A blog brought to you by: Grace Kilpatrick
Hi everyone! Today, February 13, 2014, we had yet another wonderful day of chemistry class. Here's what went down:
1) Mrs. Friedmann began class by wrapping up yesterday's notes titled "Heating Curves". These notes can be found in the Unit 8 Notes folder (titled 2.12_Notes_on_Energy_and_Heating_Curves.pdf) or at the above link . These are very important notes so be sure to check they are in your journal!
2) Mrs. Friedmann went over last night's homework titled "Phase Changes". These notes can be found in the Unit 8 Handouts folder (titled 2.12_Heating_Curve_and_Phase_Changes_Ws.pdf) or at the above link. Part C, Phase Change Calculations, proved to be a bit of a challenge. This is how to solve them:
-My keyboard lacks a delta key, so this will be indicated in equations by a "*" symbol.
29.
Step 1: Heat required to raise temperature -15.5 C to 0 C
q= m x c x *t
q=32.9 g x 2.03 J/g C x 15.5 C= 1040 J
Step 2: Heat required to convert 0 C ice to 0 C liquid
q= n x *Hfus
q=1.83 moles x 6.0 KJ/mole=11000J
TO FIND # of MOLES: USE MOLAR MASS CONVERSION
Step 3: Heat required to raise temperature 0 C to 32.9 C
q= m x c x *t
q=32.9 g x 4.18 J/g C x 32.9 C=4520 J
Sum of Steps 1-3: 1040 J + 11000 J + 4520 J= 16560 J
30. This problem is essentially a continuation of question 29,. Since we have already found the amount of heat required to change 32.9 g of water at -15.5 C to liquid water at 0 C, we now only need to find the amount of heat required to change the liquid water at 0 C to gaseous water at 125 C.
Step 4: Heat required to raise temperature 0 C to 100 C
q=m x c x *t
q= 32.9 g x 4.18 J/g C x 100 C = 13800 J
Step 5: Heat required to convert 100 C water to 100 C gas
q= n x Hvap
q=1.83 moles x 40.7 KJ/mole =74500 J
Step 6: Heat required to convert 100 C gas to 125 C gas
q= m x c x *t
q= 32.9 g x 2.01 J/g C x (125 C-100 C)= 1700 J
Sum of Steps 1, 2, 4, 5, 6: 1040 J + 1100 J + 13800 J + 74500 J + 1700 J = 102.04 KJ
3) We spent the rest of the class period working on our homework which includes:
-8.2 Webassign due at 11:59 TONIGHT!
-Specific Heat of a Metal Lab due TOMORROW!
-Heating Curve Calculations Worksheets (x3) due TOMORROW! (see link)
-Heating Curve Calculations Practice worksheet due TOMORROW!- This is not available in the handouts folder currently, so you will have to pick it up in class!
4) On behalf of the whole class, we wish Jordan a quick and speedy recovery. We miss you and we hope you get better soon!
If you have been absent or are simply confused, here is a video that might assist you in understanding this unit:
The next blogger will be Kevin Mihelic.
2) Mrs. Friedmann went over last night's homework titled "Phase Changes". These notes can be found in the Unit 8 Handouts folder (titled 2.12_Heating_Curve_and_Phase_Changes_Ws.pdf) or at the above link. Part C, Phase Change Calculations, proved to be a bit of a challenge. This is how to solve them:
-My keyboard lacks a delta key, so this will be indicated in equations by a "*" symbol.
29.
Step 1: Heat required to raise temperature -15.5 C to 0 C
q= m x c x *t
q=32.9 g x 2.03 J/g C x 15.5 C= 1040 J
Step 2: Heat required to convert 0 C ice to 0 C liquid
q= n x *Hfus
q=1.83 moles x 6.0 KJ/mole=11000J
TO FIND # of MOLES: USE MOLAR MASS CONVERSION
Step 3: Heat required to raise temperature 0 C to 32.9 C
q= m x c x *t
q=32.9 g x 4.18 J/g C x 32.9 C=4520 J
Sum of Steps 1-3: 1040 J + 11000 J + 4520 J= 16560 J
30. This problem is essentially a continuation of question 29,. Since we have already found the amount of heat required to change 32.9 g of water at -15.5 C to liquid water at 0 C, we now only need to find the amount of heat required to change the liquid water at 0 C to gaseous water at 125 C.
Step 4: Heat required to raise temperature 0 C to 100 C
q=m x c x *t
q= 32.9 g x 4.18 J/g C x 100 C = 13800 J
Step 5: Heat required to convert 100 C water to 100 C gas
q= n x Hvap
q=1.83 moles x 40.7 KJ/mole =74500 J
Step 6: Heat required to convert 100 C gas to 125 C gas
q= m x c x *t
q= 32.9 g x 2.01 J/g C x (125 C-100 C)= 1700 J
Sum of Steps 1, 2, 4, 5, 6: 1040 J + 1100 J + 13800 J + 74500 J + 1700 J = 102.04 KJ
3) We spent the rest of the class period working on our homework which includes:
-8.2 Webassign due at 11:59 TONIGHT!
-Specific Heat of a Metal Lab due TOMORROW!
-Heating Curve Calculations Worksheets (x3) due TOMORROW! (see link)
-Heating Curve Calculations Practice worksheet due TOMORROW!- This is not available in the handouts folder currently, so you will have to pick it up in class!
4) On behalf of the whole class, we wish Jordan a quick and speedy recovery. We miss you and we hope you get better soon!
If you have been absent or are simply confused, here is a video that might assist you in understanding this unit:
The next blogger will be Kevin Mihelic.
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