Heat in Changes of
State
Solids, Liquids, and Solutions
Introduction
•
When substances change state, there is a transfer of
energy.
Introduction
•
When substances change state, there is a transfer of
energy.
•
The evaporation of sweat helps to regulate our body
temperature in warm weather or when we exercise.
Introduction
•
When substances change state, there is a transfer of
energy.
•
The evaporation of sweat helps to regulate our body
temperature in warm weather or when we exercise.
Introduction
•
When substances change state, there is a transfer of
energy.
•
•
The evaporation of sweat helps to regulate our body
temperature in warm weather or when we exercise.
The solidification of a liquid results when there is a
transfer of heat from the liquid to the surroundings.
Introduction
•
When substances change state, there is a transfer of
energy.
•
•
The evaporation of sweat helps to regulate our body
temperature in warm weather or when we exercise.
The solidification of a liquid results when there is a
transfer of heat from the liquid to the surroundings.
Introduction
•
When substances change state, there is a transfer of
energy.
•
•
The evaporation of sweat helps to regulate our body
temperature in warm weather or when we exercise.
The solidification of a liquid results when there is a
transfer of heat from the liquid to the surroundings.
Heats of Fusion and Solidification
•
When we place an ice cube on a table in a warm room
Heats of Fusion and Solidification
•
When we place an ice cube on a table in a warm room
Heats of Fusion and Solidification
•
When we place an ice cube on a table in a warm room, it
begins to melt.
Heats of Fusion and Solidification
•
•
When we place an ice cube on a table in a warm room, it
begins to melt.
The ice cube is the system and the table, room, and the air
in the room are the surroundings.
Heats of Fusion and Solidification
•
•
•
When we place an ice cube on a table in a warm room, it
begins to melt.
The ice cube is the system and the table, room, and the air
in the room are the surroundings.
The system takes in energy from the surroundings to
convert the solid ice to liquid water.
Heats of Fusion and Solidification
•
•
•
When we place an ice cube on a table in a warm room, it
begins to melt.
The ice cube is the system and the table, room, and the air
in the room are the surroundings.
The system takes in energy from the surroundings to
convert the solid ice to liquid water.
Heats of Fusion and Solidification
•
•
•
When we place an ice cube on a table in a warm room, it
begins to melt.
The ice cube is the system and the table, room, and the air
in the room are the surroundings.
The system takes in energy from the surroundings to
convert the solid ice to liquid water.
Heats of Fusion and Solidification
•
•
•
When we place an ice cube on a table in a warm room, it
begins to melt.
The ice cube is the system and the table, room, and the air
in the room are the surroundings.
The system takes in energy from the surroundings to
convert the solid ice to liquid water.
Heats of Fusion and Solidification
•
•
•
•
When we place an ice cube on a table in a warm room, it
begins to melt.
The ice cube is the system and the table, room, and the air
in the room are the surroundings.
The system takes in energy from the surroundings to
convert the solid ice to liquid water.
There is no change in temperature until all of the solid is
converted to liquid.
Heats of Fusion and Solidification
•
•
•
•
•
When we place an ice cube on a table in a warm room, it
begins to melt.
The ice cube is the system and the table, room, and the air
in the room are the surroundings.
The system takes in energy from the surroundings to
convert the solid ice to liquid water.
There is no change in temperature until all of the solid is
converted to liquid.
This is an endothermic process ...
Heats of Fusion and Solidification
•
•
•
•
•
When we place an ice cube on a table in a warm room, it
begins to melt.
The ice cube is the system and the table, room, and the air
in the room are the surroundings.
The system takes in energy from the surroundings to
convert the solid ice to liquid water.
There is no change in temperature until all of the solid is
converted to liquid.
This is an endothermic process ...
•
that is, ∆H is positive.
Heats of Fusion and Solidification
•
When we make ice cream ...
Heats of Fusion and Solidification
•
When we make ice cream ...
Heats of Fusion and Solidification
•
When we make ice cream ...
•
the ice cream mix is the system and the salt/ice mixture
and the ice cream maker are the surroundings
Heats of Fusion and Solidification
•
When we make ice cream ...
•
•
the ice cream mix is the system and the salt/ice mixture
and the ice cream maker are the surroundings
energy is transferred from the system to the surroundings
Heats of Fusion and Solidification
•
When we make ice cream ...
•
•
•
the ice cream mix is the system and the salt/ice mixture
and the ice cream maker are the surroundings
energy is transferred from the system to the surroundings
Once the liquid gets to the freezing point, there is no further
change in temperature until all of the mix has frozen.
Heats of Fusion and Solidification
•
When we make ice cream ...
•
•
•
•
the ice cream mix is the system and the salt/ice mixture
and the ice cream maker are the surroundings
energy is transferred from the system to the surroundings
Once the liquid gets to the freezing point, there is no further
change in temperature until all of the mix has frozen.
This is an exothermic process ...
Heats of Fusion and Solidification
•
When we make ice cream ...
•
•
•
•
the ice cream mix is the system and the salt/ice mixture
and the ice cream maker are the surroundings
energy is transferred from the system to the surroundings
Once the liquid gets to the freezing point, there is no further
change in temperature until all of the mix has frozen.
This is an exothermic process ...
•
that is, ∆H is negative.
Heats of Fusion and Solidification
•
The amount of heat absorbed by one mol of a solid to
become one mol of a liquid is the molar heat of fusion,
∆Hfus.
Heats of Fusion and Solidification
•
•
The amount of heat absorbed by one mol of a solid to
become one mol of a liquid is the molar heat of fusion,
∆Hfus.
The amount of heat given up by one mol of a liquid to
become one mol of a solid is the molar heat of solidification,
∆Hsolid.
Heats of Fusion and Solidification
•
•
•
The amount of heat absorbed by one mol of a solid to
become one mol of a liquid is the molar heat of fusion,
∆Hfus.
The amount of heat given up by one mol of a liquid to
become one mol of a solid is the molar heat of solidification,
∆Hsolid.
For any given substance, ∆Hfus = −∆Hsolid.
Heats of Fusion and Solidification
•
•
•
The amount of heat absorbed by one mol of a solid to
become one mol of a liquid is the molar heat of fusion,
∆Hfus.
The amount of heat given up by one mol of a liquid to
become one mol of a solid is the molar heat of solidification,
∆Hsolid.
For any given substance, ∆Hfus = −∆Hsolid.
Liquid
∆Hfus
∆Hsolid
Solid
Heats of Fusion and Solidification
•
•
•
The amount of heat absorbed by one mol of a solid to
become one mol of a liquid is the molar heat of fusion,
∆Hfus.
The amount of heat given up by one mol of a liquid to
become one mol of a solid is the molar heat of solidification,
∆Hsolid.
For any given substance, ∆Hfus = −∆Hsolid.
•
Liquid
∆Hfus
∆Hsolid
Solid
For example, for H2O, going from
ice to water, ∆Hfus = 6.01 kJ/mol
Heats of Fusion and Solidification
•
•
•
The amount of heat absorbed by one mol of a solid to
become one mol of a liquid is the molar heat of fusion,
∆Hfus.
The amount of heat given up by one mol of a liquid to
become one mol of a solid is the molar heat of solidification,
∆Hsolid.
For any given substance, ∆Hfus = −∆Hsolid.
•
Liquid
∆Hfus
∆Hsolid
Solid
•
For example, for H2O, going from
ice to water, ∆Hfus = 6.01 kJ/mol
For H2O to go from water to ice,
∆Hsolid = −6.01 kJ/mol
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
How many grams of ice at 0°C will melt if 2.25 kJ of heat are
added?
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
How many grams of ice at 0°C will melt if 2.25 kJ of heat are
added?
Known: ∆Hfus = 6.01 kJ/mol
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
How many grams of ice at 0°C will melt if 2.25 kJ of heat are
added?
Known: ∆Hfus = 6.01 kJ/mol
∆H = 2.25 kJ/mol
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
How many grams of ice at 0°C will melt if 2.25 kJ of heat are
added?
Known: ∆Hfus = 6.01 kJ/mol
∆H = 2.25 kJ/mol
Tinitial = Tfinal = 0°C
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
How many grams of ice at 0°C will melt if 2.25 kJ of heat are
added?
Known: ∆Hfus = 6.01 kJ/mol
∆H = 2.25 kJ/mol
Tinitial = Tfinal = 0°C
MH2O = 18.0 g/mol
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
How many grams of ice at 0°C will melt if 2.25 kJ of heat are
added?
Known: ∆Hfus = 6.01 kJ/mol
∆H = 2.25 kJ/mol
Find:
mH2O = ? g
Tinitial = Tfinal = 0°C
MH2O = 18.0 g/mol
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
How many grams of ice at 0°C will melt if 2.25 kJ of heat are
added?
Known: ∆Hfus = 6.01 kJ/mol
∆H = 2.25 kJ/mol
Find:
mH2O = ? g
Calculate:
Tinitial = Tfinal = 0°C
MH2O = 18.0 g/mol
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
How many grams of ice at 0°C will melt if 2.25 kJ of heat are
added?
Known: ∆Hfus = 6.01 kJ/mol
∆H = 2.25 kJ/mol
Find:
mH2O = ? g
Calculate:
2.25 kJ
1
∆H
Tinitial = Tfinal = 0°C
MH2O = 18.0 g/mol
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
How many grams of ice at 0°C will melt if 2.25 kJ of heat are
added?
Known: ∆Hfus = 6.01 kJ/mol
∆H = 2.25 kJ/mol
Find:
mH2O = ? g
Calculate:
2.25 kJ
1
×
1 mol
6.01 kJ
∆Hfus
Tinitial = Tfinal = 0°C
MH2O = 18.0 g/mol
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
How many grams of ice at 0°C will melt if 2.25 kJ of heat are
added?
Known: ∆Hfus = 6.01 kJ/mol
Tinitial = Tfinal = 0°C
∆H = 2.25 kJ/mol
Find:
MH2O = 18.0 g/mol
mH2O = ? g
Calculate:
2.25 kJ
1
×
1 mol
6.01 kJ
×
18.0 g
1 mol
MH2O
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
How many grams of ice at 0°C will melt if 2.25 kJ of heat are
added?
Known: ∆Hfus = 6.01 kJ/mol
Tinitial = Tfinal = 0°C
∆H = 2.25 kJ/mol
Find:
MH2O = 18.0 g/mol
mH2O = ? g
Calculate:
2.25 kJ
1
×
1 mol
6.01 kJ
×
18.0 g
1 mol
= 6.74 g
mH2O
Heats of Fusion and Solidification
Sample Problem 17.4 (page 521)
How many grams of ice at 0°C will melt if 2.25 kJ of heat are
added?
Known: ∆Hfus = 6.01 kJ/mol
Tinitial = Tfinal = 0°C
∆H = 2.25 kJ/mol
Find:
MH2O = 18.0 g/mol
mH2O = ? g
Calculate:
2.25 kJ
1
×
1 mol
6.01 kJ
×
18.0 g
1 mol
= 6.74 g
mH2O
Heats of Vaporization and Condensation
•
The amount of heat absorbed by one mol of a liquid to
become one mol of a gas is the molar heat of vaporization,
∆Hvap.
Heats of Vaporization and Condensation
•
•
The amount of heat absorbed by one mol of a liquid to
become one mol of a gas is the molar heat of vaporization,
∆Hvap.
The amount of heat given up by one mol of a gas to become
one mol of a liquid is the molar heat of condensation,
∆Hcond.
Heats of Vaporization and Condensation
•
•
•
The amount of heat absorbed by one mol of a liquid to
become one mol of a gas is the molar heat of vaporization,
∆Hvap.
The amount of heat given up by one mol of a gas to become
one mol of a liquid is the molar heat of condensation,
∆Hcond.
For any given substance, ∆Hvap = −∆Hcond.
Heats of Vaporization and Condensation
•
•
•
The amount of heat absorbed by one mol of a liquid to
become one mol of a gas is the molar heat of vaporization,
∆Hvap.
The amount of heat given up by one mol of a gas to become
one mol of a liquid is the molar heat of condensation,
∆Hcond.
For any given substance, ∆Hvap = −∆Hcond.
Gas
∆Hvap
∆Hcond
Liquid
Heats of Vaporization and Condensation
•
•
•
The amount of heat absorbed by one mol of a liquid to
become one mol of a gas is the molar heat of vaporization,
∆Hvap.
The amount of heat given up by one mol of a gas to become
one mol of a liquid is the molar heat of condensation,
∆Hcond.
For any given substance, ∆Hvap = −∆Hcond.
•
Gas
∆Hvap
∆Hcond
Liquid
For example, for H2O, going from
water to steam, ∆Hvap = 40.7
kJ/mol
Heats of Vaporization and Condensation
•
•
•
The amount of heat absorbed by one mol of a liquid to
become one mol of a gas is the molar heat of vaporization,
∆Hvap.
The amount of heat given up by one mol of a gas to become
one mol of a liquid is the molar heat of condensation,
∆Hcond.
For any given substance, ∆Hvap = −∆Hcond.
•
Gas
∆Hvap
∆Hcond
Liquid
•
For example, for H2O, going from
water to steam, ∆Hvap = 40.7
kJ/mol
For H2O to go from steam to
water, ∆Hcond = −47.0 kJ/mol
Heats of Vaporization and Condensation
•
If we look at the change in temperature of a substance as
we add heat at a constant rate to a solid until it becomes a
liquid and then a gas, we get the following heating curve.
Heats of Vaporization and Condensation
If we look at the change in temperature of a substance as
we add heat at a constant rate to a solid until it becomes a
liquid and then a gas, we get the following heating curve.
Temperature
•
Heat Supplied
Heats of Vaporization and Condensation
•
If we look at the change in temperature of a substance as
we add heat at a constant rate to a solid until it becomes a
liquid and then a gas, we get the following heating curve.
Temperature
boiling
point
melting
point
Heat Supplied
Heats of Vaporization and Condensation
•
If we look at the change in temperature of a substance as
we add heat at a constant rate to a solid until it becomes a
liquid and then a gas, we get the following heating curve.
Temperature
boiling
point
∆Hfus
melting
point
Heat Supplied
Heats of Vaporization and Condensation
•
If we look at the change in temperature of a substance as
we add heat at a constant rate to a solid until it becomes a
liquid and then a gas, we get the following heating curve.
Temperature
boiling
point
∆Hfus
melting
point
Heat Supplied
Heats of Vaporization and Condensation
•
If we look at the change in temperature of a substance as
we add heat at a constant rate to a solid until it becomes a
liquid and then a gas, we get the following heating curve.
∆Hvap
Temperature
boiling
point
∆Hfus
melting
point
Heat Supplied
Heats of Vaporization and Condensation
•
If we look at the change in temperature of a substance as
we add heat at a constant rate to a solid until it becomes a
liquid and then a gas, we get the following heating curve.
∆Hvap
Temperature
boiling
point
∆Hfus
melting
point
Heat Supplied
Heat of Solution
•
During the formation of a solution, heat is either released or
absorbed.
Heat of Solution
•
During the formation of a solution, heat is either released or
absorbed.
•
This is the molar heat of solution, ∆Hsoln.
Heat of Solution
•
During the formation of a solution, heat is either released or
absorbed.
•
•
This is the molar heat of solution, ∆Hsoln.
This is the basis of heat packs
Heat of Solution
•
During the formation of a solution, heat is either released or
absorbed.
•
•
This is the molar heat of solution, ∆Hsoln.
This is the basis of heat packs
Heat of Solution
•
During the formation of a solution, heat is either released or
absorbed.
•
•
This is the molar heat of solution, ∆Hsoln.
This is the basis of heat packs and cold packs.
Heat of Solution
•
During the formation of a solution, heat is either released or
absorbed.
•
•
This is the molar heat of solution, ∆Hsoln.
This is the basis of heat packs and cold packs.
Heat of Solution
•
During the formation of a solution, heat is either released or
absorbed.
•
•
•
This is the molar heat of solution, ∆Hsoln.
This is the basis of heat packs and cold packs.
The dissolution of calcium chloride in water is exothermic.
•
•
CaCl2(s) → Ca2+(aq) + 2 Cl−(aq)
∆Hsoln = −82.8 kJ/mol
Heat of Solution
•
During the formation of a solution, heat is either released or
absorbed.
•
•
•
This is the molar heat of solution, ∆Hsoln.
This is the basis of heat packs and cold packs.
The dissolution of calcium chloride in water is exothermic.
•
•
•
CaCl2(s) → Ca2+(aq) + 2 Cl−(aq)
∆Hsoln = −82.8 kJ/mol
The dissolution of ammonium nitrate in water is
endothermic.
•
•
NH4NO3(s) → NH4+(aq) + NO3−(aq)
∆Hsoln = +25.7 kJ/mol
Heat of Solution
•
•
Sample Problem:
How much heat is released when 2.50 mol of NaOH is
dissolved in water. ∆Hsoln = −445.1 kJ/mol
Heat of Solution
•
•
•
•
Sample Problem:
How much heat is released when 2.50 mol of NaOH is
dissolved in water. ∆Hsoln = −445.1 kJ/mol
Knowns: n = 2.50 mol
∆Hsoln = −445.1 kJ/mol
Heat of Solution
•
•
•
•
•
Sample Problem:
How much heat is released when 2.50 mol of NaOH is
dissolved in water. ∆Hsoln = −445.1 kJ/mol
Knowns: n = 2.50 mol
∆Hsoln = −445.1 kJ/mol
Find:
∆H = ? kJ
Heat of Solution
•
•
•
•
•
•
Sample Problem:
How much heat is released when 2.50 mol of NaOH is
dissolved in water. ∆Hsoln = −445.1 kJ/mol
Knowns:
n = 2.50 mol
∆Hsoln = −445.1 kJ/mol
Find:
Calculation:
∆H = ? kJ
∆H = n × ∆Hsoln
Heat of Solution
•
•
•
•
•
•
Sample Problem:
How much heat is released when 2.50 mol of NaOH is
dissolved in water. ∆Hsoln = −445.1 kJ/mol
Knowns:
n = 2.50 mol
∆Hsoln = −445.1 kJ/mol
Find:
Calculation:
∆H = ? kJ
∆H = n × ∆Hsoln = (2.50 mol)(−445.1 kJ/mol)
Heat of Solution
•
•
•
•
•
•
Sample Problem:
How much heat is released when 2.50 mol of NaOH is
dissolved in water. ∆Hsoln = −445.1 kJ/mol
Knowns:
n = 2.50 mol
∆Hsoln = −445.1 kJ/mol
Find:
Calculation:
∆H = ? kJ
∆H = n × ∆Hsoln = −1113 kJ
Summary
•
•
•
•
•
The amount of heat absorbed by one mol of a solid to become one
mol of a liquid is the molar heat of fusion, ∆Hfus.
The amount of heat given up by one mol of a liquid to become one
mol of a solid is the molar heat of solidification, ∆Hsolid.
•
For any given substance, ∆Hfus = −∆Hsolid.
The amount of heat absorbed by one mol of a liquid to become one
mol of a gas is the molar heat of vaporization, ∆Hvap.
The amount of heat given up by one mol of a gas to become one
mol of a liquid is the molar heat of condensation, ∆Hcond.
•
For any given substance, ∆Hvap = −∆Hcond.
During the formation of a solution, heat is either released or
absorbed.
•
This is the molar heat of solution, ∆Hsoln.