Thermodynamic Entropy Clarification

[bpaterni]

Homework Statement

We mix 0.6 kg of water at a temperature of 25°C with 0.8 kg of alcohol at 30.5°C. They
come to equilibrium. (Here's some useful data: The specific heat capacity for water is 1
kcal/kg°C = 4186 J/kg°C, and for alcohol it is 0.58 kcal/kg°C = 2430 J/kg°C

What is the change in entropy (in J/K) of the system when the two are mixed?

Homework Equations

∆S = ∆Sw + ∆Sa = Mw*Cw*ln(Tf /TW) + Ma*Ca*ln(Tf /Ta)

The Attempt at a Solution

I've calculated the final temperature to be 300.55 K or 27.4 C

so now the equation looks like:
.6*4186*Ln(Tf/Tw) + .6*2430*Ln(Tf/Ta) = dS

The problem is, do I use Kelvin or Celsius for the entropy calculation? I get two totally different values if I do either one... .0794 with K or 21.87 with C.

 

[rl.bhat]

You have to use Kelvin.

 

[thomate1]

I would first clarify with the instructor or whoever gave the homework assignment whether the change in entropy should be calculated in Kelvin or Celsius. This is important because the units of entropy are Joules per Kelvin (J/K), so using the wrong temperature scale can result in incorrect values.

Assuming that the change in entropy should be calculated in Kelvin, I would proceed with the calculation using the final temperature of 300.55 K. However, if the change in entropy should be calculated in Celsius, then I would convert the final temperature to 27.4°C before plugging it into the equation.

It is also important to note that the change in entropy for this system is a positive value, indicating an increase in disorder or randomness. This is expected since the two substances are mixing and reaching equilibrium, which is a state of maximum disorder.