1. Do now 1) Take out graphic organizers (turn in any late work)
2) Work on the following problem
55 grams of Compound X has 50 kJ of energy before decomposing into elements Y and Z, which have a combined energy of 20 kJ.
What is the combined mass of elements Y and Z?
Physical or chemical change?
Is this an exothermic or endothermic reaction?
What is the ΔH (heat of reaction)?

2. DRESS APPROPRIATELY!
We will be doing labs pretty frequently this next week so wear proper attire (hair tied back, no open toed shoes, etc…)
If you do not have your safety forms AND your safety lab turned in, you will not be allowed to do the labs…

3. Calorimetry; Measuring heat flow
Ever wonder how they find out how many calories are in your food?
The answer; Calorimetry
Calo-what?

4. Calorimetry; Measuring heat flow
When a reaction occurs, potential energy is either gained (+ ΔH) or lost (- ΔH)
Can’t use a thermometer to quantify and measure this, so we need to use something else…

5. Things we know
1 gram of water needs 4.18 Joules of energy to increase its temperature 1 degree celsius
we call this its specific heat
4.18 joules / (grams ocelsius)
During a reaction, energy (joules) is going to either be gained from or lost to the surroundings
How can we use this to figure out the heat change?

6. What about this? Get the stuff to react
Energy could be absorbed by water…
The temperature change and amount of water would tell us how much energy was absorbed…

8. Calorimetry What will we need to know to figure out heat exchange?
How much energy is needed to change water’s temperature  4.18 J/(g*degree celsius)
Mass of water
Change in temperature (ΔT)
Put it together…

9. Formula J / (grams * ocelsius) How can we get just joules?
Need grams and degree celsius to cancel out…
Multiply by grams and degree celsius!

10. Where the equation came from
J/(g* ocelsius) * g * ocelsius  J
Rearrange
Joules = g * J/(g*ocelsius) * ocelsius
Heat change = mass * specific heat * Δ temp
q = m * C * ΔT
Wishing there was a song to remember all of this???
YOU GOT IT!

11. An example
How many joules are absorbed by grams of water if the temperature is increased from 35.0 oC to 50.0 oC?
q = m c ΔT
q= g * 4.18 J/(goC) * 15.0 oC
q= 6270 Joules

12. What if we had to solve for the mass of water?
A sample of water is heated by 20.0 oC by the addition of 80.0 J of energy. What is the mass of the water?
Rearrange the formula to solve for m…
q = m c ΔT………Divide each side by c ΔT
q / (c ΔT) = (m c ΔT) / (c ΔT)
m= q / (c ΔT)

13. m= q / (c ΔT)
A sample of water is heated by 20.0 oC by the addition of 80.0 J of energy. What is the mass of the water?
m= 80.0 J / (4.18 J/(goC) * 20.0 oC)
m= grams
m= g

14. How about a tough one…
200. J of energy is absorbed by an 80.0 g sample of water in a calorimeter at oC. What will the final temperature be?
Hmm…wants final temperature and tells us the initial…
Need to calculate ΔT… then we need to rearrange the formula
Δ T = q / m C

15. Δ T = q / m C
200. J of energy is absorbed by an 80.0 g sample of water in a calorimeter at oC. What will the final temperature be?
Δ T = q / m C
Δ T = 200. J / [80.0* 4.18 J/(goC)]
Δ T = oC
Δ T = 0.598oC
NOT DONE YET!!!

16. Δ T = 0.598oC
200. J of energy is absorbed by an 80.0 g sample of water in a calorimeter at oC. What will the final temperature be?
Says energy is absorbed by the water, so the temperature is going to…increase
Tfinal = Tinitial + Δ T
Tfinal = oC oC
Tfinal = oC

17. Reference Table I
‘Heats of reaction’; known heats of reaction are listed
C(s) + O2 (g)  CO2 (g) ∆H= –393.5 kJ
Is this exothermic or endothermic?
Exothermic

18. C(s) + O2 (g) CO2 (g) ∆H= –393.5 kJ
What if the reverse reaction happened, what would the heat of reaction be?
CO2 (g)  C(s) + O2 (g) ∆H=???
If it is – one way…then it would be the opposite the other way
∆H= kJ
Exothermic or endothermic?
Endothermic!

19. With your neighbor
Work on the calorimetry homework