Calculating Energy Required for Temperature Change of Substance

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The discussion revolves around calculating the energy required to raise the temperature of a 2.0g substance from 5.0°C to 15.0°C, using a specific heat formula that varies with temperature. Participants highlight the need to integrate the specific heat equation correctly to find the heat energy (Q) and clarify the integration limits. There is confusion regarding the calculations, particularly about the integration process and the resulting values of Q. After reevaluating, one participant recalculates Q to be 9894.36, but doubts its accuracy, prompting further verification of the calculations. The conversation emphasizes the importance of correctly applying calculus in thermodynamic equations.
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Homework Statement


The specific heat of a substance varies with temperature according to:
c = 0.20 + 0.14T + 0.023T^{2},​
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\DeltaT
K= C + 273.15

The Attempt at a Solution


Given:
m= 2.0g
Q=?

Q=mc\DeltaT
Q=m\int(0.2 + 1.4T + .023T^{2})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^{2})] -
[2.0g(0.2 + 1.4(278.15) + 0.23(278.15^{2})]

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
 
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Raihn said:

Homework Statement


The specific heat of a substance varies with temperature according to:
c = 0.20 + 0.14T + 0.023T^{2},​
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\DeltaT
K= C + 273.15

The Attempt at a Solution


Given:
m= 2.0g
Q=?

Q=mc\DeltaT
Q=m\int(0.2 + 1.4T + .023T^{2})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^{2})] -
[2.0g(0.2 + 1.4(278.15) + 0.23(278.15^{2})]

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.
 
Vuldoraq said:
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.
 
Raihn said:

Homework Statement


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

This is the value of the heat capacity?
Raihn said:
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\DeltaT
K= C + 273.15

The Attempt at a Solution


Given:
m= 2.0g
Q=?

Q=mc\DeltaT
Q=m\int(0.2 + 1.4T + .023T^{2})dt

This is your formula for Q

Raihn said:
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^{2})] -
[2.0g(0.2 + 1.4(278.15) + 0.23(278.15^{2})]

Here you have just subbed the value of c into Q=mc\delta T I see no evidence that c has been integrated?
 
Raihn said:
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),

\int_{5}^{15}

and you divide by two because,

\int x dx=\frac{x^{2}}{2}
 
Oh ok, I see thanks alot. Didn't notice I skipped that part. Thanks again.
 
Your welcome. Does the answer come out ok? If you need anymore help just ask.
 
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.
 
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
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
Or is Q negative, where T1 = 278.15 and T2 = 288.15?
 
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