# Specific Heat/Energy problem

#### Raihn

1. 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.

2. Homework 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

Related Introductory Physics Homework Help News on Phys.org

#### Vuldoraq

1. 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.

2. Homework 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

-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

1. 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.

2. Homework 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
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 alot. 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?

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving