1. 1. What is the specific heat of a 10. g sample of a substance
that requires 950 J of heat to raise its temp from 18 C to 42 C?
2. Hg has a specific heat of 0.14 J/gC. How much energy
does a 22.8 g sample of mercury lose as its temp goes from
33 C to 16 C ?

2. V. Calorimetry
Calorimeter – an instrument that measures the gain or loss of heat. (See pg. 66)

3. A calorimeter can be used to determine the specific heat of
a substance, or to determine the heat given off or absorbed
by a chemical rxn.
In calorimetry, the heat change for
a process is determined by
measuring the heat change in
the water surrounding the process
in an insulated container called
a calorimeter.

4. According to the law of conservation of energy:
the heat lost by the process is gained by the water
(or the heat absorbed by the process is lost by the water.)
In equation form this looks like:
qprocess = -qwater
(The specific heat of water is a
constant: J/gC)

5. Determining the specific heat of an object
using calorimetry:
heat lost by object = - heat gained by water
m x cp x T = - m x cp x T
(object) (water)
Solve for cp of the object.
*The right side of a calorimetry problem
always starts with a negative sign.

6. EX1: A 28.5 g sample of metal with a temperature of 100.0 C
is placed in a calorimeter holding g of water at 0.00 C.
The temperature of the water rises to 2.00 C.
What is the specific heat of the metal?
What metal is it?
(use appendix A; Table A-3 in back of text)
(For these problems, the final temperature
of the water and the object are the same.)
(Remember that the specific heat of water
is always J/gºC)

7. EX2: A 12.5 g sample of metal with a temperature of
100.0 C is placed in a calorimeter holding 75.0 g of
water at 22.0 C.
The temperature of the water rises to 24.0 C.
What is the specific heat of the metal?
What metal is it?

8. B. Determining the heat given off or absorbed by a rxn
using calorimetry:
H = heat given off or absorbed by a chemical rxn
H of a rxn = - heat gained or lost by the water
In a calorimeter:
H = - m x cp x T
(water)
*Signs for H:
+ H endothermic rxn
- H exothermic rxn

9. EX1: A chemical rxn causes 150.g of water in a calorimeter to
go from 25.0 C to C .
How much heat was given off or absorbed by the rxn?
b) Was the rxn exothermic or endothermic?
EX2: A chemical rxn causes 200.g of water in a calorimeter to
go from 30.0 C to C .
What was H for the rxn?
b) Was the rxn exothermic or endothermic?

10. Thermochemical Equations
Heats of Reaction and
Thermochemical Equations

11. Heat of Reaction (H) –
the amount of heat given off or
absorbed in a chemical rxn.
(H stands for “enthalpy” – the
heat content of a system at
constant pressure;
H = Hproducts - Hreactants )
Signs for H:
+ H endothermic rxn
- H exothermic rxn

12. B. Thermochemical Equations
2 P(s) + 3 Br2(l)  2 PBr3(l) kJ
Exothermic or Endothermic?
Usually the equation would be written as:
2 P(s) + 3 Br2(l)  2 PBr3(l) H = -278 kJ
The above are examples of thermochemical equations –
equations that include the heat given off or absorbed.

13. Write the equation:
H2(g) + I2(g) + 53 kJ  2 HI(g)
using H.
H2(g) + I2(g)  2 HI(g) H = +53 kJ
***The H in an equation is only for the number of moles
of reactants and products indicated by the coefficients
in the equation.
The actual H of an individual rxn depends on how much
reactant was used.***
2 H2(g) + O2(g)  2 H2O(l) H = -572 kJ

14. Example Problems:
2 H2(g) + O2(g)  2 H2O(l) H = -572 kJ
1. If 2.00 mol of O2 react, how much heat is produced?
2. If 5.00g of H2 is burned, how much heat is released?
3. How much heat is produced in the production of 100.g of H2O?
4. How many grams of H2 are needed to produce
1.00 x 104 kJ of heat?

15. 2 PBr3(l)  2 P(s) + 3 Br2(l) H = +278 kJ
1. If 1200 kJ of heat was absorbed by the rxn, how many grams of Br2 was produced?
2. How much heat is needed to produce 25.0g of Br2?

16. End College Prep

17. Additional Facts About
Thermochemical Equations
Often fractions are used as coefficients:
C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(l) H = kJ
2. The H of a reverse rxn has the same value but opposite sign.
2 H2(g) + O2(g)  2 H2O(l) H = -572 kJ
2 H2O(l)  2 H2(g) + O2(g) H =
+572 kJ

18. 3. The value of H depends on what physical state the reactants and products are in:
2 H2(g) + O2(g)  2 H2O(l) H = -572 kJ
2 H2(g) + O2(g)  2 H2O(g) H = -484 kJ
(The difference is due to: H2O(g)  H2O(l) H = -88 kJ)

19. END