How much new entropy have you created by mixing the water

[mikee]

Homework Statement

In order to take a warm bath, you mix 50 liters of hot water at 55 degrees Celsius with 25 liters of cold water at 10 degrees Celsius, How much new entropy have you created by mixing the water

Homework Equations

S = Integral(Ti to Tf) (Cv / T) dT

The Attempt at a Solution

Ok i am doing the question, let's call the 50 liters system A and 25 liters system B, so i calculated the entropy of each respective system and then getting the total entropy by Stot = Sa + Sb, and I am getting a negative answer and i do not understand this at at all because i thought entropy always increased. Anyways I am pretty sure I am doing the calculations correct but i will show what i did nonetheless.


50(55) + 25(10) = 50(Tf) + 25(Tf) where Tf was found to be 33.3 degrees celsuis.

Cv for system A is 50*1000*4.186 therefore Sa = Cv * ln((273+33.3)/(273+55)) = -1.432*10^4

Cv for system B is 25*1000*4.186 therefore Sb = Cv * ln((273 + 33)/(273+10)) = 8.27*10^3 and when you add them you get a negative entropy ?

 

[diazona]

Sa = Cv * ln((273+33.3)/(273+55)) = -1.432*10^4

Sb = Cv * ln((273 + 33)/(273+10)) = 8.27*10^3

Why are you using those formulas?

 

[mikee]

Because S = Integral(Ti to Tf) (Cv / T) dT, and Cv is not dependant on temperature for the problem so integral 1/T T = Ln(Tf/Ti).. and then i just plugged in the values.

 

[diazona]

Ah, yes... something looked wrong about it but I think I was just mixing it up with another equation.

50(55) + 25(10) = 50(Tf) + 25(Tf) where Tf was found to be 33.3 degrees celsuis.

Check that calculation again.