Question

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?

Answer 1

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.