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Change s in State

Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

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Within One State warm solid warm liquid warm gas Time -> Temperature -> This is measured by the specific heat (heat capacity): the calories needed to raise the temperature of 1 gram of a substance by 1°C. Rising temperature indicates a change in kinetic energy: molecules are moving faster.

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Page 1: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

Changes in

State

Page 2: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

Warming Curve

warming the solid

melting

boiling

warming the liquid

warming the gas

Time ->

Tem

pera

ture

->Consider what happens to an ice cube which is heated.

Page 3: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

Within One State

warm solid

warm liquid

warm gas

Time ->

Tem

pera

ture

->

This is measured by the specific heat (heat capacity): the calories needed to raise the temperature of 1 gram of a substance by 1°C.

Rising temperature indicates a change in kinetic energy: molecules are moving faster.

Page 4: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

Changing States

melt

boil

Time ->

Tem

pera

ture

->

This is measured by the heat of vaporization (gas/liquid) or heat of fusion (solid/liquid): the calories needed to change the state of 1 gram.

A temperature plateau indicates a change in potential energy: bonds between molecules are changing, and energy can be stored in them.

Page 5: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

SUMMARY

warm so

lidmelt

boil

warm

liqui

d

warm gas

Time ->

Tem

pera

ture

->

sp. heat solid

sp. heat liquid

sp. heat gas

heat of fusion

heat of vaporatization

Page 6: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

Energy

Amount

cal or kcal

Joules or kJ

grams or kg

moles

You will see a variety of units for both energy and amount of material in these problems.

Page 7: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

W happening at each stage of this graph?

Time ->

Tem

pera

ture

-> 1 23

45

When is kinetic energy affected? When is potential energy affected? Answers are in the notes.

What states are present at each stage?

What property describes each stage?

Page 8: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

Calculations with Specific Heat

1. How many calories are needed to warm up 3.45 g of water from 18.5°C to 60.0°C?

The specific heat of water is 1 cal/g°C.Heat = (1 cal/g°C)(3.45 g)(41.5°C) = 143 cal

2. How many calories are removed (released) when 1.29 kg of H2O are cooled from 98.5°C to 20.0°C?Heat = (1 cal/g°C)(1,290 g)(78.5°C) = 101,000 calTechnically, this is negative because heat is removed. We will not be using this convention, leaving all energy without signs and using works to decribe addition/loss of energy.

Page 9: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

More Calculations with Specific Heat

3. What is the final temperature when 85.0 J of heat energy are added to 23.5 g of iron (specific heat 0.125 J/g°C) at 19°C?

(0.125 J/g°C)* (23.5 g) * (T) = 85.0 JT = 28.9 °C

Heat is being added, so Tfinal = 19°C + 28.9°C Tfinal = 47.9 °C

Note: If the problem stated that heat was being removed the iron, then T would still be 28.9 °C but the final temperature would be -9.9 °C. Do you see why?

Page 10: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

Heats of Vaporization & Fusion1. How much heat must be released to freeze 25.0 g of

water? The heat of fusion of water is 1.14 kcal/mole.

(1.14 kcal/1 mole) x [25.0 g x (1 mole/18.0 g)]= 1.77 kcal

2. How much heat is needed to boil 125 g of alcohol? The heat of vaporization of alcohol is 102.4 J/g.

(102.4J/1 g) x 125 g = 12,800 J

1.14 kcal = x kcal alternative proportional solution 18.0 g 25.0 g x = 1.77 kcal

102.4 J = x J alternative proportional solution 1 g 125 g x = 12,800 J

Page 11: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

How much heat energy must be removed from 35.0 g of water at 80.°C to turn it into ice at -25 °C?

Heat of fusion H2O = 1.14 kcal/molSp. Heat of Ice = 0.50 cal/g°C

FIRST: Sketch a heating or cooling curve and mark the plateaus and your starting/ending points.

This tells you there are 3 stages to consider: (a) cooling the liquid, (b) freezing, (c) cooling the solid.

Time ->

Tem

pera

ture

->

starts at 80 °C; liquid

ends at -25 °C; solid

100 °C

0 °C

Page 12: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

How much heat energy must be removed from 35.0 g of water at 80.°C to turn it into ice at -25 °C?

Heat of fusion H2O = 1.14 kcal/molSp. Heat of Ice = 0.50 cal/g°C

Second: Find the properties that control each stage of the problem.(a) cooling the liquid = specific heat of liquid: 1 cal/g°C(b) freezing = heat of fusion: 1.14 kcal/mol(c) cooling the solid = specific heat of solid: 0.50 cal/g°C

The units of each property show you how to do the math!

Page 13: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

How much heat energy must be removed from 35.0 g of water at 80.°C to turn it into ice at -25 °C?

1. Cool the water to its freezing point. (1 cal/g°C)(35.0 g)(80.-0.0°C) = 2800 cal

Heat of fusion H2O = 1.14 kcal/molSp. Heat of Ice = 0.50 cal/g°C

2. Freeze the water into ice. (1.14 kcal/1 mole) (35.0 g)(1 mole/18.0 g) = 2.22 kcal3. Cool the ice to its final temperature. (0.50 cal/g°C)(35.0 g)(0-[-25°C]) = 440 cal

Last, add all the heats together. Make sure units agree!2800 cal + 2220 cal + 440 cal = 5460 cal

Page 14: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

crystalline solid

highly ordered

minimum entropy

liquid

some order

some entropy

gas

very random

maximum entropy

Page 15: Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an

can affect

is seen as a

has units of

can affect

is seen as a

includes

is characterized

by

has units of

includes

ischaracterized

by

has units of

is characterized by

includes

ischaracterized

by

includes

is characterized by

Heating & Cooling

kinetic energy

change in temperature

specific heat

cal/g°C

potential energy

change of state

freezing

heat of fusion

cal/g ORkcal/mole

boiling

heat of vaporization

condensingmelting

cal/g°CJ/g°C

cal/g, cal/molekcal/g, kcal/mole

J/g, J/mole