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## Homework Statement

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?

## Homework Equations

The formula for entropy, which is [itex]\Delta Entropy = \frac{Q}{T}[/itex]

and the formula for Entropy of the universe which is [itex]\Delta Entropy_{universe} = \Delta Entropy_{1} + \Delta Entropy_{2} ... [/itex]

## The Attempt at a Solution

I know that the entropy of the bag will be :

[itex]\Delta Entropy_{bag} = \frac{3.17 x 10^8 J}{531 K}[/itex]

[itex]\Delta Entropy_{bag} = 596986.8173 \frac{J}{K}[/itex]

Then the entropy of the table would be :

[itex]\Delta Entropy_{table} = \frac{-3.17 x 10^8 J}{338 K}[/itex]

[itex]\Delta Entropy_{table} = -937453.7927 \frac{J}{K}[/itex]

From this, the entropy of the universe would be :

[itex]\Delta Entropy_{universe} = 596986.8173 \frac{J}{K} + -937453.7927 \frac{J}{K}[/itex]

[itex]\Delta Entropy_{universe} = -340466.9754 \frac{J}{K}[/itex]

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!

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