# Entropy of Objects and the Universe

1. Feb 29, 2012

### frozonecom

1. The problem statement, all variables and given/known data
A bag of mass 2000 kg at 531 K is on the table at a constant temperature of 65 degrees Celsius. The bag absorbs 3.17 x 10^8 J of heat. What is the change in entropy of the bag, the table and the universe?

2. Relevant equations
The formula for entropy, which is $\Delta Entropy = \frac{Q}{T}$
and the formula for Entropy of the universe which is $\Delta Entropy_{universe} = \Delta Entropy_{1} + \Delta Entropy_{2} ...$

3. The attempt at a solution

I know that the entropy of the bag will be :
$\Delta Entropy_{bag} = \frac{3.17 x 10^8 J}{531 K}$
$\Delta Entropy_{bag} = 596986.8173 \frac{J}{K}$

Then the entropy of the table would be :
$\Delta Entropy_{table} = \frac{-3.17 x 10^8 J}{338 K}$
$\Delta Entropy_{table} = -937453.7927 \frac{J}{K}$

From this, the entropy of the universe would be :
$\Delta Entropy_{universe} = 596986.8173 \frac{J}{K} + -937453.7927 \frac{J}{K}$
$\Delta Entropy_{universe} = -340466.9754 \frac{J}{K}$

Oh, and I put negative for Heat in the table because I assumed that it is where the bag absorbed the heat from.
But, I know that Entropy of the universe can't be negative, since CHANGE IN ENTROPY CANNOT DECREASE, it can only be maintained, or increased. Right?? So, Change in Entropy of the universe can only be ZERO or any POSITIVE NUMBER...

So from this point, I know I'm doing something wrong. Please help!

Last edited: Mar 1, 2012
2. Mar 1, 2012

### Andrew Mason

Your entropy calculation is correct if you assume that the heat flows from the table to the bag.

The problem is with your assumption. The second law says that heat cannot spontaneously flow a colder to a hotter body: ie from the table (T = 338K) to the bag (T=531K).

I am not sure where the heat comes from or how the table is kept at 65 C. But one thing is certain, the heat flow into the bag did not come from the table.

AM

3. Mar 1, 2012

### frozonecom

It says here in my book that "There can be no process whose only final result is to transfer thermal energy from cooler object to a hotter one". --- which agrees with what you said that heat cannot sponataneously flow from cooler to hotter body.

I guess I was tricked by the homework question. :)

So, the process cannot occur, and therefore, there will be no change in entropy??

Last edited: Mar 1, 2012
4. Mar 1, 2012

### Andrew Mason

No. The process cannot occur spontaneously - that is, without supplying work - if the heat flow into the bag comes from a colder body such as the table.

Since there is no indication where the heat flow into the bag comes from - and since it cannot come from the table - the correct answer would be that you do not have sufficient information to answer the question. You can determine the ΔS for the table (assuming it remains at 65C) and for the bag. But you cannot determine the ΔS of the rest of the universe.

AM

5. Mar 1, 2012

### frozonecom

Thanks for the help!! :)