Help with this thermodynamics and entropy question please

[takelight2]

Homework Statement

A 124.4 g insulated aluminum cup at 18.46 ∘C is filled with 130.0 g of water at 46.91∘C. After a few minutes, equilibrium is reached.

a. Determine the final temperature. (completed)

b. Determine the total change in entropy.

Relevant Equations

Q=mcT

So I've answered the first question and I got a final temp of 42.06 Celsius.

Now for this second one, I don't know why I am getting it wrong:

Im doing 0.215*ln(315.06/291.46) + 1*ln(315.06/319.91)

But it says I am wrong. What about my process is faulty?

 

[haruspex]

Homework Statement:: A 124.4 g insulated aluminum cup at 18.46 ∘C is filled with 130.0 g of water at 46.91∘C. After a few minutes, equilibrium is reached.

a. Determine the final temperature. (completed)

b. Determine the total change in entropy.
Relevant Equations:: Q=mcT

So I've answered the first question and I got a final temp of 42.06 Celsius.

Now for this second one, I don't know why I am getting it wrong:

Im doing 0.215*ln(315.06/291.46) + 1*ln(315.06/319.91)

But it says I am wrong. What about my process is faulty?

Please fill in the details of how you get the factors 0.215 and 1.

 

[mjc123]

Also, to be pedantic, if you are working to hundredths of a degree, 0°C is 273.15K, not 273.00, though doubtless it makes very little difference to your answer.

 

[Chestermiller]

Please fill in the details of how you get the factors 0.215 and 1.

Yes, I differ from these values (and more importantly, from their ratio). Instead of the 0.215, I get 112 J/K, and, instead of the 1, I get 544 J/K.

 

[rude man]

My numbers agree with @Chestermiller.
(But I don't get the 'ratio' part).

 

[Chestermiller]

My numbers agree with @Chestermiller.
(But I don't get the 'ratio' part).

Thanks Rudy. I don't either. I was mistaken about that.

 

[rude man]

Thanks Rudy. I don't either. I was mistaken about that.

Actually, I thought you meant the limits of integration which turn out to be ratios after the integration, but separately for the water and the cup.

 

[Chestermiller]

Actually, I thought you meant the limits of integration which turn out to be ratios after the integration, but separately for the water and the cup.

Nope. I mistakenly meant the ratio of the MC's.