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.

Delta S= Q/T (Where T is the temperature in Kelvins)

First we need to convert the 2000 cal of energy into Joules

1kcal-----------> 4.2 kJ

2 kcal ----> x

x= (2 kCal * 4.2 kJ)/ 1kcal

x= 8.4 J

Delta S= 8.4 J / (37+273) -----> Delta S= 0,027

b.

The entropy change between two reservoirs is the sum of the entropy change of each. If the high temperature reservoir is at TH and the low temperature reservoir is at TL, the total entropy change is:

Delta S= [(-Q)/TH]+[(Q)/TL] = [Q(TH-TL)]/(TH*TL)

----> Delta S= [8.4J(37+273)(23+273)]/[(37+273)(23+273)]

Delta S= 1,28x10^-3

c. Same case as part a.

Delta S= Q/T

Delta S= 8.4 J / 3000 K

Delta S= 2,8 x 10^-3

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