   Chapter 11, Problem 44P

Chapter
Section
Textbook Problem

The thermal conductivities of human tissues vary greatly. Fat and skin have conductivities of about 0.20 W/m · K and 0.020 W/m · K, respectively, while other tissues inside the body have conductivities of about 0.50 W/m · K. Assume that between the core region of the body and the skin sin face lies a skin layer of 1.0 mm, fat layer of 0.50 cm, and 3.2 cm of other tissues. (a) Find the R-factor for each of these layers, and the equivalent R-factor for all layers taken together, retaining two digits. (b) Find the rate of energy loss when the core temperature is 37°C and the exterior temperature is 0°C. Assume that both a protective layer of clothing and an insulating layer of unmoving air a absent, and a body area of 2.0 m2.

(a)

To determine
The R factor for skin, fat, and tissue.

Explanation

Section1:

To determine: TheR factor for skin.

Answer: TheR factor for skin is 5.0×102m2K/W

Explanation:

Given info: Thickness of the skin is 1.0 mm and thermal conductivity is 0.20W/mK .

Formula to calculate R factor  for skin is,

Rskin=Lskinkskin

• Rskin is the R factor for skin,
• Lskin is the thickness of the skin,
• kskin is the thermal conductivity of the skin,

Substitute 1.0 mm for Lskin and  0.20W/mK for kskin to find Rskin .

Rskin=1.0mm(1m103mm)0.20W/mK=5.0×102m2K/W

Therefore, the R factor for the skin is 5.0×102m2K/W

Section2:

To determine: TheR factor for fat layer.

Answer: TheR factor for fat layer is 2.5×102m2K/W

Explanation:

Given info: Thickness of the fat layer is 0.50 cm and thermal conductivity of fat layer is 0.02W/mK .

Formula to calculate R factor for fat layer is,

Rfat=Lfatkfat

• Rfat is the R factor for fat layer,
• Lfat is the thickness of the fat layer,
• kfat is the thermal conductivity of the fat layer,

Substitute 0.50 cm for Lfat and  0.20W/mK for kfat to find Rskin .

Rskin=0.50cm(1m102cm)0

(b)

To determine
The rate of energy loss from the body.

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