Answer
$ \Delta S = -5.80 J/K$
Work Step by Step
To calculate the change in entropy of the human body when eating 2.50 g of butter, the heat produced, $Q_H = 3.12 \times 10^4 J/g$ at the internal body's temperature of $T_H = 37^oC = 310 K$. The heat leaves the body at $T_C = 30^o C = 303K $
Before we use the entropy equation. note that the given $Q_H$ is in $J/g$ unit so we need to multiply the value with the mass of butter $2.50g$
$ \Delta S = \frac{Q_H}{T_H} + \frac{Q_C}{T_C}$
$ \Delta S = [\frac{ 3.12 \times 10^4 J/g}{310 K} - \frac{ 3.12 \times 10^4 J/g}{303K} ] 2.50 g$
$ \Delta S = -5.80 J/K$
The negative sign indicates that entropy of the body decreases.
