1. Quiz, Sept 24: physical property
Identify the following as physical or chemical property:
porosity
boiling point
When acid spills on your skin, what is the first aid procedure?
Convert cal to Joules.
Which state of matter has no definite shape, but has a definite volume?
physical property
Wash it off immediately. Then notify the teacher.
10 cal (4.184 J/cal) = Joule
liquid.
-1 = = 16

2. Homework 6 Help

3. How are Joules and cal related?
5.0 Calories  Joules
How are Joules and cal related?
1000 cal
Joules
5.0 Cal
4.184 1
Cal
1 cal.
= 20,900 Joules

4. Convert from 0.06 Calorie  Joule
= 251
Joules
Conversion factor!

5. 200 calories  joules 200 calories 4.184 Joules = 800 Joules 1
Always start
with known

6. 6.8 joules  Calories e = 197 joules 0.00162 Cal or 1.62 x 10-3 Cal
4.184
joules
1000 calories
Cal or
1.62 x 10-3 Cal
Always start
with known
e = 197 joules

7. For “story problems” like 11-16 Underline knowns: Then plug them into the equation.
q = heat
m = mass
T = change in temperature = Tf - Ti
Cp = specific heat; for liquid water it is always J/goC (1 cal/goC)
q = m (T2-T1) Cp

8. q = m  T Cp T2 mass of WATER T1 q = m  T Cp Cp = 4.184 J/goC
T = T2 – T1
q = m  T Cp

9. 11. How much heat is required to raise the temperature of 854 g H2O from 23.5oC to 85.0oC?
q = m  T Cp
T2= 85.0 oC
T1=23.5 o C
q = m  T Cp
T = T – T1
85.0oC
23.5oC
q = m  T Cp
(854 g)
(61.5oC)
4.184J/goC

10. 12. Phosphorous trichloride, PCl3, is a compound used…
12. Phosphorous trichloride, PCl3, is a compound used….. How much heat is needed to raise the temperature of 96.7g PCl3 from 31.7oC to 69.20C? The specific heat of PCl3 is J/g oC
q = m  T Cp
 T= (69.2 – 31.7)oC

11. q = m  T Cp (96.7g) ((69.2 – 31.7)oC) (0.874 J/g oC)
12. Phosphorous trichloride, PCl3, is a compound used….. How much heat is needed to raise the temperature of 96.7g PCl3 from 31.7oC to 69.20C? The specific heat of PCl3 is J/g oC
(96.7g)
q = m  T Cp
((69.2 – 31.7)oC)
(0.874 J/g oC)
96.7g x 37.5oC x J/goC = 3170 J

12. Begin by underlining important information. q = m T Cp = -(m T Cp)
Example (Two variables and two unknowns) Suppose a piece of iron with a mass of 21.5 g at a temperature of 100oC is dropped into an insulated container of water. The mass of the water is 132 g and its temperature before adding the iron is 20oC. What is the final temperature of the system?
Begin by underlining important information.
q = m T Cp = (m T Cp)
q = m (Tf – Ti) Cp = m (Tf – Ti) Cp

13. q = m (Tf – Ti) Cp = -m (Tf – Ti) Cp
Example (Two variables and two unknowns) Suppose a piece of iron with a mass of 21.5 g at a temperature of 100oC is dropped into an insulated container of water. The mass of the water is 132 g and its temperature before adding the iron is 20oC. What is the final temperature of the system?
q = m (Tf – Ti) Cp = m (Tf – Ti) Cp
miron = 21.5 g
Ti-iron = 100oC
Tf = same as Tf for water
Cp-iron = look up in book
0.44 J/goC
mwater = 132 g
Ti-water = 20oC
Tf = same as Tf for iron
Cp-water = J/goC
Plug in numbers
21.5g( Tf – 100oC) (0.44 j/goC)= 132g (Tf – 20oC) 4.18j/goC

14. Example (Two variables and two unknowns) Suppose a piece of iron with a mass of 21.5 g at a temperature of 100oC is dropped into an insulated container of water. The mass of the water is 132 g and its temperature before adding the iron is 20oC. What is the final temperature of the system?
21.5g( Tf – 100oC) (0.44 j/goC)= 132g (Tf – 20oC) 4.18j/goC
9.46J/oC*Tf J = 552J/oC*Tf – J
9.46J/oC*Tf = 552J/oC*Tf – J J
9.46J/oC*Tf J/oC*Tf = – J J
-543 J/oC * Tf = – J
-543 J/oC = – 543 J/oC
= 18.4oC = final temperature!

15.  Study For Test X Safety Branches of Chemistry
Organic= hydrocarbons
Inorganic= other compounds
Biochemistry = Biological Compounds
Analytical = ID
Physical = interrelationships & math
Study For Test
Physical: melts, boils, condenses, crystallizes,…
Chemical: reacts w/, oxidizes, decomposes,…
Color, Mass, Volume, Density = phys. prop.
Reacts w/, stability = chem.
Safety
Branches of Chemistry
Physical and Chemical Properties
Physical and Chemical Changes
Energy
Basic Problem Solving (converting Joules, Calories, calories)
4.184 J = 1 cal
1000 cal = 1 Cal
1Cal = 1 kcal X 

16. Homework Homework #5: Do #6-9 page 62 Lab # 5 page 29 in lab manual.
Homework #6: Re-read pages 63-68; Do # 10 a  e; page 64; #11  18 page 69;
Due 9/20
Due 9/28
Due by Thursday
Homework #7: Do # 28  36 page 72 (evens only) Due Friday
TEST MONDAY
Homework #5: Do #6-9 page 62
Lab # 5 page 29 in lab manual.
Homework #6: Re-read pages 63-68; Do # 10 a  e; page 64; #11  18 page 69