Given:
The sun radiates energy at a rate of
[tex]W=3.80\times10^{26}\text{ W}[/tex]The temperature in the empty space is,
[tex]\begin{gathered} T_H=5500+273 \\ =5773\text{ K} \end{gathered}[/tex]The temperature at the deep space is,
[tex]\begin{gathered} T_c=-270+273 \\ =3\text{ K} \end{gathered}[/tex]To find:
the increase in entropy (in J/K) in one day
Explanation:
The value of heat is,
[tex]\begin{gathered} Q_H=Q_c=Q=3.80\times10^{26}\times24\times3600 \\ =3.28\times10^{31}\text{ J} \end{gathered}[/tex]The change in entropy is,
[tex]\begin{gathered} \Delta s=-\frac{Q_H}{T_H}+\frac{Q_c}{T_c} \\ =-\frac{3.28\times10^{31}}{5773}+\frac{3.28\times10^{31}}{3} \\ =1.08\times10^{31}\text{ J/K} \end{gathered}[/tex]Hence, the increase in entropy is
[tex]1.08\times10^{31}\text{ J/K}[/tex]