Cooling Curve Calculations EQ Why is an ideal heating curve not a straight line graph Ideal HeatingCooling Curve 1 Heating a solid 2 Change from a solid to liquid 3 Heating a liquid ID: 529975
Download Presentation The PPT/PDF document "Heating" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
Slide1
Heating / Cooling Curve Calculations
EQ: Why is an ideal heating curve not a straight line graph?Slide2
Ideal Heating/Cooling Curve
1) Heating a solid
2) Change from a solid to liquid
3) Heating a liquid
4) Change from a liquid to a gas5) Heating a gas
solid present
solid & liquid
liquid
liquid & gas
gasSlide3
On the back of you notes write:Slide4
1) Diagonal lines: KE changes
Flat lines: Potential Energy changes
∆H
vap
> ∆Hfus because it requires more energySlide5
y= mx + b
q= mc
∆T
The slope of the diagonal line is determined by the specific heatSlide6
Example
How much heat is needed to convert 50.0 grams of water from a temperature of 70.0 degrees to 110.0 degree steam?
C ice = 2.03 J/gºC
C
water = 4.184 J/gºCC steam = 2.01 J/gºCHfus = 6.01 kJ/molHvap = 40.7 kJ/molSlide7
1st: sketch part of the graph you need
0.0-
100.0-
70.0-
110.0-
1
2
3
Heating
: Draw graph going up
∆T= 30.0
C
∆T= 10.0
CSlide8
1) Diagonal (q = m c
D
T)
q = 50.0 g x 4.184 J/g
oC x (30.0oC )
q = 6280 J2nd: Perform calculation for each line segment
= 6.28 kJ
1000 J
1 KJ
xSlide9
1st: sketch part of the graph you need
0.0-
100.0-
70.0-
110.0-
1
2
3
∆T= 30.0
C
∆T= 10.0
CSlide10
50.0 g H
2
O
18.02 g H
2
O
1 mol H2O
1 mol H2O
40.7 kJ= 114 kJ
2) Flat line (
H
vap
= 40.7 kJ/mol)
x
xSlide11
1st: sketch part of the graph you need
0.0-
100.0-
70.0-
110.0-
1
2
3
∆T= 30.0
C
∆T= 10.0
CSlide12
q = (50.0 g)(2.01 J/g
o
C)(10.0
oC)q = 1010 J
= 1.01 kJ
3) Diagonal (q= mc∆T)
1000 J
1 KJ
xSlide13
3
rd
: Add them up
+
+
6.28 kJ
114 kJ
1.01 kJ
= 121 kJSlide14
Summary
Use
q = m x c x
D
T (Diagonal lines)
Use
H(Flat lines)Slide15
When dealing with a cooling curve…
draw the curve backwards
∆
H solidification & condensation are negative Slide16
Notebook Problems
Calculate the amount of heat needed to convert 10.0 grams of ice from a temperature of -23.0
oC to water at 27.0oC.
Calculate the amount of heat released when 50.0 grams of steam at a temperature of 123.0oC cools into water at 77.0oC.Slide17
Answers
1) 4.94 kJ
2) 120. kJ