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 m 2 .

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Answer 1

Textbook Question

Chapter 11, Problem 44P

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)

Expert Solution

To determine

The R factor for skin, fat, and tissue.

Answer to Problem 44P

The R factor for skin, fat, and tissue is

$5.0×10−2 m2⋅K/W$ ,

$2.5×10−2 m2⋅K/W$ , and

$6.4×10−2 m2⋅K/W$ respectively.

Explanation of Solution

Section1:

To determine: TheR factor for skin.

Answer: TheR factor for skin is $5.0×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the skin is 1.0 mm and thermal conductivity is $0.20 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kskin$ to find $Rskin$ .

$Rskin=1.0 mm(1 m103 mm)0.20 W/m⋅K=5.0×10−2 m2⋅K/W$

Therefore, the R factor for the skin is $5.0×10−2 m2⋅K/W$

Section2:

To determine: TheR factor for fat layer.

Answer: TheR factor for fat layer is $2.5×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the fat layer is 0.50 cm and thermal conductivity of fat layer is $0.02 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kfat$ to find $Rskin$ .

$Rskin=0.50 cm(1 m102 cm)0.02 W/m⋅K=2.5×10−2 m2⋅K/W$

Therefore, the R factor for the fat layer is $2.5×10−2 m2⋅K/W$

Section3:

To determine: TheR factor for tissue.

Answer: TheR factor for tissue is $6.4×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the tissue is 3.2 cm and thermal conductivity of tissue is

$0.50 W/m⋅K$ .

Formula to calculate R factor for tissue is,

$Rtissue=Ltissuektissue$

$Rtissue$ is the R factor for tissue,
$Ltissue$ is the thickness of the tissue,
$ktissue$ is the thermal conductivity of the tissue,

Substitute 3.2 cm for $Ltissue$ and $0.50 W/m⋅K$ for $ktissue$ to find $Rtissue$ .

$Rtissue=3.2 cm(1 m102 cm)0.50 W/m⋅K=6.4×10−2 m2⋅K/W$

Therefore, the R factor for the tissue is $6.4×10−2 m2⋅K/W$

Conclusion:

Therefore, the R factor for skin, fat, and tissue is $5.0×10−2 m2⋅K/W$ , $2.5×10−2 m2⋅K/W$ , and $6.4×10−2 m2⋅K/W$ respectively.

(b)

Expert Solution

To determine

The rate of energy loss from the body.

Answer to Problem 44P

The rate of energy loss from the body is

$5.3×102 W$

Explanation of Solution

Given info: The surface area of the body is $2.0 m2$ , temperature of the core is $37°C$ , and exterior temperature is $0°C$ .

The R factor of the body is the sum of R factor of skin, fat layer and tissue.

$Rbody=Rskin+Rfat+Rtissue$

Formula to calculate the rate at which energy is transferred by conduction from the body is,

$P=A(Th−Tc)Rbody$

$P$ is the rate at which energy is transferred by conduction from the body,
$Rbody$ is the R factor of the body,
$A$ is the surface area of the body,
$Th$ is core temperature,
$Tc$ is exterior temperature,

Use $Rskin+Rfat+Rtissue$ for $Rbody$ in $P=A(Th−Tc)Rbody$ to rewrite P.

$P=A(Th−Tc)Rskin+Rfat+Rtissue$

Substitute $2.0 m2$ for A, $0°C$ for $Tc$ , $5.0×10−2 m2⋅K/W$ for $Rskin$ , $2.5×10−2 m2⋅K/W$ for $Rfat$ , and $6.4×10−2 m2⋅K/W$ for $Rtissue$ and $37°C$ for $Th$ to find P.

$P=(2.0 m2)[(37+273)K−(0+273)K]5.0×10−2 m2⋅K/W+2.5×10−2 m2⋅K/W+6.4×10−2 m2⋅K/W=5.3×102 W$

Conclusion:

Therefore, the rate at which energy is transferred through body is $5.3×102 W$ .

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Make sure to draw a Free Body Diagram as well

Make sure to draw a Free Body Diagram please as well

Answer 2

Textbook Question

Chapter 11, Problem 44P

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)

Expert Solution

To determine

The R factor for skin, fat, and tissue.

Answer to Problem 44P

The R factor for skin, fat, and tissue is

$5.0×10−2 m2⋅K/W$ ,

$2.5×10−2 m2⋅K/W$ , and

$6.4×10−2 m2⋅K/W$ respectively.

Explanation of Solution

Section1:

To determine: TheR factor for skin.

Answer: TheR factor for skin is $5.0×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the skin is 1.0 mm and thermal conductivity is $0.20 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kskin$ to find $Rskin$ .

$Rskin=1.0 mm(1 m103 mm)0.20 W/m⋅K=5.0×10−2 m2⋅K/W$

Therefore, the R factor for the skin is $5.0×10−2 m2⋅K/W$

Section2:

To determine: TheR factor for fat layer.

Answer: TheR factor for fat layer is $2.5×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the fat layer is 0.50 cm and thermal conductivity of fat layer is $0.02 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kfat$ to find $Rskin$ .

$Rskin=0.50 cm(1 m102 cm)0.02 W/m⋅K=2.5×10−2 m2⋅K/W$

Therefore, the R factor for the fat layer is $2.5×10−2 m2⋅K/W$

Section3:

To determine: TheR factor for tissue.

Answer: TheR factor for tissue is $6.4×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the tissue is 3.2 cm and thermal conductivity of tissue is

$0.50 W/m⋅K$ .

Formula to calculate R factor for tissue is,

$Rtissue=Ltissuektissue$

$Rtissue$ is the R factor for tissue,
$Ltissue$ is the thickness of the tissue,
$ktissue$ is the thermal conductivity of the tissue,

Substitute 3.2 cm for $Ltissue$ and $0.50 W/m⋅K$ for $ktissue$ to find $Rtissue$ .

$Rtissue=3.2 cm(1 m102 cm)0.50 W/m⋅K=6.4×10−2 m2⋅K/W$

Therefore, the R factor for the tissue is $6.4×10−2 m2⋅K/W$

Conclusion:

Therefore, the R factor for skin, fat, and tissue is $5.0×10−2 m2⋅K/W$ , $2.5×10−2 m2⋅K/W$ , and $6.4×10−2 m2⋅K/W$ respectively.

(b)

Expert Solution

To determine

The rate of energy loss from the body.

Answer to Problem 44P

The rate of energy loss from the body is

$5.3×102 W$

Explanation of Solution

Given info: The surface area of the body is $2.0 m2$ , temperature of the core is $37°C$ , and exterior temperature is $0°C$ .

The R factor of the body is the sum of R factor of skin, fat layer and tissue.

$Rbody=Rskin+Rfat+Rtissue$

Formula to calculate the rate at which energy is transferred by conduction from the body is,

$P=A(Th−Tc)Rbody$

$P$ is the rate at which energy is transferred by conduction from the body,
$Rbody$ is the R factor of the body,
$A$ is the surface area of the body,
$Th$ is core temperature,
$Tc$ is exterior temperature,

Use $Rskin+Rfat+Rtissue$ for $Rbody$ in $P=A(Th−Tc)Rbody$ to rewrite P.

$P=A(Th−Tc)Rskin+Rfat+Rtissue$

Substitute $2.0 m2$ for A, $0°C$ for $Tc$ , $5.0×10−2 m2⋅K/W$ for $Rskin$ , $2.5×10−2 m2⋅K/W$ for $Rfat$ , and $6.4×10−2 m2⋅K/W$ for $Rtissue$ and $37°C$ for $Th$ to find P.

$P=(2.0 m2)[(37+273)K−(0+273)K]5.0×10−2 m2⋅K/W+2.5×10−2 m2⋅K/W+6.4×10−2 m2⋅K/W=5.3×102 W$

Conclusion:

Therefore, the rate at which energy is transferred through body is $5.3×102 W$ .

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Answer 3

Textbook Question

Chapter 11, Problem 44P

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)

Expert Solution

To determine

The R factor for skin, fat, and tissue.

Answer to Problem 44P

The R factor for skin, fat, and tissue is

$5.0×10−2 m2⋅K/W$ ,

$2.5×10−2 m2⋅K/W$ , and

$6.4×10−2 m2⋅K/W$ respectively.

Explanation of Solution

Section1:

To determine: TheR factor for skin.

Answer: TheR factor for skin is $5.0×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the skin is 1.0 mm and thermal conductivity is $0.20 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kskin$ to find $Rskin$ .

$Rskin=1.0 mm(1 m103 mm)0.20 W/m⋅K=5.0×10−2 m2⋅K/W$

Therefore, the R factor for the skin is $5.0×10−2 m2⋅K/W$

Section2:

To determine: TheR factor for fat layer.

Answer: TheR factor for fat layer is $2.5×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the fat layer is 0.50 cm and thermal conductivity of fat layer is $0.02 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kfat$ to find $Rskin$ .

$Rskin=0.50 cm(1 m102 cm)0.02 W/m⋅K=2.5×10−2 m2⋅K/W$

Therefore, the R factor for the fat layer is $2.5×10−2 m2⋅K/W$

Section3:

To determine: TheR factor for tissue.

Answer: TheR factor for tissue is $6.4×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the tissue is 3.2 cm and thermal conductivity of tissue is

$0.50 W/m⋅K$ .

Formula to calculate R factor for tissue is,

$Rtissue=Ltissuektissue$

$Rtissue$ is the R factor for tissue,
$Ltissue$ is the thickness of the tissue,
$ktissue$ is the thermal conductivity of the tissue,

Substitute 3.2 cm for $Ltissue$ and $0.50 W/m⋅K$ for $ktissue$ to find $Rtissue$ .

$Rtissue=3.2 cm(1 m102 cm)0.50 W/m⋅K=6.4×10−2 m2⋅K/W$

Therefore, the R factor for the tissue is $6.4×10−2 m2⋅K/W$

Conclusion:

Therefore, the R factor for skin, fat, and tissue is $5.0×10−2 m2⋅K/W$ , $2.5×10−2 m2⋅K/W$ , and $6.4×10−2 m2⋅K/W$ respectively.

(b)

Expert Solution

To determine

The rate of energy loss from the body.

Answer to Problem 44P

The rate of energy loss from the body is

$5.3×102 W$

Explanation of Solution

Given info: The surface area of the body is $2.0 m2$ , temperature of the core is $37°C$ , and exterior temperature is $0°C$ .

The R factor of the body is the sum of R factor of skin, fat layer and tissue.

$Rbody=Rskin+Rfat+Rtissue$

Formula to calculate the rate at which energy is transferred by conduction from the body is,

$P=A(Th−Tc)Rbody$

$P$ is the rate at which energy is transferred by conduction from the body,
$Rbody$ is the R factor of the body,
$A$ is the surface area of the body,
$Th$ is core temperature,
$Tc$ is exterior temperature,

Use $Rskin+Rfat+Rtissue$ for $Rbody$ in $P=A(Th−Tc)Rbody$ to rewrite P.

$P=A(Th−Tc)Rskin+Rfat+Rtissue$

Substitute $2.0 m2$ for A, $0°C$ for $Tc$ , $5.0×10−2 m2⋅K/W$ for $Rskin$ , $2.5×10−2 m2⋅K/W$ for $Rfat$ , and $6.4×10−2 m2⋅K/W$ for $Rtissue$ and $37°C$ for $Th$ to find P.

$P=(2.0 m2)[(37+273)K−(0+273)K]5.0×10−2 m2⋅K/W+2.5×10−2 m2⋅K/W+6.4×10−2 m2⋅K/W=5.3×102 W$

Conclusion:

Therefore, the rate at which energy is transferred through body is $5.3×102 W$ .

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Answer 4

Textbook Question

Chapter 11, Problem 44P

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)

Expert Solution

To determine

The R factor for skin, fat, and tissue.

Answer to Problem 44P

The R factor for skin, fat, and tissue is

$5.0×10−2 m2⋅K/W$ ,

$2.5×10−2 m2⋅K/W$ , and

$6.4×10−2 m2⋅K/W$ respectively.

Explanation of Solution

Section1:

To determine: TheR factor for skin.

Answer: TheR factor for skin is $5.0×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the skin is 1.0 mm and thermal conductivity is $0.20 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kskin$ to find $Rskin$ .

$Rskin=1.0 mm(1 m103 mm)0.20 W/m⋅K=5.0×10−2 m2⋅K/W$

Therefore, the R factor for the skin is $5.0×10−2 m2⋅K/W$

Section2:

To determine: TheR factor for fat layer.

Answer: TheR factor for fat layer is $2.5×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the fat layer is 0.50 cm and thermal conductivity of fat layer is $0.02 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kfat$ to find $Rskin$ .

$Rskin=0.50 cm(1 m102 cm)0.02 W/m⋅K=2.5×10−2 m2⋅K/W$

Therefore, the R factor for the fat layer is $2.5×10−2 m2⋅K/W$

Section3:

To determine: TheR factor for tissue.

Answer: TheR factor for tissue is $6.4×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the tissue is 3.2 cm and thermal conductivity of tissue is

$0.50 W/m⋅K$ .

Formula to calculate R factor for tissue is,

$Rtissue=Ltissuektissue$

$Rtissue$ is the R factor for tissue,
$Ltissue$ is the thickness of the tissue,
$ktissue$ is the thermal conductivity of the tissue,

Substitute 3.2 cm for $Ltissue$ and $0.50 W/m⋅K$ for $ktissue$ to find $Rtissue$ .

$Rtissue=3.2 cm(1 m102 cm)0.50 W/m⋅K=6.4×10−2 m2⋅K/W$

Therefore, the R factor for the tissue is $6.4×10−2 m2⋅K/W$

Conclusion:

Therefore, the R factor for skin, fat, and tissue is $5.0×10−2 m2⋅K/W$ , $2.5×10−2 m2⋅K/W$ , and $6.4×10−2 m2⋅K/W$ respectively.

(b)

Expert Solution

To determine

The rate of energy loss from the body.

Answer to Problem 44P

The rate of energy loss from the body is

$5.3×102 W$

Explanation of Solution

Given info: The surface area of the body is $2.0 m2$ , temperature of the core is $37°C$ , and exterior temperature is $0°C$ .

The R factor of the body is the sum of R factor of skin, fat layer and tissue.

$Rbody=Rskin+Rfat+Rtissue$

Formula to calculate the rate at which energy is transferred by conduction from the body is,

$P=A(Th−Tc)Rbody$

$P$ is the rate at which energy is transferred by conduction from the body,
$Rbody$ is the R factor of the body,
$A$ is the surface area of the body,
$Th$ is core temperature,
$Tc$ is exterior temperature,

Use $Rskin+Rfat+Rtissue$ for $Rbody$ in $P=A(Th−Tc)Rbody$ to rewrite P.

$P=A(Th−Tc)Rskin+Rfat+Rtissue$

Substitute $2.0 m2$ for A, $0°C$ for $Tc$ , $5.0×10−2 m2⋅K/W$ for $Rskin$ , $2.5×10−2 m2⋅K/W$ for $Rfat$ , and $6.4×10−2 m2⋅K/W$ for $Rtissue$ and $37°C$ for $Th$ to find P.

$P=(2.0 m2)[(37+273)K−(0+273)K]5.0×10−2 m2⋅K/W+2.5×10−2 m2⋅K/W+6.4×10−2 m2⋅K/W=5.3×102 W$

Conclusion:

Therefore, the rate at which energy is transferred through body is $5.3×102 W$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The R factor for skin, fat, and tissue.

Answer to Problem 44P

The R factor for skin, fat, and tissue is

$5.0×10−2 m2⋅K/W$ ,

$2.5×10−2 m2⋅K/W$ , and

$6.4×10−2 m2⋅K/W$ respectively.

Explanation of Solution

Section1:

To determine: TheR factor for skin.

Answer: TheR factor for skin is $5.0×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the skin is 1.0 mm and thermal conductivity is $0.20 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kskin$ to find $Rskin$ .

$Rskin=1.0 mm(1 m103 mm)0.20 W/m⋅K=5.0×10−2 m2⋅K/W$

Therefore, the R factor for the skin is $5.0×10−2 m2⋅K/W$

Section2:

To determine: TheR factor for fat layer.

Answer: TheR factor for fat layer is $2.5×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the fat layer is 0.50 cm and thermal conductivity of fat layer is $0.02 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kfat$ to find $Rskin$ .

$Rskin=0.50 cm(1 m102 cm)0.02 W/m⋅K=2.5×10−2 m2⋅K/W$

Therefore, the R factor for the fat layer is $2.5×10−2 m2⋅K/W$

Section3:

To determine: TheR factor for tissue.

Answer: TheR factor for tissue is $6.4×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the tissue is 3.2 cm and thermal conductivity of tissue is

$0.50 W/m⋅K$ .

Formula to calculate R factor for tissue is,

$Rtissue=Ltissuektissue$

$Rtissue$ is the R factor for tissue,
$Ltissue$ is the thickness of the tissue,
$ktissue$ is the thermal conductivity of the tissue,

Substitute 3.2 cm for $Ltissue$ and $0.50 W/m⋅K$ for $ktissue$ to find $Rtissue$ .

$Rtissue=3.2 cm(1 m102 cm)0.50 W/m⋅K=6.4×10−2 m2⋅K/W$

Therefore, the R factor for the tissue is $6.4×10−2 m2⋅K/W$

Conclusion:

Therefore, the R factor for skin, fat, and tissue is $5.0×10−2 m2⋅K/W$ , $2.5×10−2 m2⋅K/W$ , and $6.4×10−2 m2⋅K/W$ respectively.

(b)

Expert Solution

To determine

The rate of energy loss from the body.

Answer to Problem 44P

The rate of energy loss from the body is

$5.3×102 W$

Explanation of Solution

Given info: The surface area of the body is $2.0 m2$ , temperature of the core is $37°C$ , and exterior temperature is $0°C$ .

The R factor of the body is the sum of R factor of skin, fat layer and tissue.

$Rbody=Rskin+Rfat+Rtissue$

Formula to calculate the rate at which energy is transferred by conduction from the body is,

$P=A(Th−Tc)Rbody$

$P$ is the rate at which energy is transferred by conduction from the body,
$Rbody$ is the R factor of the body,
$A$ is the surface area of the body,
$Th$ is core temperature,
$Tc$ is exterior temperature,

Use $Rskin+Rfat+Rtissue$ for $Rbody$ in $P=A(Th−Tc)Rbody$ to rewrite P.

$P=A(Th−Tc)Rskin+Rfat+Rtissue$

Substitute $2.0 m2$ for A, $0°C$ for $Tc$ , $5.0×10−2 m2⋅K/W$ for $Rskin$ , $2.5×10−2 m2⋅K/W$ for $Rfat$ , and $6.4×10−2 m2⋅K/W$ for $Rtissue$ and $37°C$ for $Th$ to find P.

$P=(2.0 m2)[(37+273)K−(0+273)K]5.0×10−2 m2⋅K/W+2.5×10−2 m2⋅K/W+6.4×10−2 m2⋅K/W=5.3×102 W$

Conclusion:

Therefore, the rate at which energy is transferred through body is $5.3×102 W$ .

Answer 8

(a)

Expert Solution

To determine

The R factor for skin, fat, and tissue.

Answer to Problem 44P

The R factor for skin, fat, and tissue is

$5.0×10−2 m2⋅K/W$ ,

$2.5×10−2 m2⋅K/W$ , and

$6.4×10−2 m2⋅K/W$ respectively.

Explanation of Solution

Section1:

To determine: TheR factor for skin.

Answer: TheR factor for skin is $5.0×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the skin is 1.0 mm and thermal conductivity is $0.20 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kskin$ to find $Rskin$ .

$Rskin=1.0 mm(1 m103 mm)0.20 W/m⋅K=5.0×10−2 m2⋅K/W$

Therefore, the R factor for the skin is $5.0×10−2 m2⋅K/W$

Section2:

To determine: TheR factor for fat layer.

Answer: TheR factor for fat layer is $2.5×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the fat layer is 0.50 cm and thermal conductivity of fat layer is $0.02 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kfat$ to find $Rskin$ .

$Rskin=0.50 cm(1 m102 cm)0.02 W/m⋅K=2.5×10−2 m2⋅K/W$

Therefore, the R factor for the fat layer is $2.5×10−2 m2⋅K/W$

Section3:

To determine: TheR factor for tissue.

Answer: TheR factor for tissue is $6.4×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the tissue is 3.2 cm and thermal conductivity of tissue is

$0.50 W/m⋅K$ .

Formula to calculate R factor for tissue is,

$Rtissue=Ltissuektissue$

$Rtissue$ is the R factor for tissue,
$Ltissue$ is the thickness of the tissue,
$ktissue$ is the thermal conductivity of the tissue,

Substitute 3.2 cm for $Ltissue$ and $0.50 W/m⋅K$ for $ktissue$ to find $Rtissue$ .

$Rtissue=3.2 cm(1 m102 cm)0.50 W/m⋅K=6.4×10−2 m2⋅K/W$

Therefore, the R factor for the tissue is $6.4×10−2 m2⋅K/W$

Conclusion:

Therefore, the R factor for skin, fat, and tissue is $5.0×10−2 m2⋅K/W$ , $2.5×10−2 m2⋅K/W$ , and $6.4×10−2 m2⋅K/W$ respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The R factor for skin, fat, and tissue.

Answer 13

Answer to Problem 44P

The R factor for skin, fat, and tissue is

$5.0×10−2 m2⋅K/W$ ,

$2.5×10−2 m2⋅K/W$ , and

$6.4×10−2 m2⋅K/W$ respectively.

Answer 14

Explanation of Solution

Section1:

To determine: TheR factor for skin.

Answer: TheR factor for skin is $5.0×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the skin is 1.0 mm and thermal conductivity is $0.20 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kskin$ to find $Rskin$ .

$Rskin=1.0 mm(1 m103 mm)0.20 W/m⋅K=5.0×10−2 m2⋅K/W$

Therefore, the R factor for the skin is $5.0×10−2 m2⋅K/W$

Section2:

To determine: TheR factor for fat layer.

Answer: TheR factor for fat layer is $2.5×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the fat layer is 0.50 cm and thermal conductivity of fat layer is $0.02 W/m⋅K$ .

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 $0.20 W/m⋅K$ for $kfat$ to find $Rskin$ .

$Rskin=0.50 cm(1 m102 cm)0.02 W/m⋅K=2.5×10−2 m2⋅K/W$

Therefore, the R factor for the fat layer is $2.5×10−2 m2⋅K/W$

Section3:

To determine: TheR factor for tissue.

Answer: TheR factor for tissue is $6.4×10−2 m2⋅K/W$

Explanation:

Given info: Thickness of the tissue is 3.2 cm and thermal conductivity of tissue is

$0.50 W/m⋅K$ .

Formula to calculate R factor for tissue is,

$Rtissue=Ltissuektissue$

$Rtissue$ is the R factor for tissue,
$Ltissue$ is the thickness of the tissue,
$ktissue$ is the thermal conductivity of the tissue,

Substitute 3.2 cm for $Ltissue$ and $0.50 W/m⋅K$ for $ktissue$ to find $Rtissue$ .

$Rtissue=3.2 cm(1 m102 cm)0.50 W/m⋅K=6.4×10−2 m2⋅K/W$

Therefore, the R factor for the tissue is $6.4×10−2 m2⋅K/W$

Conclusion:

Therefore, the R factor for skin, fat, and tissue is $5.0×10−2 m2⋅K/W$ , $2.5×10−2 m2⋅K/W$ , and $6.4×10−2 m2⋅K/W$ respectively.

Answer 15

(b)

Expert Solution

To determine

The rate of energy loss from the body.

Answer to Problem 44P

The rate of energy loss from the body is

$5.3×102 W$

Explanation of Solution

Given info: The surface area of the body is $2.0 m2$ , temperature of the core is $37°C$ , and exterior temperature is $0°C$ .

The R factor of the body is the sum of R factor of skin, fat layer and tissue.

$Rbody=Rskin+Rfat+Rtissue$

Formula to calculate the rate at which energy is transferred by conduction from the body is,

$P=A(Th−Tc)Rbody$

$P$ is the rate at which energy is transferred by conduction from the body,
$Rbody$ is the R factor of the body,
$A$ is the surface area of the body,
$Th$ is core temperature,
$Tc$ is exterior temperature,

Use $Rskin+Rfat+Rtissue$ for $Rbody$ in $P=A(Th−Tc)Rbody$ to rewrite P.

$P=A(Th−Tc)Rskin+Rfat+Rtissue$

Substitute $2.0 m2$ for A, $0°C$ for $Tc$ , $5.0×10−2 m2⋅K/W$ for $Rskin$ , $2.5×10−2 m2⋅K/W$ for $Rfat$ , and $6.4×10−2 m2⋅K/W$ for $Rtissue$ and $37°C$ for $Th$ to find P.

$P=(2.0 m2)[(37+273)K−(0+273)K]5.0×10−2 m2⋅K/W+2.5×10−2 m2⋅K/W+6.4×10−2 m2⋅K/W=5.3×102 W$

Conclusion:

Therefore, the rate at which energy is transferred through body is $5.3×102 W$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The rate of energy loss from the body.

Answer 20

Answer to Problem 44P

The rate of energy loss from the body is

$5.3×102 W$

Answer 21

Explanation of Solution

Given info: The surface area of the body is $2.0 m2$ , temperature of the core is $37°C$ , and exterior temperature is $0°C$ .

The R factor of the body is the sum of R factor of skin, fat layer and tissue.

$Rbody=Rskin+Rfat+Rtissue$

Formula to calculate the rate at which energy is transferred by conduction from the body is,

$P=A(Th−Tc)Rbody$

$P$ is the rate at which energy is transferred by conduction from the body,
$Rbody$ is the R factor of the body,
$A$ is the surface area of the body,
$Th$ is core temperature,
$Tc$ is exterior temperature,

Use $Rskin+Rfat+Rtissue$ for $Rbody$ in $P=A(Th−Tc)Rbody$ to rewrite P.

$P=A(Th−Tc)Rskin+Rfat+Rtissue$

Substitute $2.0 m2$ for A, $0°C$ for $Tc$ , $5.0×10−2 m2⋅K/W$ for $Rskin$ , $2.5×10−2 m2⋅K/W$ for $Rfat$ , and $6.4×10−2 m2⋅K/W$ for $Rtissue$ and $37°C$ for $Th$ to find P.