m^{2}.

________ kJ/day

How does this compare with the average heat transfer rate to the body resulting from an energy intake of about 2400 kcal per day? (No exercise is included.)

Fraction of
(Q/t)_{conduction and} |

(Q/t)_{food intake. _____________} |

Answer #1

Rate of heat transfer is given by:

H = Q/t = k*A*dT/dx

Using given values:

k = thermal conductivity of tissue = 0.2

A= 1.35 m^2

dT = 2.04 C

dx = thickness = 2.90 cm

So,

H = 0.2*1.35*2.04/(2.90*10^-2) = 19.0 J/sec

H = 19.0 J/sec = 19*10^-3 kJ/sec)*(86400 sec/day)

**H = 1641.6 kJ/day**

Part B.

food energy intake = 2400 kcal/day

1 k-cal = 4186 J

So, Energy intake = (2400 kcal/day)*(4186 J/1 kcal) = 10046.4 kJ/day

Fraction = Due to conduction/food intake= 1641.6/10046.4= 0.1634

If answer is required in percentage than, 16.34%

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