135k views
4 votes
In a vacuum bottle, 320 g of water and 120 g of ice are initially in equilibrium at 0.00∘∘C. The bottle is not a perfect insulator. Over time, its contents come to thermal equilibrium with the outside air at 25.0∘∘C. 1) How much does the entropy of universe increase in the process?

User Omnidan
by
8.7k points

1 Answer

3 votes

Answer:

Δ
S_(univ)=267.7J/K

Step-by-step explanation:

Hello,

To find the change in the entropy of the universe, we must take into account the following entropy balance:

Δ
S_(univ)=(Q_(water)+Q_(ice))/(T_(univ))

We can stand the universe as the surroundings so the
T_(univ) 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:


Q_(water)=320g*4.18J/g^0C*(25-0)^0C= 33440J


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

Finally, we obtain the change (increasing) in the entropy of the universe as follows:

Δ
S_(univ)=(33440J+46374J)/((25+273.15)K)

Δ
S_(univ)=267.7J/K

* All the data were extracted from Cengel's thermodynamics book.

Best regards.

User Steve Jorgensen
by
8.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.