## Change in Entropy for the Surroundings vs. System

PeterTran1C
Posts: 30
Joined: Thu Jul 13, 2017 3:00 am

### Change in Entropy for the Surroundings vs. System

Is there a conceptual reason why $\Delta S_{surround} = -\frac{\Delta H_{system}}{T}$ is negative compared to $\Delta S_{system} = \frac{\Delta H_{system}}{T}$ which is a positive value?

Joshua Xian 1D
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am
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### Re: Change in Entropy for the Surroundings vs. System

∆S(system) = ∆H(system)/T since (assuming constant pressure) ∆H(system)=q(system), and if heat were added to the system, then entropy would increase. (+q(system) --> +∆S(system)). ∆S(surroundings) = -∆H(system)/T since if heat from the surroundings were added to the system, then the entropy of the surroundings would decrease (+q(system) --> -∆S(surroundings)).

mayapartha_1D
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: Change in Entropy for the Surroundings vs. System

The equations are opposite for the system vs the surroundings because heat is being transferred from one to the other, so the entropies must be opposite values to reflect this.

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