Calculating Energy Required for Temperature Change of Substance

[Raihn]

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

 

[Vuldoraq]

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.

 

[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.

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.

 

[Vuldoraq]

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?

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

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?

 

[Vuldoraq]

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}

 

[Raihn]

Oh ok, I see thanks a lot. Didn't notice I skipped that part. Thanks again.

 

[Vuldoraq]

Your welcome. Does the answer come out ok? If you need anymore help just ask.

 

[Raihn]

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.

 

[Vuldoraq]

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).

 

[Raihn]

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

 

[Raihn]

Or is Q negative, where T1 = 278.15 and T2 = 288.15?