Specific Heat Problem

[Dovahkiin]

For the life of me I cannot remember how to do this type of problem.

 

c = (100g) (4.180j/gc) (8.8c) / (127.50g) (64.5)c

 

I changed the numbers around so that they do not match with any assignment I have. Could someone guide me here with an example? (You can change the numbers if you do not believe me)

[studiot]

It would help if you stated the problem as written.

 

[Sensei]

c = (100g) (4.180j/gc) (8.8c) / (127.50g) (64.5)c

 

In numerator I see calculation of energy needed to heat 100 grams of water for 8.8 degrees.

But what is in denominator?

There will be unit issue this way.. J / (g*K) won't match with K...

It would help if you stated the problem as written.

 

 

[latex]4.1855 \frac{J}{g*K}[/latex] is energy needed to heat 1 gram of water for 1 C (1 calorie XIX century energy unit).

[Dovahkiin]

They are using it as 4.180J/gC (Celsius for chemistry)

 

100g * 4.180J/gC * 8.8c (our rate of change)

-------------------------------

127.50g (mass) * 64.5c (final temp)

 

I think I figured it out. I knew the first one was iron so I just played around with the problem.

I did it two ways but it seemed that dividing like terms first:

100g/127.50g and then 8.8c/64.5c = (.78g) and (.14c)

Then multiplying that to get: .0192g/c

 

The multiplying that by the 4.180Jgc gives us .45 which is the Specific Heat of Iron which is the object we were working with for this problem.

 

Thanks anyway guys!