State Changes
Heat of Fusion
Heat energy required to convert a solid at its melting
point to a liquid
Value for water: 333 J/g
Heat of Vaporization
Heat energy required to convert a liquid at its boiling
point to a solid
Value for water: 2260 J/g
Vaporization
endothermic process (H positive)
Condensation
exothermic process (H negative)
Melting
endothermic process (H positive)
Freezing
exothermic process (H negative)
Finding the Energy to Change the
State of a Given Mass of a Substance Pause the video and solve the problems
below.
J = (mass)(heat of vaporization or heat of fusion)
Find the heat energy needed to change 50.0 g of
water from liquid to gas.
How much heat energy will be released when 50.0 g
of water freeze (change from liquid to solid)?
Finding the Energy to Change the
State of a Given Mass of a Substance Pause the video and solve the problems
below.
J = (mass)(heat of vaporization or heat of fusion)
Find the heat energy needed to change 50.0 g of
water from liquid to gas.
(2260J/g)(50.0g) = 1.13 x 106 J
How much heat energy will be released when 50.0 g
of water freeze (change from liquid to solid)?
(-333J/g)(50.0g) = -1.67 x 105 J
Graphing Temperature v. Time During a
State Change
Temperature (C)
120
-10
Time (min)
Why do the plateaus occur in this graph even though
heat was steadily added over time?
Cpgas =
1.86 J/gK
Heat of vaporization = 2260J/g
Specific heat of water = 4.184J/gK
heat of fusion = 333 J/g
Specific heat of ice = 2.06 J/gK
Find the energy needed to convert 500.g of
ice at -50.0 C to steam at 200. C.
Pause the video while you solve the problem.
Solve this in 5 steps by finding energy needed to:
1. Heat ice from -50.0 to 0 using specific heat of ice
2.
Melt ice using heat of fusion of ice
3.
Heat water from 0 to 100 using specific heat of water
4.
Boil water using heat of vaporization of water
5.
Heat steam from 100 to 200. using specific heat of steam
Then add all of these q values together to find the total amount of energy needed.
Find the energy needed to convert 500.g of
ice at -50.0C to steam at 200. C.
Pause the video while you solve the problem.
Solve this in 5 steps by finding energy needed to:
1. Heat ice from -50.0 to 0 using specific heat of ice
q = (2.06J/gK)(50.0g)(0 C - -50.0C) = 5.15 x 104J
1. Melt ice using heat of fusion of ice
q = (333J/g)(50.0g) = 1.67 x 105J
1. Heat water from 0 to 100 using specific heat of water
q = (4.184J/gK)(50.0g)(100 C - 0C) = 2.09 x 105J
1. Boil water using heat of vaporization of water
q = (2260J/g)(50.0g)= 1.13 x 106J
1. Heat steam from 100 to 200. using specific heat of steam
q = (1.86J/gK)(50.0g)(200. C – 100.0C) = 9.30 x 104J
Then add all of these q values together to find the total amount of energy needed.
1650kJ
Let’s try something harder!!!
Pause the video while you solve the problem.
Suppose you pour 250. mL of tea (density 1g/mL) at 18.2C into a
glass with 5 ice cubes (each weighing 15g). What quantity of ice will
melt and what mass will remain? (Hint: the tea will cool until it is at
thermal equilibrium with the ice at 0 C.) Assume the specific heat of
tea is 4.2J/gK.