1. Thermochemistry
Part 2: Calorimetry

2. Questions to ponder…
If you leave your keys and your chemistry book sitting in the sun on a hot summer day, which one is hotter?
Why is there a difference in temperature between the two objects?

3. Because…
Different substances have different specific heats (amount of energy needed to raise the temperature of 1 g of a substance by 1 degree Celsius).

4. Questions to ponder…
How much heat does it take to melt an ice-cube?

5. Questions to ponder… How much heat does it take to melt an ice-cube?
Q=mc∆T, but ∆T=0

6. Questions to ponder… How much heat does it take to melt an ice-cube?
But I
KNOW
q ≠ 0

7. Questions to ponder… How much heat does it take to melt an ice-cube?
How can I solve this???

8. Calorimetry!!!
Heat required to melt ice (a.k.a. latent heat of fusion) cannot be measured directly, but calorimetry provides an experimental method allowing this heat transfer to be measured indirectly.

9. Calorimetry!!!
Calorimetry: measurement of the amount of heat evolved or absorbed in a chemical reaction, change of state or formation of a solution.

10. CALORIMETRY
The enthalpy change associated with a chemical reaction or process can be determined experimentally.
Measure the heat gained or lost during a reaction at CONSTANT pressure
A calorimeter is a device used to measure the heat absorbed or released during a chemical or physical process

11. Coffee Cup Calorimeter
The cup is filled with water, which absorbs the heat evolved by the reaction, so:
qice = -qwater
OR qrxn = -qcal

12. Coffee Cup Calorimeter
A more “high tech” drawing…
Styrofoam cup

13. What happens in a calorimeter?
One object will LOSE heat, and the other will ABSORB the heat
System loses heat to surroundings = EXO = -q
System absorbs heat from surroundings = ENDO = +q
When a hot chunk of metal is dropped in a cool glass of water, the metal cools off. Where did the heat from the metal go?
Did the metal lose more heat then the water gained?
Magnitude of HEAT GAINED = HEAT LOST (ALWAYS!)

14. To do calorimetry problems…
The numbers in these two boxes are always the same, but with different signs (+/-). What heat one lost, the other gained.
Make a chart:
measurement
“Cal” (often water)
Object/”Rxn”
Heat (q)
Mass (m)
Specific Heat (c)
Final Temp (Tf)
Initial Temp(Ti)

15. Water (cal) Pebble (rxn) Heat Mass Specific Heat 4.184 Final Temp
EXAMPLE 1: A small pebble is heated and placed in a foam cup calorimeter containing 25.0 g of water at 25.0 C. The water reaches a maximum temperature of 26.4 C. How many joules of heat were released by the pebble? The specific heat of water is J/g C.
Water (cal)
Pebble (rxn)
Heat
Mass
Specific Heat
4.184
Final Temp
Initial Temp
The numbers in these two boxes are always the same, but with different signs (+/-). What heat one lost, the other gained.
25.0 g
26.4 oC
25.0 oC

16. Hints:
The pebble lost heat because the water heated up from 25.0 C to 26.4 C.
Pebble loses heat (-q, exothermic) while water gains heat (+q, endothermic)
Do you calculation based on water (since the problem gave all the water’s information)

17. Water (cal) Pebble (rxn) Heat Mass Specific Heat 4.184 Final Temp
Initial Temp
25.0 g
26.4 oC
25.0 oC
qcal = mwatercwaterTwater
qcal = (25.0g)(4.184J/goC)(26.4oC-25.0oC)
qcal = 150 J
qrxn = J
If the water ABSORBED 150 J of heat,
then the pebble RELEASED 150 J of heat.

18. qcal = mwcwt qcal = - 351 J qrxn = - qcal
Example 2: When 1.00 g of ammonium nitrate, NH4NO3, is added to 50.0 g of water in a coffee cup calorimeter, it dissolves, NH4NO3 (s)  NH4+(aq) + NO3-(aq), and the temperature of the water drops from 25.00C to 23.32C. Calculate q for the reaction system.
cal
rxn q m c
4.184 Tf Ti
qcal = mwcwt
qcal = (50.0g)(4.18 J/gC)(-1.68C)
50.0 g
1.00 g
qcal = J
23.32 oC
25.00 oC
qrxn = - qcal
qrxn = 351 J (endothermic)
When one substance absorbs heat, the other substance releases heat (energy)

19. Example 3 (similar to what you will do in tomorrow’s lab)
Suppose that g of water at 22.4 °C is placed in a calorimeter. A g sample of Al is removed from boiling water at a temperature of 99.3 °C and quickly placed in a calorimeter. The substances reach a final temperature of 32.9 °C . Determine the SPECIFIC HEAT of the metal. The specific heat of water is J/gC.
 MAKE YOUR CHART

20. Example 3: Suppose that 100. 00 g of water at 22
Example 3: Suppose that g of water at 22.4 °C is placed in a calorimeter. A g sample of Al is removed from boiling water at a temperature of 99.3 °C and quickly placed in a calorimeter. The substances reach a final temperature of 32.9 °C . Determine the SPECIFIC HEAT of the metal. The specific heat of water is J/g C.
cal (H2O)
rxn (Al) q m c
4.184 Tf Ti
Make chart
Calculate q for water
Q for Al is the same (but with different sign) as q for metal.
Using q metal, calculate c metal g
75.25 g
32.9 oC
32.9 oC
22.4 oC
99.3 oC

21. Example 3 cont’d: specific heat of aluminum…
qcal = mwcwT
cal (H2O)
rxn (Al) q m g
75.25 g c
4.184 J/gC Tf
32.9 C Ti
22.4 C
99.3 C
qcal = (100.00g)(4.184J/gC)(10.5C)
4,393.2 J
-4,393.2 J
qcal = 4,393.2 J
qrxn = - qcal
qrxn = -4,393.2 J
qrxn = mAlcAlT
-4,393.2 J = (75.25 g)(cAl)(-66.4C)
cAl = J/gC

22. What if you wanted to measure the heat of a reaction or process that couldn’t be measured in a simple coffee cup calorimeter (e.g., heat of combustion of Mg)?

23. You would need something like this…

24. Bomb Calorimeter

25. Another type of calorimeter is a Bomb Calorimeter
NOTE: In a bomb calorimeter, heat is transferred from the sample to the oxygen-enriched chamber, to the metal that makes up the chamber, to the water… thus we cannot just use the specific heat of water; instead heat capacity of the calorimeter, Ccal, can be used or calculated.

26. If Heat Capacity (C) is known
It is possible to calculate the amount of heat absorbed or evolved by the reaction if you know the heat capacity, Ccal, and the temp. change, ΔT, of the calorimeter:
qcal = CcalΔT
Everything else is the same (remember, the heat lost from the reaction goes into the calorimeter)

27. c qcal = CcalΔT qcal = (9.33 kJ/C)(9.82 C) qcal = 91.6 kJ
EXAMPLE 4: The reaction between hydrogen and chlorine, H2 + Cl2  2HCl, can be studied in a bomb calorimeter. It is found that when a 1.00 g sample of H2 reacts completely, the temp. rises from 20.00C to 29.82C. Taking the heat capacity of the calorimeter to be 9.33 kJ/C, calculate the amount of heat evolved in the reaction.
qcal = CcalΔT
qcal = (9.33 kJ/C)(9.82 C)
qcal = 91.6 kJ
qrxn = kJ
Cal
Rxn q c Tf Ti
9.33 kJ/oC
29.82 oC
20.00 oC
*** must have a negative sign if “heat evolved in the reaction” = released

28. EXAMPLE 5: When 1. 00 mol of caffeine (C8H10N4O2) is burned in air, 4
EXAMPLE 5: When 1.00 mol of caffeine (C8H10N4O2) is burned in air, 4.96 x 103 kJ of heat is evolved. Five grams of caffeine is burned in a bomb calorimeter. The temperature is observed to increase by 11.37C. What is the heat capacity of the calorimeter in J/C?
4.96 x 103 kJ is for 1.00 mol of caffeine. We are burning only 5.00 g of caffeine to increase the temp by oC. We FIRST need to figure out how much heat energy is given off by just 5 grams.
Cal
Rxn q c DT
+128 kJ
-128 kJ ?
11.37 oC

29. c qcal = CcalΔt qcal = (10.34 kJ/C)(64.2 C) qcal = 664 kJ
EXAMPLE 6: When twenty milliliters of ethyl ether, C4H10O. (d=0.714 g/mL) is burned in a bomb calorimeter, the temperature rises from 24.7C to 88.9C. The calorimeter heat capacity is kJ/C. (a) What is q for the calorimeter? (b) What is q when 20.0 mL of ether is burned? (c) What is q for the combustion of one mole of ethyl ether?
qcal = CcalΔt
qcal = (10.34 kJ/C)(64.2 C)
qcal = 664 kJ
Cal
Rxn q c Tf Ti
10.34 kJ/oC
88.9 oC
24.7 oC

30. EXAMPLE 6: When twenty milliliters of ethyl ether, C4H10O. (d=0
EXAMPLE 6: When twenty milliliters of ethyl ether, C4H10O. (d=0.714 g/mL) is burned in a bomb calorimeter, the temperature rises from 24.7C to 88.9C. The calorimeter heat capacity is kJ/C. (a) What is q for the calorimeter? (b) What is q when 20.0 mL of ether is burned? (c) What is q for the combustion of one mole of ethyl ether?
qrxn = -qcal
qrxn = -664 kJ

31. EXAMPLE 6: When twenty milliliters of ethyl ether, C4H10O. (d=0
EXAMPLE 6: When twenty milliliters of ethyl ether, C4H10O. (d=0.714 g/mL) is burned in a bomb calorimeter, the temperature rises from 24.7C to 88.9C. The calorimeter heat capacity is kJ/C. (a) What is q for the calorimeter? (b) What is q when 20.0 mL of ether is burned? (c) What is q for the combustion of one mole of ethyl ether?