(a) Explanation Given information: The rate of energy transferred to the skin by metabolic reaction is 280 W , the rate of heat transferred to the skin by convection from the outside air is k ′ A skin ( T air − T skin ) , the value of k ′ is 54 J / h ⋅ ° C ⋅ m 2 , the exposed skin area is 1.5 m 2 , the air temperature is 47 ° C and the skin temperature is 36 ° C , the rate of radiant energy from the sun absorb by the skin is 1400 W / m 2 , the temperature of the environment is 47 ° C . Section 1: To determine: The rate of heat transferred to the skin by convection. Answer: The rate of heat transferred to the skin by convection is 0.248 W . Explanation: Given information: The rate of energy transferred to the skin by metabolic reaction is 280 W , the rate of heat transferred to the skin by convection from the outside air is k ′ A skin ( T air − T skin ) , the value of k ′ is 54 J / h ⋅ C ° ⋅ m 2 , the exposed skin area is 1.5 m 2 , the air temperature is 47 ° C and the skin temperature is 36 ° C , the rate of radiant energy from the sun absorb by the skin is 1400 W / m 2 , the temperature of the environment is 47 ° C . The formula to calculate the heat transferred to the skin by convection is, P conv = k ′ A skin ( T air − T skin ) k ′ is a constant. A skin is the exposed skin area. T air is the air temperature. T skin is the skin temperature. P conv is the heat transferred to the skin by convection. Substitute 54 J / h ⋅ C ° ⋅ m 2 for k ′ , 1.5 m 2 for A skin , 47 ° C for T air and 36 ° C for T skin to find P conv . P conv = ( 54 J / h ⋅ ° C ⋅ m 2 ) ( 1 hour 3600 sec ) ( 1.5 m 2 ) ( 47 ° C − 36 ° C ) = 0.248 W Thus, the rate of heat transferred to the skin by convection is 0.248 W . Section 2: To determine: The rate of heat transferred to the skin by the sun. Answer: The rate of heat transferred to the skin by the sun is 2.10 × 10 3 W . Explanation: Given information: The rate of energy transferred to the skin by metabolic reaction is 280 W , the rate of heat transferred to the skin by convection from the outside air is k ′ A skin ( T air − T skin ) , the value of k ′ is 54 J / h ⋅ C ° ⋅ m 2 , the exposed skin area is 1.5 m 2 , the air temperature is 47 ° C and the skin temperature is 36 ° C , the rate of radiant energy from the sun absorb by the skin is 1400 W / m 2 , the temperature of the environment is 47 ° C . The formula to calculate the rate of the heat transferred to the skin by the sun is, P sun = I × A skin I is the intensity of sun radiation . A skin is the exposed skin area. P sun is the heat transferred to the skin by the sun. Substitute 1400 W / m 2 for I and 1.5 m 2 for A skin to find P sun . P sun = ( 1400 W / m 2 ) ( 1 (b) To determine The rate of perspiration evaporates from the person to maintain a constant skin temperature. (c) To determine The required rate of perspiration when emissivity is zero and the exposed skin area is only 0.45 m 2 .

14th Edition

ISBN: 9780133969290

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Human Metabolism

Being described as the most complicated machine ever encountered by humans, the human body and the cell metabolism is nothing less than a wonder. Metabolism is the set of life-sustaining chemical reactions within organisms.

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Chapter 17, Problem 17.114CP

(a)

To determine

The net rate (in Watts) at which the person’s skin is heated by all the four mechanism given in the question.

(b)

To determine

The rate of perspiration evaporates from the person to maintain a constant skin temperature.

(c)

To determine

The required rate of perspiration when emissivity is zero and the exposed skin area is only 0.45 m2 .

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Chapter 17, Problem 17.113P

Chapter 17, Problem 17.115CP

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The net rate (in Watts) at which the person’s skin is heated by all the four mechanism given in the question.

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The net rate (in Watts) at which the person’s skin is heated by all the four mechanism given in the question.

The rate of perspiration evaporates from the person to maintain a constant skin temperature.

The required rate of perspiration when emissivity is zero and the exposed skin area is only 0.45 m2 .

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Chapter 17 Solutions