Answer
$ \Delta S = -4.8 \times 10^{-3} J/K$
Work Step by Step
The heat that enters the body is 80% from the total metabolism rate. Which is $Q_H = 0.8 \times 80 J = 64J$ at $T_H = 37^oC = 310K$
The heat that leaves the body on the cooler side is $Q_C = -64 J $ at $T_L = 30^oC = 303K$
Now plug in the entropy equation
$ \Delta S = \frac{Q_H}{T_H} + \frac{Q_C}{T_C}$
$ \Delta S = \frac{64J}{310K} + \frac{-64J}{303K}$
$ \Delta S = 0.20645 J/K - 0.21122 J/K$
$ \Delta S = -0.0048 J/K = -4.8 \times 10^{-3} J/K$
The negative sign indicates that the entropy of the body decreases. It does not violate second law of thermodynamics since the person is not an isolated system.
