Specific Heat/Energy problem

  • Thread starter Raihn
  • Start date
  • #1
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Homework Statement


The specific heat of a substance varies with temperature according to:
c = 0.20 + 0.14T + 0.023T[tex]^{2}[/tex],​
with T in deg-C and c in cal/gK. Find the energy required to raise the temperature of 2.0g of this substance from 5.0C to 15C.

Homework Equations


Q=mc[tex]\Delta[/tex]T
K= C + 273.15

The Attempt at a Solution


Given:
m= 2.0g
Q=?

Q=mc[tex]\Delta[/tex]T
Q=m[tex]\int[/tex](0.2 + 1.4T + .023T[tex]^{2}[/tex])dt

Convert C to Kelvin:
T1: K = 5.0 + 273.15 = 278.15K
T1: K = 15 + 273.15 = 288.15K

Solve for heat:
Q = [2.0g(0.2 + 1.4(288.15) + 0.23(288.15[tex]^{2}[/tex])] -
[2.0g(0.2 + 1.4(278.15) + 0.23(278.15[tex]^{2}[/tex])]

Q=2792.98

I don't think I did the math correctly can anyone please check it for me and tell me what I did wrong? Thank you in advanced.

-Raihn
 

Answers and Replies

  • #2
272
0

Homework Statement


The specific heat of a substance varies with temperature according to:
c = 0.20 + 0.14T + 0.023T[tex]^{2}[/tex],​
with T in deg-C and c in cal/gK. Find the energy required to raise the temperature of 2.0g of this substance from 5.0C to 15C.

Homework Equations


Q=mc[tex]\Delta[/tex]T
K= C + 273.15

The Attempt at a Solution


Given:
m= 2.0g
Q=?

Q=mc[tex]\Delta[/tex]T
Q=m[tex]\int[/tex](0.2 + 1.4T + .023T[tex]^{2}[/tex])dt

Convert C to Kelvin:
T1: K = 5.0 + 273.15 = 278.15K
T1: K = 15 + 273.15 = 288.15K

Solve for heat:
Q = [2.0g(0.2 + 1.4(288.15) + 0.23(288.15[tex]^{2}[/tex])] -
[2.0g(0.2 + 1.4(278.15) + 0.23(278.15[tex]^{2}[/tex])]

Q=2792.98

-Raihn

Looking at this it doesn't look like you did the integral for the heat capacity? You should have 0.2T+1.4T^2/2 etc.
 
  • #3
7
0
Looking at this it doesn't look like you did the integral for the heat capacity? You should have 0.2T+1.4T^2/2 etc.

Why divide by 2? And I thought I did take the integral? I couldn't figure out how to add the 15 and 5 onto the integral sign, but I thought I did solve it in that manner? Maybe not? Now I've confused myself lol.
 
  • #4
272
0

Homework Statement


The specific heat of a substance varies with temperature according to:
c = 0.20 + 0.14T + 0.023T[tex]^{2}[/tex],​

This is the value of the heat capacity?
with T in deg-C and c in cal/gK. Find the energy required to raise the temperature of 2.0g of this substance from 5.0C to 15C.

Homework Equations


Q=mc[tex]\Delta[/tex]T
K= C + 273.15

The Attempt at a Solution


Given:
m= 2.0g
Q=?

Q=mc[tex]\Delta[/tex]T
Q=m[tex]\int[/tex](0.2 + 1.4T + .023T[tex]^{2}[/tex])dt

This is your formula for Q

Convert C to Kelvin:
T1: K = 5.0 + 273.15 = 278.15K
T1: K = 15 + 273.15 = 288.15K

Solve for heat:
Q = [2.0g(0.2 + 1.4(288.15) + 0.23(288.15[tex]^{2}[/tex])] -
[2.0g(0.2 + 1.4(278.15) + 0.23(278.15[tex]^{2}[/tex])]

Here you have just subbed the value of c into [tex]Q=mc\delta T[/tex] I see no evidence that c has been integrated?
 
  • #5
272
0
Why divide by 2? And I thought I did take the integral? I couldn't figure out how to add the 15 and 5 onto the integral sign, but I thought I did solve it in that manner? Maybe not? Now I've confused myself lol.

You put the limits on like this (just click on the equation to see the source),

[tex]\int_{5}^{15}[/tex]

and you divide by two because,

[tex]\int x dx=\frac{x^{2}}{2}[/tex]
 
  • #6
7
0
Oh ok, I see thanks alot. Didn't notice I skipped that part. Thanks again.
 
  • #7
272
0
Your welcome. Does the answer come out ok? If you need anymore help just ask.
 
  • #8
7
0
I did it again and I got Q = 9894.36. I don't think that's right. This question is so simple I don't know why I'm not getting it.
 
  • #9
272
0
Don't worry, it's only easy once you know it. Check your calculation again and post it here before you enter it, that way I can check it and hopefully you won't lose another attempt. I seem to be getting quite a big number.

Edit: Why don't you think it's right? Also I got confused with someone else, thought you were using a mastering physics package (me tired, sorry).
 
  • #10
7
0
It's probably not clicking in my head since I'm studying for organic chemistry II and doing this at the same time, but this is what I did:

c = [0.20 + 1.4(288.15) + 0.023(288.152))] - [0.20 + 0.14(278.15) + 0.023(278.152)]
c = 494.78

Then:

Q=(2.0g)(494.718(288.15-278.15)
Q = 9894.36
 
  • #11
7
0
Or is Q negative, where T1 = 278.15 and T2 = 288.15?
 

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