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Solids, Liquids, and Gases
UNIT 8
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Calculate the energy needed to heat a substance from one temperature to another (Specific Heat).
q=mCpΔT
+Homework check
1. 0.46 J/g*C
2. 13794 J
3. 0.14 J/g*C
4. .240 J/g*C
5. 31.94 C
1.Fluidity , Diffuse
2.FALSE
3.IT WILL DIFFUSE TO FILL THE CONTAINER
4.GAS
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Recall,q=mCpΔT
■ q = energy ( in units of joules)
■ m = mass (in units of grams)
■ Cp = specific heat (from reference table!)
■ ΔT = change in temperature (final temp – initial temp)
+TEMPERATURE AND STATE OF WATER
Temp < 0oC Water is a SOLID (ice)
Temp = 0oC SOLID and LIQUID ARE BOTH PRESENT
0oC < Temp < 100oC Water is a LIQUID
Temp = 100oC LIQUID and GAS ARE BOTH PRESENT
Temp > 100oC Water is a GAS (steam)
KNOW THIS!!!!!
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Water has a DIFFERENT specific heat for each state of matter:Solid water (ICE) Cpice = 2.05 J/goCLiquid water Cpliquid = 4.18 J/goCGaseous Water (STEAM) Cpgas = 2.02 J/goC
THIS IS ON YOUR REFERENCE TABLE….SEE PAGE 1These values are used when the temperature of the water
sample is changing
Heat Calculations for Heating and Cooling Curves of Water
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Use 2.05 for
Temp < 0oC
Use 4.18 for
0oC < Temp < 100oC
Use 2.02 for
100oC < Temp
+Ex 1: Phase Change Calculations (Water)
How much energy is required to change 5.00 g of water from 15o to 100o C?
q=mCpΔT
Given: m = 5.00 g H2O, Cp = 4.18 J/goC∆ T = 100 - 15 = 85oC
q = x
x = (5.00 g) (4.18 J/goC) ( 100-15)
x = (5.00 g) (4.18 J/goC) ( 85)
x =1777 J
In this
temperature
range, water is
a LIQUID, thus
we use the
value of 4.18 for
the specific
heat.
+Ex 2: Phase Change Calculations (Water)
■ How much energy is required to change the temperature 23.5 g of water from -15.6 oC to 0 oC ?
q=mCpΔT
q= (23.5 g)(?)[0 - (-15.6)] C
What is the value for
the Specific Heat if
Water’s temperature
is between -15.6 and
0oC?
(Hint: In what state of
matter is water at this
temperature range?)
+Phase Change Calculations (Water)
■ How much energy is required to change 23.5 g of water (ICE!) from -15.6oC to 0oC ?
Given:m = 23.5gCp = 2.05J/goCΔT = [0 - (-15.6 C)]q = x
q=mCpΔTq= (23.5 g) ( 2.05 J/goC) [0 - (-15.6 C)]q=(23.5 ) ( 2.05) (15.6)q = 752 J
In this temperature range...water is a solid
(ICE)...thus we use this specific heat
value.
+You Try It!!
■ How much energy is required to change 23.5 g of water from 100.5oC to 125oC?
+You Try It!!
■ How much energy is required to change 23.5 g of water from 100.5oC to 125oC?
q=mCpΔTq= (23.5 g) (2.02 J/goC) ( 125 - 100 C)q = 1187 J
In this temperature range, water is a gas
(STEAM), thus we use this specific heat value.
+Phase Changes withHeating and Cooling
https://www.youtube.com/watch?v=ndw9XYA4iF0
+Phase Change Calculations -
WATER ONLY!!!
You need to remember the
FREEZING/MELTING
POINT OF WATER (0oC)
and the BOILING POINT
OF WATER (100oC)!!!!!
WHAT IS THE
TEMPERATURE
RANGE FOR
WATER IN ITS
LIQUID STATE?
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+ Calculate the energy needed to melt/freeze or vaporize/condense a substance.
q= mHf and q= mHv
WHAT HAPPENS WHEN A SUBSTANCE CHANGES PHASES.
(Melts/Freezes/Evaporates/Condenses)
• We will only be dealing with water for these.
• You need to know the energy required to
cause a phase change.
• This information is on your reference tables!
Don’t Copy….Just Read This !
CALCULATING THE ENERGY REQUIRED TO MELT OR FREEZE
WATER
q= m Hf
• q = energy (joules)
• m = mass (grams)
• Hf =heat of fusion (334 J/g for water) From Reference Table
We are doing calculations at
this point on the graph
CALCULATING THE ENERGY REQUIRED TO VAPORIZE OR
CONDENSE WATER
q= m Hv
• q - energy (joules)
• m – mass (grams)
• Hv - heat of vaporization (2260 J/g for water)
We are doing
calculations at this point
on the graph
From Reference Table
EXAMPLE
• HOW MUCH ENERGY IS REQUIRED TO MELT 15.0
GRAMS OF WATER?
q = m Hf
= (15) (334)
= 5010 J
EXAMPLE
• HOW MUCH ENERGY IS REQUIRED TO VAPORIZE
25.0 GRAMS OF WATER?
q = m Hv
= (25) (2260)
= 56500 J
EXAMPLE• HOW MUCH ENERGY IS REQUIRED TO FREEZE 255.0
GRAMS OF WATER?
q = m Hf
= (255) ( 334)
= 85170 J
YOU TRY THEM
(1)HOW MUCH ENERGY IS NEEDED TO MELT 75 GRAMS OF
WATER?
(2)HOW MUCH ENERGY IS NEEDED TO VAPORIZE 5.89 GRAMS
OF WATER?
(3)HOW MUCH ENERGY IS NEEDED TO CONDENSE 15.6
GRAMS OF WATER?
+Put it all Together
+CLASSWORK / HOMEWORK:
Water’s Heating and Cooling Curve q Calculations
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