Self-Test 9.2 B


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lizzygaines1D
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Joined: Thu Jul 13, 2017 3:00 am

Self-Test 9.2 B

Postby lizzygaines1D » Sun Jan 28, 2018 5:10 pm

the question:
5.5 g of stainless steel is increased from 20 C to 100 C. what is the change in entropy? The specific heat capacity of stainless steel is .51 J/Cxg
Obviously you plug it into Cln(T2/T1), but then you have to multiply by grams? Why do you do this and when do you know when to do this?

Akash_Kapoor_1L
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Joined: Thu Jul 13, 2017 3:00 am
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Re: Self-Test 9.2 B

Postby Akash_Kapoor_1L » Sun Jan 28, 2018 5:14 pm

This is a great question! I was lowkey confused about this too. The easiest way to know is based on the units that they give you and the units you want to end up with. You know that you want the entropy, which is in J/K... But the specific heat capacity value that they give you is in J/C*g. The only way that you'll be able to take that g out of the denominator is by multiplying the standard formula by the grams of the molecule.

Yixin Angela Wang 2H
Posts: 51
Joined: Thu Jul 27, 2017 3:00 am

Re: Self-Test 9.2 B

Postby Yixin Angela Wang 2H » Sun Jan 28, 2018 11:33 pm

I agree. Also, since the specific heat is given per gram, then that means the more grams of substance there are, the more C there must be. Thus, multiplying by grams makes sense.


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