CALORIMETRY
Calorimetry
 Calorimetry, the measurement of heat released or
absorbed by a chemical reaction.
 calorimeter is the device used to measure heat.
 The quantity of heat transferred by the reaction
causes a change in temperature of the solution.
q = c x m x DT
q = heat lost or gained
c = specific heat capacity
m = mass in g
DT = Tfinal - Tinitial
The specific heat capacity (c) of a substance is the
quantity of heat required to change the temperature of
1 gram of the substance by 1 K.
Table: Specific Heat Capacities (c) of Some Elements, Compounds,
and Materials
Substanc
e
Specific Heat
Capacity (J/g∙K)
Elements
Substanc
e
Specific Heat
Capacity (J/g∙K)
Solid materials
aluminum, Al
0.900
wood
1.76
graphite,C
0.711
cement
0.88
iron, Fe
0.450
glass
0.84
copper, Cu
0.387
granite
0.79
gold, Au
0.129
steel
0.45
Compounds
water, H2O(l)
4.184
ethyl alcohol, C2H5OH(l)
2.46
ethylene glycol, (CH2OH)2(l)
2.42
carbon tetrachloride, CCl4(l)
0.862
Finding the Quantity of Heat from a Temperature Change
PROBLEM: A layer of copper welded to the bottom of a skillet weighs
125 g. How much heat is needed to raise the
temperature of the copper layer from 25°C to 300°C?
The specific heat capacity (c) of Cu is 0.387 J/g∙K.
PLAN:
We know the mass (125 g) and c (0.387 J/g∙K) of Cu
and can find DT in °C, which equals DT in K. We can
use the equation q = cmDT to calculate the heat.
SOLUTION
:
DT = Tfinal – Tinitial = 300 – 25 = 275°C = 275 K
q = cmDT = 0.387 J x 125 g x 275 K
g∙K
= 1.33x104 J
Coffee-cup calorimeter.
This device measures the heat transferred at constant pressure
(qP).
Determining the Specific Heat Capacity of a Solid
PROBLEM: A 22.05 g solid is heated in a test-tube to 100.00°C and
added to 50.00 g of water in a coffee-cup calorimeter. The
water temperature changes from 25.10°C to 28.49°C.
Find the specific heat capacity of the solid.
PLAN: Since the water and the solid are in contact, heat is transferred
from the solid to the water until they reach the same Tfinal. In
addition, the heat given out by the solid (-qsolid) is equal to the
heat absorbed by the water (qwater).
SOLUTION:
DTwater = Tfinal – Tinitial = (28.49°C – 25.10°C) = 3.39°C = 3.39 K
DTsolid = Tfinal – Tinitial = (28.49°C – 100.00°C) = -71.51°C = -71.51 K
c
csolid =
=
H2O
x mass
H2O
x DT
H2O
masssolid x DTsolid
4.184 J/g∙K x 50.00 g x 3.39 K
22.05 g x (-71.51 K)
= 0.450 J/g∙K
Determining the Enthalpy Change of an Aqueous Reaction
PROBLEM: 50.0 mL of 0.500 M NaOH is placed in a coffee-cup
calorimeter at 25.00 oC and 25.0 mL of 0.500 M HCl is
carefully added, also at 25.00 oC. After stirring, the final
temperature is 27.21 oC. Calculate qsoln (in J) and the change
in enthalpy, DH, (in kJ/mol of H2O formed).
Assume that the total volume is the sum of the individual
volumes, that d = 1.00 g/mL and c = 4.184 J/g∙K
PLAN Heat flows from the reaction (the system) to its surroundings (the
:
solution). Since –qrxn = qsoln, we can find the heat of the reaction
by calculating the heat absorbed by the solution.
SOLUTION:
(a) To find qsoln:
Total mass (g) of the solution = (25.0 mL + 50.0 mL) x 1.00 g/mL = 75.0 g
DTsoln = 27.21°C – 25.00°C = 2.21°C = 2.21 K
qsoln = csoln x masssoln x DTsoln = (4.184 J/g∙K)(75.0 g)(2.21 K)
(b) To find DHrxn we first need a balanced equation:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
= 693 J
For HCl:
25.0 mL HCl x
1L
0.500 mol
x
103 mL
1L
x 1 mol H2O
1 mol HCl
= 0.0125 mol H2O
For NaOH:
50.0 mL NaOH x 1 L
x
3
10 mL
0.500 mol x 1 mol H2O
= 0.0250 mol H2O
1L
1 mol NaOH
HCl is limiting, and the amount of H2O formed is 0.0125 mol.
DHrxn =
qrxn
mol H2O
=
-693 J
x 1 kJ
0.0125 mol 103J
= -55.4 kJ/mol H2O
Practice Problem
When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0oC is mixed with 1.00
L of 1.00 M Na2SO4 solution at 25oC in a calorimeter, the white solid
BaSO4 forms and the temperature of the mixture increases to 28.1oC.
Assuming that the calorimeter absorbs only a negligible quantity of
heat, that the specific heat capacity of the solution is 4.18 J/oC g, and
that the density of the final solution is 1.0 g/mL, calculate the
enthalpy change per mole of BaSO4 formed.
Soln cont
Exercise
 A metal pellet with mass 100.0 g, originally at 88.4 °C, is
dropped into 125 g of water originally at 25.1 °C. The final
temperature of both the pellet and the water is 31.3 °C.
Calculate the heat capacity C (in J/°C) and specific heat
capacity Cs (in J/g•°C) of the pellet. The specific heat of water
is 4.184 J/g•°C.
 When 200.0 g of a AgNO3 solution mixes with 150.0 g of NaI
solution, 2.93 g of AgI precipitates, and the temperature of the
solution rises by 1.34oC. Assume 350. g of solution and a
specific heat capacity of 4.184 J/g•oC. Calculate ΔH for the
following: