Calculating Energy Required for Temperature Change of Substance

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In summary, the specific heat of a substance is given by the equation c = 0.20 + 0.14T + 0.023T^{2} where T is in degrees Celsius and c is in cal/gK. To find the energy required to raise the temperature of 2.0g of the substance from 5.0C to 15C, we can use the formula Q=mc\DeltaT, where m is the mass, c is the specific heat, and \DeltaT is the change in temperature. Converting the temperatures to Kelvin, T1 = 278.15K and T2 = 288.15K, we can solve for c by plugging in the values and integrating
  • #1
Raihn
7
0

Homework Statement


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

The Attempt at a Solution


Given:
m= 2.0g
Q=?

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

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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  • #2
Raihn said:

Homework Statement


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

The Attempt at a Solution


Given:
m= 2.0g
Q=?

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

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.
 
  • #3
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.
 
  • #4
Raihn said:

Homework Statement


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

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[tex]\Delta[/tex]T
K= C + 273.15

The Attempt at a Solution


Given:
m= 2.0g
Q=?

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

Here you have just subbed the value of c into [tex]Q=mc\delta T[/tex] I see no evidence that c has been integrated?
 
  • #5
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),

[tex]\int_{5}^{15}[/tex]

and you divide by two because,

[tex]\int x dx=\frac{x^{2}}{2}[/tex]
 
  • #6
Oh ok, I see thanks alot. Didn't notice I skipped that part. Thanks again.
 
  • #7
Your welcome. Does the answer come out ok? If you need anymore help just ask.
 
  • #8
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
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?
 

1. What is specific heat and why is it important in energy problems?

Specific heat is the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius. It is important in energy problems because it helps determine how much energy is needed to heat or cool a substance, and how much energy a substance can store.

2. How is specific heat different from heat capacity?

Specific heat is the amount of heat energy required to raise the temperature of a substance by a certain amount, while heat capacity is the amount of heat energy required to raise the temperature of an entire object by a certain amount. Specific heat is an intensive property while heat capacity is an extensive property.

3. How do you calculate the specific heat of a substance?

The specific heat of a substance can be calculated by dividing the amount of heat energy transferred to the substance by the mass of the substance and the change in temperature (Q = mcΔT). This will give the specific heat in units of J/g°C or J/kgK.

4. Can specific heat vary for different substances?

Yes, specific heat can vary for different substances. This is because different substances have different molecular structures and properties, which affect how much energy is needed to raise their temperature. For example, water has a higher specific heat than most other substances, which is why it takes longer to heat up or cool down.

5. How does specific heat affect the transfer of thermal energy?

Specific heat plays a role in the transfer of thermal energy because it determines how much energy is needed to change the temperature of a substance. This means that substances with higher specific heat will require more energy to change their temperature, and will also release more energy when their temperature changes. This can impact processes such as heating and cooling systems and cooking methods.

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