Changes in State

Warming Curve

warm solid

melting

boiling

warm liquidwarm gas

Time ->

Tem

pera

ture

->

Within One State

warm solid

warm liquid

warm gas

Time ->

Te

mp

era

ture

->

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

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.

Calculations with Sp. 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

1. How many calories are released when 1.29 kg of water are cooled from 98.5°C to 20.0°C?

Heat = (1 cal/g°C)(1,290 g)(78.5°C) = -101,000 cal

This is negative because heat is removed.

More Calculations with Sp. Heat

3. What is the final temperature when 20.0 cal of heat energy are added to 0.235 g of iron (specific heat 0.030 cal/g°C) at 19°C?

Heat = (sp. heat)(mass)(T)

20.0 cal = (0.0300 cal/g°C)(T);

667 °C = T

Heat is being added, so Tfinal = 19°C + 667 °C

Tfinal = 686 °C

Changing States

melt

boil

Time ->

Te

mp

era

ture

->

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

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.

Calculations with Heats of Vap/Fusion

1. How much heat is needed to freeze 25.0 g of water? The heat of fusion of water is 1.14 kcal/mole.

25.0 g x (1 mole/18.0 g) x (1.14 kcal/1 mole)

= 1.77 kcal

2. How much heat is needed to boil 125 g of alcohol? The heat of vaporization of alcohol is 24.5 cal/g.

125.0 g x x (24.5 cal/1 g) = 3060 cal

Can you identify what’s happening at each stage of this graph?

Time ->

Te

mp

era

ture

->

1 2

34

5

When is kinetic energy affected? When is potential energy affected?

Can you identify what states are present at each stage?

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

First, cool the water to its freezing point.Heat = (1 cal/g°C)(35.0 g)(80.0-0.0°C) = - 2800 cal

Heat of fusion H2O = 1.14 kcal/mol

Sp. Heat of Ice = 0.50 cal/g°C

Second, freeze the water into ice.35.0 g (1 mole/18.0 g)(1.14 kcal/1 mole) = 2.22 kcal

Third, cool the ice to its final temperature.Heat = (0.50 cal/g°C)(35.0 g)(-25°C) = - 438 cal

Last, add all the heats together. Make sure units agree!2800 cal + 2,220 cal + 438 cal = -5460 cal

crystalline solid

highly ordered

minimum entropy

liquid

some order

some entropy

gas

very random

maximum entropy