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Topic: Specific Heat  (Read 2781 times)

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Offline chrisjohns

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Specific Heat
« on: April 27, 2007, 09:44:31 AM »
For the following problem:  A 20.00 g slug of Fe is heated to 95.0°C was placed into 100.00g of water at 25°C. The specific heat of the Fe is 0.108cal/g°C. The specific heat of water is 1.00cal/g°C. What is the final temperature of the Fe?
If I set up the equation as:
-[ (20.00gFe)(0.108cal/g°CFe)(Tf-95°C)] = [ (100.00g H2O) (1.00cal/g°C H2O)(Tf-25°C H2O)]
the next step:
 -[ (2.16)(95.0°C - Tf] = [ (100.00(25°-Tf]
Is that last step correct, and can someone help me with the math here I'm very poor mathematically?

Offline enahs

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Re: Specific Heat
« Reply #1 on: April 27, 2007, 04:31:46 PM »
This is nothing but application of the "laws of algebra". They do not teach these enough, as all algebra is is these few "laws", even the complicated ones that can take pages of work:
http://mentorproducts.com/laws.html


- ( 2.16 * (95 - X) ) = 100 * (25 - X)

First, use your distributive law:
=
 
(-2.16 * 95) - (-2.16 * X)  = (100 * 25) - (100 * X)

=

-205.2 - (-2.16 * X) = 2500 - (100 * X)

rearrange, and simplify the negative signs (to negatives equals a positive)

2500 + 205.2 = (2.16 * X) + (100 * X)
 =
2705.2 =  (2.16 * X) + ( 100 * X)

Apply the distributive law again (factor out)

2705.2 =  X * (2.16+ 100)
=
2705.2 = X * (102.16)

Divide both sides by 102.16
2705.2/102.16 = X
X = 26.48


Now, you can also then THINK about the math you just did. Does the number seem correct? Does that at least seem the right change in temperature? Did it get hotter or colder, and was it supposed to get hotter or colder?








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