[smileshaikh]

I calculated them but got negative numbers for each

do I just change them to positive?

you don't need to read on but if you want to see my data table to tell me its

______________________________________…

we did a lab on calorimeter constant

let me give you my data table just for trial 1,2,3

mass (volume) of cold water: t1,t2,t3= 50.0ml
temperature of cold water: t1= 23.8 degrees C, t2= 24.2 degrees C, t3= 25.0 degrees C
mass (volume) of hot water: t1,t2,t3= 50.0ml
temperature of hot water: t1=78.8 degrees C, t2=75.1 degrees C, t3= 77.7 degrees C
final temperature reached: t1=48.3 degrees C, t2= 47.0 degrees C, t3=48.3 degrees C

temperature change: t1= 24.5 degrees C (cold) -30.5 degrees C (hot), t2= 22.8 degrees C (cold), -28.1 degrees C (hot), t3= 23.3 degrees C (cold), -29.4 degrees (hot)

calorimeter constant:t1= -51.2 j/c, t2= -48.6 j/c, t3= -54.7 j/c

mean value of calorimeter constant: -51.5 j/c

standard deviation: 3.06 j/c

My question is, are my calorimeter constants wrong because they are negative numbers, perhaps I calculated wrong ?

And what about my standard deviation, is it correct?

[Unknown008]

Hmm, your methodology is not good. You need to find the expected final temperature, and find the difference when compared to the actual final temperature. This is where the mass will be important.

t1, final expected temperature is 51.3 C
Difference from actual = 3.0 C

So, calorimeter absorbed 3 C. Now, you need to use the specific heat capacity of water, the mass and the difference in temperature to find the energy that 100 mL of water gives off when cooled by 3 C. This is the energy for 3 C, find that for 1 C.

Repeat for t2 and t3, then find the average from here.

In case you didn't know, the formula to be used is



To find the expected final temperature, you use a simple derivation from that formula;



As m and c are equal, this gives

[smileshaikh]

Hmm, your methodology is not good. You need to find the expected final temperature, and find the difference when compared to the actual final temperature. This is where the mass will be important.

t1, final expected temperature is 51.3 C
Difference from actual = 3.0 C

So, calorimeter absorbed 3 C. Now, you need to use the specific heat capacity of water, the mass and the difference in temperature to find the energy that 100 mL of water gives off when cooled by 3 C. This is the energy for 3 C, find that for 1 C.

Repeat for t2 and t3, then find the average from here.

In case you didn't know, the formula to be used is



To find the expected final temperature, you use a simple derivation from that formula;



As m and c are equal, this gives

Thanks unknown, I was able to correct the calculations.

[Unknown008]

You're welcome :)

[Jongar]

I got negative calorimeter constant even though I'm sure my calculations were right
What does it signify?

Im using styrofoam cup as the calorimeter by the way
Would really appreciate your help ;)

[Unknown008]

Being sure doesn't necessarily mean that it's good. You should have made another thread for your problem instead of piggy-backing this one. Anyway, show what you did, you may have overlooked something...

[Jongar]

My apologies, I've thought its better to comment on a thread with a related topic than open a new thread. Anyway back to my topic:

Tc Tw
Exp. 1 19.5○C 38.5○C
Exp. 2 21○C 44○C
Exp. 3 21○C 44.5○C

The equation I used is: mwCs(Tw-Tm) = mcCs(Tm -Tc) + Ccal(Tm-Tc)
Where Tc stands temp. of cold water, Tw for warm and Tm for mixture temperature.
The Values I got were for Exp1: Ccal = -42 J○C-1
Exp2: Ccal = 0
Exp3: Ccal = 18.2 J○C-1
Funny thing is if I take the average Ccal which is negative and use it for further calculations it gives a more accurate result than If I just take the last positive value. When measuring ΔHneutralization; using the negative Ccal constant gives a final result of -54kj/mol which is close to the reference ΔHneutralization (-57kj/mol) while if I use the positive value I get a much higher ΔH

[Unknown008]

My apologies, I've thought its better to comment on a thread with a related topic than open a new thread. Anyways back to my topic:

Tc Tw
Exp. 1 19.5○C 38.5○C
Exp. 2 21○C 44○C
Exp. 3 21○C 44.5○C

The equation I used is: mwCs(Tw-Tm) = mcCs(Tm -Tc) + Ccal(Tm-Tc)
Where Tc stands temp. of cold water, Tw for warm and Tm for mixture temperature.
The Values I got were for Exp1: Ccal = -42 J○C-1
Exp2: Ccal = 0
Exp3: Ccal = 18.2 J○C-1
Funny thing is if I take the average Ccal which is negative and use it for further calculations it gives a more accurate result than If I just take the last positive value. When measuring ΔHneutralization; using the negative Ccal constant gives a final result of -54kj/mol which is close to the reference ΔHneutralization (-57kj/mol) while if I use the positive value I get a much higher ΔH

Can you post the question too with the information you collected.

I need Tm for the mixture, mass of calorimeter (which you didn't use in your formula and is considered in calculations), and the mass of hot and cold water you used to be completely sure of what is being done...

[Jongar]

The question is we should calculate the Ccal of calorimeter which is Styrfoam then used the Ccal constant to calculate the Enthalpy of three different acid-base reactions.



TmExpected
Tc Tw TmActual
Difference
Exp. 1 19.5○C 38.5○C 29○C 29.5○C 0.5○C
Exp. 2 21○C 44○C 32.5○C 32.5○C 0○C
Exp. 3 21○C 44.5○C 32.75○C 32.5○C -0.25○C

Mass used is 100g for both Cold and hot water, and I disregarded the mass of the Styrofoam as the equation didn't I'm using didn't need it not to mention that we weren't supplied the necessary equipment at the lab.

[Unknown008]

Well, it doesn't seem to be an acid-base reaction at all... the temperatures remained practically the same (disregarding experimental errors), meaning that no heat was given off...

If you used pure water, the same temperature differences would have been obtained, and since the final temperature is the same as the expected temperature, the specific heat capacity of the calorimeter must be very high (it requires a lot of energy to get hot, and hence is an insulator, and like any insulator, it won't take much heat from its surroundings.)

[Jongar]

Actually, the results I just linked you were for the mixture of cold and hot water results, the acid and base reaction showed great change in temperature, for example the reaction of HCl with NaOH; both had a temperature of 19C at the beginning, then the final temperature had 26C

About the calorimeter part, yea that's what I thought, Styrofoam is one of the best insulators has between its walls it has spaces for air in which conduction is slowest through gases, but yet if that's the case why would they ask us to find it then use it for further calculations?
Moreover, here's something that puzzled me :S what's the difference between the specific heat capacity and the calorimeter constant. as both are defined as the heat required to increase temperature by 1C

[Unknown008]

They are the same thing, just called in two different ways:

Calorimeter constant - Wikipedia, the free encyclopedia

Your experiments have temperature differences that were too small. Over larger differences in temperature, the calorimeter would've absorbed more heat from the mixture, making it easier to calculate its specific heat capacity.

[Jongar]

Makes sense :) Thanks for your help

[Unknown008]

No problem! :)