Digesting fat produces $9.3$food calories per gram of fat, and typically 80% of this energy goes to heat when metabolized. (One food calorie is 1000 calories and therefore equals 4186 J.) The body then moves all this heat to the surface by a combination of thermal conductivity and motion of the blood. The internal temperature of the body (where digestion occurs) is normally 37°C, and the surface is usually about 30°C.By how much do the digestion and metabolism of a 2.50g pat

of butter change your body’s entropy? Does it increase or decrease?

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 gas, $\mathrm{Q}$is the heat gain by the gas and $\mathrm{T}$is the temperature of gas.

For the body, there are two sides with different heat and temperature .So, the total change in the entropy of system is sum of change in entropy of both inner and outer side of body.

So,

$$\begin{gathered} ∆\mathrm{S}=∆\mathrm{S}\left(\mathrm{Hotside}\right)+∆\mathrm{S}\left(\mathrm{Coldside}\right) \\ ∆\mathrm{S}=\frac{{\mathrm{Q}}_{\mathrm{H}}}{{\mathrm{T}}_{\mathrm{H}}}+\frac{{\mathrm{Q}}_{\mathrm{C}}}{{\mathrm{T}}_{\mathrm{C}}} \end{gathered}$$

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 and multiply it by $0.250\mathrm{g}$

$$\begin{gathered} ∆\mathrm{S}=2.50\mathrm{g}\left(\frac{3.12\times {10}^4\mathrm{J}/\mathrm{g}}{310.0\mathrm{K}}+\frac{-3.12\times {10}^4\mathrm{J}/\mathrm{g}}{303.0\mathrm{K}}\right) \\ ∆\mathrm{S}=-5.80\mathrm{J}/\mathrm{K} \end{gathered}$$

Thus, The total change in entropy of body is $-5.80\mathrm{J}/\mathrm{K}$