What Is the Correct Approach to Solve This Specific Heat Calculation Error?

In summary, a student was having trouble with a specific heat problem and consulted with a classmate for help. They both used an incorrect formula for the problem, which led to an incorrect answer. After discussing with the classmate, the student realized their mistake and corrected it. The correct formula should be (cp)(m1)(Tf- Ti) + (cp)(m2)(Tf- Ti) = 0, as stated in their textbook.
  • #1
porschedriver192
12
0
Please Help! - Specific Heat Problem...

I have been stuck for a while, it's online (Webassign), so it told me my answer was wrong.
-------
A 6.3x10^2 g sample of water at 90.0°C is mixed with 4.45x10^2 g of water at 22.0°C. Assume no heat loss to the surroundings. What is the final temperature of the mixture?
-------

The formula I know is right:

(cp)(m)(Tf- Ti) = (cp)(m)(Tf- Ti)

cp= spec. heat
m = mass
Tf = final Temp
Ti = initial Temp
--------
work:

converted grams into kilograms...

cp water, given
||
(4186) (.630) (x-90) = (4186) (.445) (x-22)

774.41x = 196365.26

x = <<<it says it's wrong>>>

Where did I slip up?

Thanks.
 
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  • #2
porschedriver192 said:
The formula I know is right:

(cp)(m)(Tf- Ti) = (cp)(m)(Tf- Ti)
I think you mean:
(cp)(m1)(Tf- Ti) + (cp)(m2)(Tf- Ti) = 0
 
  • #3
My textbook gave me the original formula...but would the one you gave yield a correct answer? Thanks.
 
  • #4
porschedriver192 said:
My textbook gave me the original formula...
Are you sure you copied it exactly? The formula you gave makes no sense. Note that the final temperature will be somewhere between 22°C and 90°C. Thus one side of your equation would be negative, the other positive. Nothing good can come of that. :smile:
 
  • #5
Doc Al,

I called a classmate, and he had used the same formula I did...and got the correct answer. After walking through my steps with him, I found that messed up here:

cp water, given
||
(4186) (.630) (x-90) = (4186) (.445) (x-22)

774.41x = 196365.26
It should have been 90-x
------
About the equation...my book is right. It is on specific heat capactity, and states how the energy absorbed by one object must be released by the other, similar to conservation of momentum.

Thanks for the help, and I was eventually led in the right direction!
 
  • #6
porschedriver192 said:
I called a classmate, and he had used the same formula I did...and got the correct answer.
Well... that formula is still wrong for the reasons I gave. (I'm guessing that you copied it incorrectly.)
After walking through my steps with him, I found that messed up here:

cp water, given
||
(4186) (.630) (x-90) = (4186) (.445) (x-22)

774.41x = 196365.26
It should have been 90-x
Right! I hope you realize that since you had to reverse things in your formula, that your formula cannot be correct.
 

Related to What Is the Correct Approach to Solve This Specific Heat Calculation Error?

1. What is specific heat and why is it important?

Specific heat is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius. It is important because it helps us understand how different substances respond to heat and how much energy is needed to change their temperature.

2. How do you calculate specific heat?

Specific heat can be calculated by dividing the amount of heat added to a substance by its mass and the change in temperature. The formula is Q = m x c x ΔT, where Q is the heat added, m is the mass, c is the specific heat, and ΔT is the change in temperature.

3. What are the units for specific heat?

The units for specific heat are J/g·°C or J/kg·K, which represent the amount of heat energy required to raise the temperature of 1 gram or 1 kilogram of a substance by 1 degree Celsius or 1 Kelvin, respectively.

4. How does specific heat differ between substances?

Specific heat can vary greatly between different substances. Generally, substances with stronger intermolecular forces, such as metals, have higher specific heat compared to substances with weaker intermolecular forces, such as gases. The specific heat of a substance also depends on its state (solid, liquid, or gas) and can change with temperature.

5. How is specific heat used in real-world applications?

Specific heat is used in various real-world applications, such as in cooking, designing thermal systems, and studying climate change. It helps determine the amount of energy needed to heat or cool a substance, and also plays a role in understanding the transfer of heat within and between different systems.

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