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## Homework Statement

An open insulated vessel is divided into two parts by a vertical non-conducting partition.On one side of this partition is 2.5 kg of water initially at 38°C, and on the other side is 5.0 kg of water initially at a temperature of 70°C. On removal of the partition and the subsequent mixing, uniform conditions in the vessel are finally attained. Assuming no external heat transfer, evaluate the increase in entropy of the system. Assume that Cp for water is constant at 4.187 kJ/kg K

## Homework Equations

Δs= m x Cp x ln(Tfinal/Tinitial)

## The Attempt at a Solution

This is a mixing process so it must be irreversible and no heat transfer so adiabatic. Δs must be bigger than zero.

m1= 2.5 kg , m2= 5.0 kg , T1= 313.15 K, T2= 345.15 K

m1CpΔT+m2CpΔT= 0 (internal energy is dependent on the temperature only, temp doesnt change like Joule's experiment)

I assumed that they would have a final temp that would be the same for both m1 and m2.

I solved the equation above and got Tfinal= 334.48 K

For m1:

Δs = 2.5 x 4.187 x ln( 334.48/313.15)= 0.6898 kJ/K

For m2:

Δs = 5.0 x 4.187 x ln( 334.48/345.15)= -0.13148 kJ/K

Δs increase for the entire system =0.6898 + (-0.13148) =+0.0331 kJ/K

This is the correct answer according to the solutions so I did the maths right. However I am confused with the entropy change for m2=5.0 kg . As it can be seen in the calculation above that we have obtained a negative change in entropy. Isn't this undefined in the sense it disobeys the second law of thermodynamics. 'Entropy of the universe is always increasing' So how come we are allowed to proceed with the calculations if it isn't possible to have this in the first place.. If someone can enlighten me on this it would be great because i'm finding the second law quite hard to understand.

Thanks a lot