Q34P - University Physics with Modern Physics

Question

An average sleeping person metabolizes at a rate of about $80 \mathrm{W}$ by digesting food or burning fat. Typically, $20\%$ of this energy goes into bodily functions, such as cell repair, pumping blood, and other uses of mechanical energy, while the rest goes to heat. Most people get rid of all this excess heat by transferring it (by conduction and the flow of blood) to the surface of the body, where it is radiated away. The normal

internal temperature of the body (where the metabolism takes place) is $37°\mathrm{C}$, and the skin is typically  $7°\mathrm{C}$ cooler. By how much does the person’s entropy change per second due to this heat transfer?

Step-by-Step Solution

Answer

The person’s entropy change per second due to heat transfer is $-0.0048\mathrm{J}/\mathrm{K}$.

Given: Temperature of internal body is  ${\mathrm{T}}_{\mathrm{H}}=37°\mathrm{C}=310\mathrm{K}$

            Temperature of outer side of body is ${\mathrm{T}}_{\mathrm{C}}=30°\mathrm{C}=303\mathrm{K}$

            The heat absorbed by internal body is  $Q_H=80\%\times 80.0\mathrm{J}=64.0\mathrm{J}$

             The heat rejected to outer skin is  $Q_H=-64.0\mathrm{J}$

Step 1: Defining entropy

Entropy can be defined as the measure of a system’s thermal energy per unit temperature that is unavailable for doing useful work. It is also the measure of randomness of system.

 

Formula of change in entropy 

 

 $∆\mathrm{S}=\frac{\mathrm{Q}}{\mathrm{T}}$

 

Where,  $∆\mathrm{S}$  is the change in entropy of substance  $\mathrm{Q}$  is the heat gain by the substance and is the temperature of substance.

Step 2: Change in entropy of body

The change in entropy of body                    

 

$∆\mathrm{S}=\frac{{\mathrm{Q}}_{\mathrm{H}}}{{\mathrm{T}}_{\mathrm{H}}}+\frac{{\mathrm{Q}}_{\mathrm{C}}}{{\mathrm{T}}_{\mathrm{C}}}$

 

Where, $∆\mathrm{S}$ is the change in entropy of body.

Step 3: Calculation of change in entropy of body

Using relation

 

$∆\mathrm{S}=\frac{{\mathrm{Q}}_{\mathrm{H}}}{{\mathrm{T}}_{\mathrm{H}}}+\frac{{\mathrm{Q}}_{\mathrm{C}}}{{\mathrm{T}}_{\mathrm{C}}}$

Now, putting the values of constants in above equation

 

$$\begin{gathered} ∆\mathrm{S}=\frac{64.0\mathrm{J}}{310\mathrm{K}}+\frac{-64.0\mathrm{J}}{303\mathrm{K}} \\ ∆\mathrm{S}=-0.0048\mathrm{J}/\mathrm{K} \end{gathered}$$

 

Thus, the person’s entropy change per second due to heat transfer is $-0.0048\mathrm{J}/\mathrm{K}$.