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 10.00 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) = 41.84 Joule liquid. -1 =18 -2 = 16

Homework 6 Help

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

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

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

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

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 4.184 J/goC (1 cal/goC) q = m (T2-T1) Cp

q = m  T Cp T2 mass of WATER T1 q = m  T Cp Cp = 4.184 J/goC T = T2 – T1 http://www.sciencebyjones.com/heating_curve1.htm q = m  T Cp

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 = T2 – T1 85.0oC 23.5oC http://www.sciencebyjones.com/heating_curve1.htm q = m  T Cp (854 g) (61.5oC) 4.184J/goC

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 0.874 J/g oC q = m  T Cp  T= (69.2 – 31.7)oC

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 0.874 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 0.874 J/goC = 3170 J

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

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 = 4.184 J/goC Plug in numbers 21.5g( Tf – 100oC) (0.44 j/goC)= 132g (Tf – 20oC) 4.18j/goC

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 - 946 J = 552J/oC*Tf – 11000 J 9.46J/oC*Tf = 552J/oC*Tf – 11000 J + 946 J 9.46J/oC*Tf - 552J/oC*Tf = – 11000 J + 946 J -543 J/oC * Tf = – 10000 J -543 J/oC = – 543 J/oC = 18.4oC = final temperature!

 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 

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