Answer:
Δ[tex]S_{univ}=267.7J/K[/tex]
Explanation:
Hello,
To find the change in the entropy of the universe, we must take into account the following entropy balance:
Δ[tex]S_{univ}=\frac{Q_{water}+Q_{ice}}{T_{univ}}[/tex]
We can stand the universe as the surroundings so the [tex]T_{univ}[/tex] equals the outside air temperature. Then, we must compute both the water's and ice's released heat due to their interaction with the "hot" air:
[tex]Q_{water}=320g*4.18J/g^0C*(25-0)^0C= 33440J[/tex]
[tex]Q_{ice}=m_{ice}*(H_{melt,ice}+Cp_{ice}*(T_{univ}-T_{initial}))\\Q_{ice}=120g*(333.7J/g+2.11J/g^0C*(25-0)^0C)\\Q_{ice}=46374J[/tex]
Finally, we obtain the change (increasing) in the entropy of the universe as follows:
Δ[tex]S_{univ}=\frac{33440J+46374J}{(25+273.15)K}[/tex]
Δ[tex]S_{univ}=267.7J/K[/tex]
* All the data were extracted from Cengel's thermodynamics book.
Best regards.
