Changes in State. Warming Curve warming the solid melting boiling warming the liquid warming the gas...

Changes in

State

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.

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.

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.

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

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.

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?

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.

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?

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

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

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!

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

crystalline solid

highly ordered

minimum entropy

liquid

some order

some entropy

gas

very random

maximum entropy

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