Energy Requirements for changing state:

In ice the water molecules are held together by strong intermolecular forces. The energy required to melt 1 gram of a substance is called the latent heat of fusion

For ice it is 334 J/g

The energy required to change 1 gram of a liquid to its vapor is called the latent heat of vaporization

For steam it is 2259 J/g

It takes more energy to vaporize water than to melt it. This is because in melting you weaken the intermolecular forces.Here about 1/6 of the hydrogen bonds are broken.

In vaporization you totally break them.All the hydrogen bonds are broken

Fusion is when a solid melts to form a liquid Vaporization is when a liquid evaporates to form a gas.

A-B = Solid ice, temperature is increasing.

Particles gain kinetic energy, vibration of particles increases.

Heating and cooling curve for water heated at a constant rates.

Ice

B-C = Solid starts to change state from solid to liquid. Temperature remains constant as energy is used to break inter-molecular bonds.

H2O (s) H2O () energy required 334 J/g

0ºC

C-D = temperature starts to rise once all the solid has melted. Particles gain kinetic energy.

Liquid water

D-E = Liquid starts to vaporize, turning from liquid to gas. The temperature remains constant as energy is used to break inter-molecular forces.

H2O () H2O (g) energy required 2259 J/g

100ºC

E-F = temperature starts to rise once all liquid is vaporized. Gas particles gain kinetic energy.

steam

Calculating Energy Changes

Temperature

Heat Added

(oC)

Hf = 334 J/g Hv= 225 J/g 3. cl = 4.18 J/g • oC 5. cv = 2.1 J/g • oC 1. cs = 2.1 J/g • oC

Heating Diagram

Calculating Energy Changes

Q(gained or lost) = m x L. H.(fusion/vaporization)

Phase change:

Temperature change:

heat = mass specific (Tf - Ti ) heat

Q(gained or lost) = m • c • T

Problem

How much energy is required to heat 25 g of liquid water from 25C to 100C and change it to steam?

Step 1: Calculate the energy needed to heat the water from 25C to 100C

Q = 25g 4.184 J/g C 75 C =

Q = m c T

Step 2: Vaporization: Use the Latent Heat to calculate the energy required to vaporize 25g of water at 100C

.25g 1mol H2O / 18g mol-1 H2O = 1.4 mol H2O

vap H (H2O) = 1.4 mol H2O 40.6kJ/mol = 57 kJ

Q = 25.0 g 2259 J/g =

Total energy change is:

18.0 g x 2.06 J/g•oC x 7oC18.0 g x 334 J/g

18.0 g x 4.184 J/g•oC x 100oC

18.0 g x 2259 J/g

18.0 g x 2.06 J/g•oC x 25oC

= 260 J= 6012 J

= 7531 J

= 40668 J

= 927 J

= 55398 J

Calculating Energy Changes

Calculate the total amount of heat needed to change 1 mole of ice at -7oC to steam at 125oC.

Tphase

T

Tphase

Calculating Energy Changes: Solid to liquid

How much energy is required to melt 8.5 g of ice at 0C? The molar heat of fusion for ice is 6.02 kJmol-1

Step 1: How many moles of ice do we have?

n = m/M n = 8.5g / 18gmol-1 = 0.47 mol H2O

Step 2: Use the equivalence statement to work the energy (6.02 kJ is required for 1 mol H2O)

kJ = 0.47 mol H2O 6.02 kJ / mol H2O

= 2.8kJ

What is specific heat capacity? The amount of energy required to change the temperature of one gram of a substance by 1C .

Another name for specific heat is a calorie (1 calorie = 4.184 Joules)Specific heat capacity of liquid water (H2O (L) ) is 4.18 J g-1C–1.

Water (s) = 2.03 J g-1 C –1 0.5 cal/g to break up ice

Water (g) = 2.0 J g-1 C –1

10 C 11 C

Calculating the energy to increase the temperature of liquid water.

Calculating specific heat using the equation:Q = ms (tf ti) or Q = energy (heat) requiredQ = ms T or s = specific heat capacityHeat (H) = ms (tf ti) m = mass of the sample

T = change in temperature in C

EXAMPLE:How much energy does it take to heat 10g of water from 50 to 100 C ?Specific heat capacity of water = 4.184 J g-1C–1

Q = m s T

Q = (10g) (4.184 J g-1 C -1) (50 C) = 2.1 10 3 J

vap H (H2O) = 1.4 mol 40.6kJ/mol = 57 kJ

Heating Diagram