Explanation Section 1: To determine: The amount of heat released in the metabolic process of 2.5 gm butter. Answer: The amount of heat released in the metabolic process of 2.5 gm butter is 7 .786 × 10 4 J . Given information: The temperature of the body is 37 ° C , the temperature of the skin is 30 ° C , the amount of butter is 2.50 gm , the digesting food produces 9.3 gm food calories per gram fat and 80 % energy is lost as heat. 1 food calorie = 4186 J . Calculate heat released in metabolic process of 2.5 gm of butter as, Q = 80 100 ( 9.3 × 2.5 × 4186 ) J = 7 .786 × 10 4 J Thus, the amount of heat released in the metabolic process of 2.5 gm butter is 7 .786 × 10 4 J . Section 2: To determine: The entropy change of the person. Answer: The entropy change of the person is − 5.8 J / K . Given information: The temperature of the body is 37 ° C , the temperature of the skin is 30 ° C , the amount of butter is 2.50 gm , the digesting food produces 9.3 gm food calories per gram fat and 80 % energy is lost as heat

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ISBN: 9781323575208

Author: YOUNG

Publisher: PEARSON C

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First law of Thermodynamics

The word thermodynamics consists of two words: thermo- and dynamics. Dynamics means the study of motion. The field of thermodynamics is all about the study of the movement of heat. In a more meaningful way, it is stated as the study of the transfer of en…

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Chapter 20, Problem 20.37P

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The entropy change of the person and whether it deceases or increases.

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The entropy change of the person and whether it deceases or increases.

Solution Summary: The author calculates the amount of heat released in the metabolic process of 2.5gm butter.

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The entropy change of the person and whether it deceases or increases.

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