Solids, Liquids, and Gasesarmorrissey.weebly.com/uploads/1/2/1/5/121599261/... · Solids, Liquids, and Gases UNIT 8 + Calculate the energy needed to heat a substance from one temperature

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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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