Specific Heat/Energy problem

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$$\Delta$$T
K= C + 273.15

The Attempt at a Solution

Given:
m= 2.0g
Q=?

Q=mc$$\Delta$$T
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

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$$\Delta$$T
K= C + 273.15

The Attempt at a Solution

Given:
m= 2.0g
Q=?

Q=mc$$\Delta$$T
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.

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.

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$$\Delta$$T
K= C + 273.15

The Attempt at a Solution

Given:
m= 2.0g
Q=?

Q=mc$$\Delta$$T
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?

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.

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).

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

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