image
Physics Forums Logo
image
image
* Register * Upgrade Blogs Library Staff Rules Mark Forums Read
image
image   image
image

Go Back   Physics Forums > Other Sciences > Chemistry


Reply

image Lab Experiment:Calculating Molar Enthalpy of Fusion of Water Share It Thread Tools Search this Thread image
Old Dec5-04, 02:50 PM                  #1
nobb

nobb is Offline:
Posts: 34
Lab Experiment:Calculating Molar Enthalpy of Fusion of Water

Hey.
I am doing a lab experiment in school to find the molar enthalpy of fusion of water. Ice and water are mixed in a calorimeter and the temperatures are recorded. Here are my results. Mass of ice cube: 6.61g. Mass of water: 100g. Initial ice temperature: 0 degrees celsius. Initial water temperature: 21 degrees celsius. Final Temperature of ice and water: 16 degrees celsius.

To calculate the molar enthaply I used mcT(ice)+nH = mcT(water)
now I substitute in my values. 6.61(4.19)(16)+(6.61/18.02)H=100(4.19)(5)
My question is about the change in temperature value for water(T). Should it be 5, or -5? Would it be 5 since heat is being lost? I've calculated using both values and it seems like using -5 will get an answer that is closer to the accepted value of molar enthalpy of fusion of water. Can someone please tell me the correct way to do this? Thanks.
  Reply With Quote
Old Dec5-04, 04:59 PM                  #2
chem_tr
 
chem_tr's Avatar

chem_tr is Offline:
Posts: 613
Recognitions:
PF Contributor PF Contributor
Science Advisor Science Advisor
In your binary ice-water mixture, ice is the one that receives energy, and water is the one that gives energy. So energy is lost from water, indicating a minus sign before. The minus has a physical meaning, not a mathematical one.
  Reply With Quote
Old Dec5-04, 05:30 PM                  #3
nobb

nobb is Offline:
Posts: 34
And that means I dont use it?
  Reply With Quote
Old Dec5-04, 06:10 PM                  #4
gerald

gerald is Offline:
Posts: 13
nobb, you use the negative sign in your calculation because it indicates a physical temperature decrease.

if you were to use the absolute value all of the time it would be hard to determine whether there was a loss of energy or gain.

also why did you choose to use that formula. I received a value of -6906 j.

finding the change in enthalpy of water(q=mct)then dividing it by the # of moles of ice (6.61g/18.02g/mol) gives me a value of -5691 J. This value from your data is closer to the molar enthalpy of fusion of ice.
  Reply With Quote
Old Dec5-04, 07:11 PM       Last edited by Gokul43201; Dec5-04 at 07:16 PM..            #5
Gokul43201
 
Gokul43201's Avatar

Gokul43201 is Offline:
Posts: 10,319
Recognitions:
PF Contributor PF Contributor
Retired Staff Retired Staff
Originally Posted by nobb
Hey.
I am doing a lab experiment in school to find the molar enthalpy of fusion of water. Ice and water are mixed in a calorimeter and the temperatures are recorded. Here are my results. Mass of ice cube: 6.61g. Mass of water: 100g. Initial ice temperature: 0 degrees celsius. Initial water temperature: 21 degrees celsius. Final Temperature of ice and water: 16 degrees celsius.

To calculate the molar enthaply I used mcT(ice)+nH = mcT(water)
now I substitute in my values. 6.61(4.19)(16)+(6.61/18.02)H=100(4.19)(5)
My question is about the change in temperature value for water(T). Should it be 5, or -5? Would it be 5 since heat is being lost? I've calculated using both values and it seems like using -5 will get an answer that is closer to the accepted value of molar enthalpy of fusion of water. Can someone please tell me the correct way to do this? Thanks.
Nobb. You must use change in T = absolute value of change = +5. Here's the reason :

LaTeX Code: Q_f = Q _i

LaTeX Code: Q_f(water) + Q_f(water from ice) = Q _i(water) + Q_i(ice)

LaTeX Code: Set~Q(ice~at~0C) = 0

Now :
LaTeX Code: Q_f(water) = m(water)*C*T_f + n(water)H

LaTeX Code: Q_f(water from ice) = m(water~from~ice)*C*T_f + n(water~from~ice)H

LaTeX Code: Q_i(water) = m(water)*C*T_i + n(water)H

LaTeX Code: Q_i(ice) = 0

Plugging in these values into the second equation gives:

LaTeX Code:  m(water)*C*T_f + n(water)H + m(water~from~ice)*C*T_f + n(water~from~ice)H =  m(water)*C*T_i + n(water)H + 0

Collecting terms, this becomes :

LaTeX Code:  m(water)*C*(T_f-T_i)  + m(water~from~ice)*C*T_f + n(water~from~ice)H = 0
Which, upon rearranging, becomes :

LaTeX Code:  m(water)*C*(T_i-T_f)  = m(water~from~ice)*C*T_f + n(water~from~ice)H

Look at the LHS and you will see that it contains T(i) - T(f) = 21 - 16 = +5, not -5.

A shorter, but less rigorous explanation of why you use +5 is in looking at the the conservation statement that says "heat lost = heat gained". Now if one side of this equation is negative, and the other side is positive, the two sides can never be equal to each other. So, stated this way, the equation only talks about absolute values of heat lost or gained.

You will end up with a number like 4.5 KJ/mol, which appears to be lower than the expected 6 KJ/mol. This is because of heat leaking in from the surroundings, raising the final temperature. A more carefully controlled experiment would have resulted in a lower final temperature.
  Reply With Quote
Old Dec5-04, 07:15 PM                  #6
Gokul43201
 
Gokul43201's Avatar

Gokul43201 is Offline:
Posts: 10,319
Recognitions:
PF Contributor PF Contributor
Retired Staff Retired Staff
Originally Posted by gerald
nobb, you use the negative sign in your calculation because it indicates a physical temperature decrease.

if you were to use the absolute value all of the time it would be hard to determine whether there was a loss of energy or gain.
No gerald, that's not correct.

also why did you choose to use that formula. I received a value of -6906 j.

finding the change in enthalpy of water(q=mct)then dividing it by the # of moles of ice (6.61g/18.02g/mol) gives me a value of -5691 J. This value from your data is closer to the molar enthalpy of fusion of ice.
gerald, this is not the correct way to find the enthalpy of fusion. Nodd has used the correct approach. The number you get may be closer to the expected number, but that is not because your approach is correct.
  Reply With Quote
Old Dec5-04, 07:29 PM       Last edited by gerald; Dec5-04 at 07:36 PM..            #7
gerald

gerald is Offline:
Posts: 13
hmm, so we always use an absolute value for temperature readings ? do we then determine a negative sign for molar enthalpy by looking at whether energy was lost or gained.

what was wrong with my approach?
  Reply With Quote
Old Dec5-04, 07:38 PM                  #8
gerald

gerald is Offline:
Posts: 13
if we always use an absolute value why is it denoted as delta t which is = t(final) - t(initial)
  Reply With Quote
Old Dec5-04, 09:14 PM                  #9
gerald

gerald is Offline:
Posts: 13
alright, i think i understand what was wrong with my approach. the energy involved in decreasing the temperature of water isn't necessarily the amount of energy involved in the fusion of ice right. am i on the right track?

i'm going to read over my textbook more thoroughly again and do some more problems.
  Reply With Quote
image image
Reply
Thread Tools


Similar Threads for: Lab Experiment:Calculating Molar Enthalpy of Fusion of Water
Thread Thread Starter Forum Replies Last Post
Hess Law: Molar Enthalpy Change for Decomposition Tomtom Other Sciences 2 Apr12-07 06:00 PM
Molar Enthalpy Change cookiefairy Other Sciences 1 Mar28-07 01:50 PM
calculating enthalpy change L²Cc Other Sciences 2 Jan28-07 11:23 AM
molar enthalpy of reactions... gstullo Chemistry 2 Mar10-05 06:29 PM
Relating Molar Enthalpy of Oil to Something: Need Lab Ideas nobb Chemistry 4 Jan17-05 04:09 PM

Powered by vBulletin Copyright ©2000 - 2010, Jelsoft Enterprises Ltd. © 2009 Physics Forums
Sciam | physorgPhysorg.com Science News Partner
image
image   image