The material conductor or insulator Area Ex a bigger window vs a smaller one Thickness Ex heat escapes a thinner styrofoam cup than a thicker one Heat and State of Matter Does the absorption or release of heat always cause a temp change ID: 649853
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Slide1
Heat and Phase ChangeSlide2
What affects heat transfer?
The material: conductor or insulator?
Area. Ex: a bigger window vs. a smaller one.
Thickness. Ex: heat escapes a thinner
styrofoam
cup than a thicker one.Slide3
Heat and State of Matter
Does the absorption or release of heat always cause a temp change?
NO
Ex:
para
-dichlorobenzene (chemical in mothballs) has a melting point of 54
o
C.
If you were to heat this to 80
o
C
, then insert it into room temperature water, you’d see the temp steadily decrease until you hit 54
o
C.
At this point it is becoming a solid.Slide4
It will stay at 54
o
C
until the last of the liquid becomes a solid. Then it will decrease again.
Was heat being transferred during that time?
YESSlide5Slide6
Phase transitions occur at a constant temperature.
Heat transfers with no change in temp are referred to as
latent heat.Slide7
For melting and freezing:
Q= m
Δ
H
Fusion
For vaporization and condensation:
Q= m
Δ
H
Vaporization
Δ
H
Fusion
= specific heat of fusion per gram
Δ
H
Vaporization
= specific heat of vaporization per gramSlide8
Solid Water:
C
p
=2.00 J/
g°C
Liquid Water:
C
p
= 4.18
J/
g°C
Gaseous Water:
C
p
= 2.01
J/
g°CSlide9
Examples
Elise places 48.2 grams of ice in her beverage. What quantity of energy would be absorbed by the ice (and released by the beverage) during the melting process? The heat of fusion of water is 333 J/g.Slide10
What is the minimum amount of liquid water at 26.5 degrees that would be required to completely melt 50.0 grams of ice? The specific heat capacity of liquid water is 4.18
J/
g°C
and the specific heat of fusion of ice is 333 J/g.Slide11
Given info about ice:
m = 50.0 g
Δ
H
fusion
= 333
J/g
Given info about water:
C = 4.18
J/
g°C
T
initial
= 26.5°C
T
final
= 0.0°C
Δ
T = -26.5°C (
T
final
-
T
initial
)Slide12
The energy gained by the ice is equal to the energy lost from the water.
Qice
= -
Qliquid
water
Qice
=
m
Δ
H
fusion
= (50.0 g
)(
333 J/g)
Qice
= 16650 JSlide13
16650 J = -
Q
liquid
water
16650 J = -
m
liquid
water
C
liquid
water
Δ
T
liquid
water
16650 J = -
m
liquid
water
(
4.18
J/
g°C
)(-
26.5°C)
16650 J = -
m
liquid
water
(-
110.77
J/g)
m
liquid
water
= 150.311 gSlide14
A 100 g sample of water’s temperature is raised from - 20
o
C to 120
o
C. What is the total heat added to the sample
over this period?Slide15