Thermochemistry Help: Finding the final temperature of a mixture.

[kirsten_2009]

Homework Statement

A 1.22 kg piece of iron at 126.5 degrees Celsius is dropped into 981 g of water at 22.1 degrees Celsius. The temperature rises to 34.4 degrees Celsius. What will be the final temperature if this same piece of iron at 99.8 degrees celsius is dropped into 350 mL of glycerol, CH2OHCHOHCH2OH, at 26.8 degrees Celsius? For glycerol, density = 1.26 g/mL, C_p = 219 J/K x mol.

Homework Equations

q = m x C x (Tf-Ti)

The Attempt at a Solution

So, I know that the specific heat of water is 4.18 J/g x C, so I could find "q" for water (50437.134) and since the heat gained by the surroundings is equal to what is lost by the system I could also say that "q" for water is the same as "q" for iron but negative (-50437.134). Now, I could calculate the specific heat of iron (0.449 J/g x C). But now I'm stuck because I don't have "q" for the iron that is then dumped into glycerol and I don't have Tf so how could I calculate it? Or, is the value of "q" for iron the same throughout the process? Help is much appreciated! :)

 

[Lok]

So, I know that the specific heat of water is 4.18 J/g x C, so I could find "q" for water (50437.134) and since the heat gained by the surroundings is equal to what is lost by the system I could also say that "q" for water is the same as "q" for iron but negative (-50437.134). Now, I could calculate the specific heat of iron (0.449 J/g x C). But now I'm stuck because I don't have "q" for the iron that is then dumped into glycerol and I don't have Tf so how could I calculate it? Or, is the value of "q" for iron the same throughout the process? Help is much appreciated! :)

If by q (for the iron) you mean energy, then you have the mass, the temperature and C_p of iron, it's all you need. It's the same as above where you found out the C_p of iron just that you need to find out the final temp.

 

[kirsten_2009]

Hello!

I think I'm not understanding what you are saying...
I found "q" for iron when it was dumped in water but I don't know "q" for iron when it was dumped in glycerol, unless, "q" for iron when it was dumped in water is the same as if when it was dumped in glycerol but how can that be if glycerol and water have different specific heat capacities?
So my problem is that I don't know how to get "q" for iron when it was dumped in glycerol because I don't have the final temperature (Tf)...am I making any sense or am I just completely lost with this whole thing...thanks so much for your help!

-K

 

[Lok]

q of iron is depending on C_p of iron and the temp it drops. q in the two cases is not the same. Find them a different name for cleanliness sake :P (e.g. Qw Qgly)

Find out C_p of iron and then you will get q of iron dropped in glycerol. Which is q of glycerol also. something like 2 equations with 2 unknowns where q is rather useless :P.

 

[rcaste]

Hey, I am having problem with this question as well. The Cp of Glycerin is given in kelvin (J/K*mol), so I changed the temperature to kelvin as well on the glycerin side, but had to keep it in Celsius for the iron side of the equation (because of the fact that the iron side is in J/g*c). But I end up having two unknowns (Final temp of iron, and final temp of glycerin).

 

[Lok]

Hey, I am having problem with this question as well. The Cp of Glycerin is given in kelvin (J/K*mol), so I changed the temperature to kelvin as well on the glycerin side, but had to keep it in Celsius for the iron side of the equation (because of the fact that the iron side is in J/g*c). But I end up having two unknowns (Final temp of iron, and final temp of glycerin).

The final temp is one and the same for both. :P

 

[kirsten_2009]

Hi!

I think I finally figured it out. It doesn't matter if the temperature on one side is in kelvin's and the other in celsius because if you think about it they are added and not multiplied
(Tf = Ti + 273). So really what one needs to do is equate both q's. This is what i came up with and it seems to be correct:

1st: If you have the specific heat of iron from a tabulated data table you can use that value, but if you are in an exam and you need to calculate the specific heat of iron you could do it in the following way (assuming that you've memorized that the specific heat of water is 4.18 J/g x Celsius):

q_water = 981 g x 4.18 J/g x Celsius x 12.3 Celsius
q_water = 50437.13 J

And we know that q_water = -q_iron so q_iron = -50437.13 J

and now to find the specific heat of iron we re-arrange the equation to isolate for spht:

q = m x spht x (Tf -Ti) -----> spht = q / m x (Tf-Ti)

Where: "q" is quantity of heat (J/g x C), "m" is mass (grams or moles), "spht" is specific heat capacity (J/g x C), "Tf" is final temperature (Celsius) and "Ti" is initial temperature (Celsius).

spht = -50437.13 J / 1220 g x (-92.1)
spht = -50437.13 J / -112362.0 g x Celsius
spht = 0.449 J/g x Celsius

Now, we are given the amount of glycerol in mL but we are also given the density which we can use to find the grams of glycerol, so...

350 mL glycerol x (1.26 grams glycerol / 1 mL) = 441 grams of glycerol

But, the specific heat of glycerol is given in units of moles, thus it is a molar specific heat and therefore the grams of glycerol must be converted to moles...

441 g glycerol x (1 mol glycerol / 92.0 g glycerol) = 4.79 moles of glycerol

So...now we have "two" unknowns: "q" and "Tf" but we know that q = -q so we can simply equate them and we are left with just one unknown. Don't worry about the units of celsius and kelvin because as I mentioned earlier they are equivalent. 1 Celsius = 1 Kelvin because they are simply being added. So...

q= 1220 g x 0.499 J/g.C x (Tf - 99.8 C) = -4.79 mol x 219 J/K.mol x (Tf - 26.8 C)
= 547.78 J/C x (Tf - 99.8 C) = -1049.01 J/K x (Tf - 26.8)
= 547.78Tf J/C x (-54668.4 C) = -1049.01Tf J/K x 28113.4 C
= 547.78Tf J/C + 1049.01Tf J/K = 28113.4 + 54668.4
= 1596.79Tf = 82781.8
= Tf = 82781.8 / 1596.79
= Tf = 51.8 Celsius

Pretty annoying but it get's the job done...anyone know a simpler way to do it? Thanks! :)

-K.