• Support PF! Buy your school textbooks, materials and every day products Here!

Entropy and heat: part 2

  • Thread starter Benzoate
  • Start date
419
0
1. 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?


2. Homework Equations
Possible equations:
dS=Q/T:
Q=m*c*delta(T)
Q=m*L

3. The Attempt at a Solution

First, I need to convert the volumes of water in to 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
 

Answers and Replies

dynamicsolo
Homework Helper
1,648
4
2. 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...
 

Related Threads for: Entropy and heat: part 2

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
2
Views
1K
Replies
7
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
3
Views
5K
Top