Generating entropy - the usage of signs

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Mara Lockhart 3J
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Generating entropy - the usage of signs

Postby Mara Lockhart 3J » Tue Jan 20, 2015 9:42 pm

When calculating the change in entropy, we use the equation deltaS=q/T. For problem 8.1, we use this equation and specifically use the rate of heat generation over temperature; however, the solutions manual uses a negative sign in front of the equation in the work, but not in the answer. Where did the negative sign come from and why is it now seen in the final answer?

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Re: Generating entropy - the usage of signs

Postby Chem_Mod » Tue Jan 20, 2015 10:48 pm

If we refer to Section 8.3, specifically page 292, we find the instance of an isothermal expansion. As we have previously discussed, isothermal means and therefore, From here, , so . We know that in a reversible, isothermal expansion, work has the formula . When we refer to our equation for entropy, , we can plug in , and with the cancellation of T, we end up with our final equation .

Amy Luong 1L
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Re: Generating entropy - the usage of signs

Postby Amy Luong 1L » Wed Jan 21, 2015 2:07 pm

When calculating entropy is it always necessary to change the temperature to Kelvin?

Niharika Reddy 1D
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Re: Generating entropy - the usage of signs

Postby Niharika Reddy 1D » Wed Jan 21, 2015 8:52 pm

I think so, since K is included in the units of entropy.

For enthalpy, when calculating q using q=mCΔT or a variation of that formula, the temperature could be left in Celsius if the heat capacity was also left in units of Celsius since relative temperatures made the change in temperature the same whether we converted it into Kelvin or left it in Celsius. The value of the heat capacity would remain the same no matter the units of temperature: i.e. for liquid water, the specific heat is 4.184 J/g°C or 4.184 J/g·K. We just needed to find heat in terms of joules, so the units of the temperature did not matter since they canceled out as long as we wrote out the same one for both the specific heat and the temperature change.

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