Creating Entropy by Mixing Hot and Cold Water

[Benzoate]

Homework Statement

In order to take a nice warm bah, you mix 50 liters of hot water at 55 degrees celsuis wti h25 liters of cold water at 10 degrees celsius . how much new entropy have you created by mixing the water?

Homework Equations

Possible equations:
dS=Q/T:
Q=m*c*delta(T)
Q=m*L

The Attempt at a Solution

First, I need to convert the volumes of water into mass by multiplying the density of water by Volume.

m(hot)=(50 L)*(1.00 g/cm^3)= 50000 grams.
deltaS(hot)=(50000 g)*(4.18 J/(K*g))*(328 K-283 K)/(328 K)= 28673 J/K
m(cold)=(25 L)*(1.00 g/cm^3)= 25000 grams
deltaS(cold)=(25000 g)*(4.18 J/(K*g))*(283 K - 328 K)/(283 K)= -16616.60 J/K

deltaS(new)=deltaS(cold)+deltaS(hot)= (-16616.60 J/K)+(28273 J/K)= 11656.4 J/K

 

[dynamicsolo]

Homework Equations

Possible equations:
dS=Q/T:
Q=m*c*delta(T)

OK, there's a little problem with one of your equations. The differential entropy is

dS = dQ/T .

If you put that together with the calorimetric equation, which will give you

dQ = mc·dT ,

you have dS = mc · (dT / T).

When the hot water is cooling to 40º C (283 K) and the cold water warming to that equilibrium temperature, the temperatures of the components are not constant. So you will have to integrate to find the change in entropy of each component. The sum should be positive, but will be different from the value you have presently...