4.a Thermochemistry
Intro
 Thermochemistry

study of transfers
of energy (E) as
heat (q) occurring
during chemical
and physical
changes
Energy
 Energy
– capacity to do some kind of
work

Energy is always involved in a chemical
or physical change
 Law
of Conservation of Energy –
energy cannot be created or destroyed

Energy is transferred in chemical or
physical changes
Energy Cont’d
 Energy

Changes
Exothermic – process that releases heat
into the environment
• Typically feels warm

Endothermic – process that absorbs heat
from the environment (uses energy)
• Typically feels cold
 Kinetic
Energy (KE) – energy of an
object that is due to the object’s
motion
Temperature
Temperature (T):
average KE of the
particles in a
sample
 As KE increases, so
does T (direct
relationship)
 Temp Scales

Farenheit
 Celsius
 Kelvin

Kelvin (K) Temp Scale
 Kelvin
– developed so that we could
have a system with no negative values



Starts at -273.15 ○C = absolute zero = 0 K
Add 273 to temp value for ○ C
Example: 10 ○C + 273 = 283 K
Heat
 Heat
- (q) energy transferred between
samples because of their difference in T
 moves spontaneously from matter with
higher T to matter with lower T until
thermal equilibrium (= temp) is reached
 heat can’t be measured directly so T is
used to track transfer of heat
 Energy transferred is measured in
Joules (J)
Heat Transfer
Heat
 measured
using a calorimeter (reaction
container that is surrounded by water)
 the
energy given off during a reaction
is equal to the energy absorbed by the
water
qabsorbed  qreleased
Two Types of Calorimeters
Specific Heat
 the
amount of energy transferred
during a temperature change depends
on 3 factors:



type of material
mass of material
size of T change
 Specific
heat: (c) the amount of energy
required to raise the T of one gram of
substance by one °C or K
Specific Heat –Math
problems 

measured under constant pressure cp
q
cP 
OR q  c P  m  ΔT
m  ΔT
cp = specific heat at constant pressure
(J/g•K or J/ °C)
q = heat (J or KJ)
m = mass (g
Δ T = change in temperature (Tfinal – Tinitial )
Example
A 4.0 g sample of glass was heated from 274 K
to 314 K and was found to have absorbed 32
J of energy.

Find the specific heat of the glass sample.
q
32 J
J
cP 

 0.20
m  ΔT (4.0g)(314 274K)
gK
Example

How much energy will the same glass
sample gain when heated from 314 K to 344
K?
q  cP  m  ΔT
J
 (0.20
)(4.0g)(344 314 K)  24 J
gK
Example

If 200. g of water at 20°C absorbs 41,840 J of
heat, what will its final T be? The specific heat
of water is 4.184 J/g°C.
q  cP  m  ΔT  cP  m  (Tfinal  Tinitial)
J

41,840J  (4.184  )  (200.g)  (Tfinal  20 C)
gC
Example
41,840J

 Tfinal  20 C
J
(4.184  )  (200.g)
gC
41,840J


Tfinal 
 20 C  70 C
J
(4.184  )  (200.g)
gC