1. Endo or Exothermic
Sign of q
(+ or -)
Melting of Ice
endo +
Evaporation
condensation
exo -
sublimination
(l)  (s)

2. s  l l  g S  g l  s g  l g  s Melting/fusion endo endo
vaporization
sublimination
endo
freezing
exo
condensation
exo
deposition
exo

3. condensation
GAS
vaporization
LIQUID
FUSION
Freezing
Melting
SOLID

4. The specific heat (c) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.
Heat capacity = the amount of energy required to raise the temperature of an object (by one degree).
Molar heat capacity = heat capacity of 1 mol of a substance.

5. Heat of fusion: The amount of energy released/required
at the solid/liquid phase change.
Heat of vaporization: The amount of energy released/required
for liquid/ gas phase change.

6. Heat is the total amount of energy possessed by the molecules in a piece of matter. This energy is both kinetic energy and potential energy.
Temperature is proportional to the average kinetic energies of the molecules

7. moves spontaneously from matter with higher T to matter with lower T
What happens when two objects of different
temperatures come into contact?
moves spontaneously from matter with higher T to matter with lower T

8. Which of the following has the greatest heat capacity?
100 g water or 1000 g water
Which of the following has the greatest specific heat?
100 g water or 100 g copper

9. Dt = 16.2 – 15.2 = 1 q = mcDt q = (205 g) (4.184 J/g C) (1 C)
How much heat is required to raise the temp of 205 g of water
from 15.2 C to 16.2 C ?
What info do we have?
Dt = 16.2 – 15.2 = 1
m = 205 g
c= J/g C
What are we looking for ? q
What is our equation ?
q = mcDt
q = (205 g) (4.184 J/g C) (1 C)
858 J

10. Dt = 25 – 0 = 25 C q = mcDt q = (25 g) (4.18 J/g C) (25 C)
Calculate the amount of heat released when 25 g of water
At 25 C is cooled to 0 C ?
What info do we have?
Dt = 25 – 0 = 25 C
m = 25 g
c= J/g C
What are we looking for ? q
What is our equation ?
q = mcDt
q = (25 g) (4.18 J/g C) (25 C)
2615 J

11. Heat lost = Heat gained q = q
What mass of 67.5 C iron must be added to 235 g of 5.00 C water
To make the final temp of both come out to be 15 C ?
What do we have?
Iron c = .444 J/g C
m = ? g
Iron initial T = 67.5 C
Iron final T = 15 C
Dt = 67.5 – 15 = 52.5
Water c = J/g C
Water m = 235 g
Water initial T = 5 C
Water final T = 15 C
Dt= 15 – 5 = 10 C
Heat lost = Heat gained
q = q
mcDt= mcDt
(?g) (.444g) (52.5) = (235g)(4.184 J/g C)(10 C)
mass = (235g)(4.184 J/g C)(10)
(.444J/g C)(52.5 C)
421.8 g Fe

12. 40,000J = (195 g) (.897 J/g C) (Tfinal – 3.00 C)
A 195 g aluminum engine part at an initial temperature of 3.00 C
absorbs 40 KJ of heat. What is the final temperature of the part?
What info do we have?
m = 195 g
c= .897 J/g C
T initial= 3.00 C
q = 40 KJ
40,000J
What are we looking for ?
T final
What is our equation ?
q = mcDt
q= mc(Tfinal-Tinit)
40,000J = (195 g) (.897 J/g C) (Tfinal – 3.00 C)
40000J = Tfinal – 3.00 C
(195g)(.897J/g C)
40000J C = Tfinal
(195g)(.897J/g C)

13. When 300 J of energy is lost from 125 g object, the temperature
decreases from 45 C to 40 C. What is the specific heat of this object?
What info do we have?
m = 125 g
T final = 40C
T initial= C
q = 300 J
What are we looking for ?
c specific heat
What is our equation ?
q = mcDt
300J = (125 g) (c) (5 C)
300J = c
(125g)( 5 C)
.48J/g C

14. q = mcDt q = (2.4 g) (.129 J/g C) (22.5C – 37.2 C)
6. The specific heat of lead (Pb) is J/g C. Find the amount of
heat released when 2.4g of lead are cooled from 37.2 C to 22.5 C?
What info do we have?
m = 2.4 g
T final = 22.5C
T initial= 37.2 C
c = .129J/g C
What are we looking for ? q
What is our equation ?
q = mcDt
q = (2.4 g) (.129 J/g C) (22.5C – 37.2 C)
4.6 J

15. q = mcDt q = (165 g) (4.184 J/g C) (47.32 C – 10.5 C) =
7. How many kJ of energy are needed to raise the temperature
of 165 g water from 10.5 C to C?
What info do we have?
m = 165 g
T final = C
T initial= 10.5 C
c = 4.18J/g C
What are we looking for ?
q in units of kJ
What is our equation ?
q = mcDt
q = (165 g) (4.184 J/g C) (47.32 C – 10.5 C) =
25395 J
25395 J x 1 kJ = 25.4 KJ
1000 J

16. 8. How much heat is absorbed by 15 g of ice being melted?
H2O (s) H2O (l) DHf = 6.01 kJ/1 mol
15 g x 1 mol = .833 mol
18 g
.833 mol x 6.01 kJ/mol = 5 kJ

17. 9. How much heat is necessary to change 5.0 g of water at 100 C to
to steam at 100 C?
H2O (l) H2O (g) DHv = 2260 J/g
5g x 2260 J/g
= J

18. Calculate Molar Mass of
Calcium Phosphate
Ammonium Sulfate
CaPO4
(40) + (31) + 4(16) = 135 g/mol
(NH4)2SO4
(18) (16) = 132 g/mol

19. How many moles of Na2CO3 are thee in 10L of 2.0M solution
2.0 mol = X mol
1 liter liter
= 20 moles

20. Find the molarity of a solution containing 59 g HCl
In 500 ml of water.
First convert grams to moles
59 g HCl x 1 mole HCL
36 g
= 1.6 mol
Molarity is moles per liter
1.6 mole x 1000 ml
500 ml liter
= 3.2M

21. What volume (in ml) of 12.o M HCl is needed to contain 3.00 moles HCl
12.0 mol = 3.0 mol
1 liter X liter
X = 12
X = .25 liter
X = 250ml

22. How many moles of gas would be present in a gas
Trapped within a 100 ml vessel at 25 C at a pressure
Of 2.50 atm
PV = nRT
n = PV RT
2.5atm (0.1L)
(0.821)(298K)

23. What volume will 1.27 moles of helium gas occupy at STP
1.27 mol x 22.4 L
mol
28.5 L