An average person eats a daily diet consisting of about 2000 Cal of energy. 1 Ca

Question

An average person eats a daily diet consisting of about 2000 Cal of energy. 1 Cal (or kcal) is equal to approximately 4.2 kJ.

a. For a body temperature of 37°C, determine the rate of entropy change for an average person.

b. For a room temperature of 23°C, determine the total rate of entropy generation for the combined, isolated system consisting of the person in Part a. and the room.

c. The temperature of a lightbulb is T = 3000 K (as opposed to 37°C for the person). Determine the rate of entropy generation for an isolated system consisting of a room containing a single lightbulb that consumes power at the same rate as an average person.

Explanation / Answer

a)

dS = (4.2*1000)/(37+273) = 13.55 J/K

(b)

dS = (4.2*1000)/(23+273) = 14.2 J/K

(c)

dS = (4.2*1000)/(3000) = 1.4 J/K

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