Chapter 2
Heat Capacity and Specific Heat
Rice
Some Basic Concepts
Kinds of Energy






Heat
Light
Chemical
Nuclear
Electrical
Mechanical
Any one of these
can be converted
into any of the
other forms.
Types of Energy
1) Kinetic Energy – energy of motion
KE = ½ mv2
m= mass in kg
v = velocity in m/s
2) Potential Energy – stored energy
1. Gravitational Potential Energy
2. Electrostatic Attraction Potential Energy
3) Total Energy of a Substance is equal to:
ETotal = PE + KE
How does KE & PE relate to
Chemistry?

KE = the motion of the molecules in
chemical substances.

PE = the sum of all the attractions in a
chemical substance, including all the
covalent bonds, ionic bonds, and
electrostatic attractions.
Temperature




A measure of the
average kinetic energy
of the particles in a
sample of matter.
It is a description of one
characteristic of the
individual particles in an
object.
How hot or cold
something is.
 K.E  Temp
Heat or Heat Energy


The sum total of the kinetic energies of the
particles in a sample of matter. The quantity
of energy in an object.
Heat always flows spontaneously from matter
at a higher temperature to matter at a lower
temperature. When the temperature of the
water is constant…heat will no longer flow in
the system.
Is there a difference between Heat and
Temperature?
Temperature relates to the average K.E. of the
individual particles.
Heat relates to the total quantity of energy
(K.E.) in the object.
Latent Heat of Phase Change
Heat Transfer

Heat flows from regions of high heat content
to regions of low heat content.
–
–
–
Conduction: heat spread by collisions between
molecules.
Convection: heat spread by movement of air
Radiation: energy moving through space in the
form of electromagnetic radiation
Heat vs Temperature

http://www.youtube.com
/watch?v=rUsPzshVnM
50°C
100°C
Which would make the water warmer?
The Measurement of Energy




Energy cannot be measured directly.
Heat is measured by determining temperature
changes caused by the release or absorbtion of
heat in a chemical process.
Heat energy was originally defined in terms of
the calorie (the amount of heat needed to raise
the temperature of 1g of water 1°C).
1 cal = 4.184 J
Heat Capacity
 The
amount of heat energy needed
to raise the temperature of a given
sample of matter by 1°C
Specific Heat
The amount of energy needed to
raise the temperature of 1g of water
1°C.

The specific heat of
water is 4.184 J/g°C
or 1 cal/g°C.
Once the specific heat of water is defined,
the specific heat of any other substance can
be determined.

One method is to immerse a hot object in a
known quantity of H2O and then measure
the temperature change.
q (gained by water) =
q (lost by metal)
Units of Temperature vs Units of Heat

Temperature °C, °F, K

Heat

cal (calorie) or Cal (food calorie)
1 Cal = 1000 cal
J (joule)
1 cal = 4.184 J
Convert the following:
1) 110 cal to J
2) 22 kJ to cal
Solutions
1) 110 cal to J
110 cal x 1 cal
= 26 J
1
4.184 J
2) 22 kJ to cal
22kJ x 1000 J
1
1 kJ
x
4.184 J = 92000 cal
1 cal
Calorimetry

Calorimetry is the science of measuring the
heat of chemical reactions or physical
changes. Calorimetry involves the use of a
calorimeter. The word calorimetry is derived
from the Latin word calor, meaning heat.
Coffee Cup Calorimeter
http://www.chem.iastate.edu/group/Greenbowe/sections/projectf
older/flashfiles/thermochem/heat_metal.html
Coffee Cup Calorimeter
Constant pressure calorimeter is called a
coffee cup calorimeter.

Water is put in the cup and the temperature is taken,
the temperature of heated metal is taken. The
heated metal is then placed in the calorimeter and
the cover is placed on the calorimeter. When the
temperature no longer changes, it is recorded.
qlost = qgained
Example
An insulated cup contains 75.0g of water at 24.00°C. A
26.00g sample of a metal at 85.25°C is added. The
final temperature of the water and metal is 28.34°C.
–
What is the specific heat of the metal?
Determining the specific heat

What is the specific heat of the metal?
 q = c x m x ΔT
q = heat
c = specific heat capacity
m = mass in grams
ΔT = Tf – Ti
Determine the specific heat of the metal
An insulated cup contains 75.0g of water at 24.00°C. A 26.00g
sample of a metal at 85.25°C is added. The final temperature of
the water and metal is 28.34°C.
qlost = qgained H2O gained heat, Metal lost heat
+ qwater = - qmetal
cH2O x mH2O x ΔTH2O = - (cmetal x mmetal x ΔTmetal)
(4.184) (75.0g) (28.34-24.00) = - c (26.00) (28.34-85.25)
- 0.920 J/g°C = cmetal
Your Turn
An insulated cup contains 250g of water at
25.0°C. A 5.0g sample of a metal at 100.0°C
is added. The final temperature of the water
and metal is 25.2°C.
–
What is the specific heat of the metal?
Solution
1)
cH2O x mH2O x ΔTH2O = - (cmetal x mmetal x ΔTmetal)
(4.184)(250g)(25.2-25.0) = - c(5.00)(25.2-100.0)
- 0.559 J/g°C = cmetal
Your Turn
The specific heat of graphite is
0.71 J/gºC. Calculate the energy
needed to raise the temperature of
75kg of graphite from 294 K to 348 K.
Solution
First convert:

75kg x 1000g = 75,000g
1
1 kg
Tf = 348K - 273 = 75ºC
Need ºC for Units
Ti = 294K - 273 = 21ºC
q = cmΔT
q = (0.71 J/gºC) (75,000g)(75-21)
q = 2,875,500 J or 2,875.5 kJ
Your Turn
A 46.2 g sample of copper is heated to
95.4ºC and then placed in a calorimeter
containing 75.0 g of water at 19.6ºC. The final
temperature of both the water and the
copper is 21.8ºC. What is the specific heat
of copper?
Solution
+qgained = -qlost

cH2O x mH2O x ΔTH2O = - (cmetal x mmetal x ΔTmetal)
(4.184)(75.0g)(21.8 -19.6) = - c (46.2)(21.8 - 95.4)
- 0.203
J/gºC = cmetal