# Specific Heat/Energy problem

1. Mar 1, 2009

### Raihn

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Q=mc$$\Delta$$T
K= C + 273.15

3. 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

2. Mar 1, 2009

### Vuldoraq

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. Mar 1, 2009

### Raihn

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. Mar 1, 2009

### Vuldoraq

This is the value of the heat capacity?
This is your formula for Q

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

5. Mar 1, 2009

### Vuldoraq

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,

$$\int x dx=\frac{x^{2}}{2}$$

6. Mar 1, 2009

### Raihn

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

7. Mar 1, 2009

8. Mar 1, 2009

### 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.

9. Mar 1, 2009

### 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).

10. Mar 1, 2009

### 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

11. Mar 1, 2009

### Raihn

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