The document explains specific heat capacity, which is the heat required to raise the temperature of 1 gram of a substance by one degree Celsius. It provides examples and problems involving calculations of specific heat, heat of fusion, and heat of vaporization for various substances. Additionally, it demonstrates how to solve practical problems related to temperature changes and phase transitions.

1. Specific Heat
Specific Heat
Capacity
Capacity

2. Imagine…
Imagine…
A hot day in Arizona…in your back yard
A hot day in Arizona…in your back yard
is a
is a metal barbeque
metal barbeque and a
and a glass of
glass of
water
water. Would you want to stick your
. Would you want to stick your
hand in the water or touch the bbq?
hand in the water or touch the bbq?
You would want to touch the
You would want to touch the water
water
because different substances have
because different substances have
different abilities to absorb heat!!
different abilities to absorb heat!!

3. Specific Heat Capacity
Specific Heat Capacity
The quantity of heat required
The quantity of heat required
to raise the temperature of 1
to raise the temperature of 1
gram of a substance by one
gram of a substance by one
degree Celsius (or Kelvin)
degree Celsius (or Kelvin)
The unit is J/g(
The unit is J/g(o
o
C)
C)

4. Some Specific Heat Capacities
Some Specific Heat Capacities
Substance
Substance Specific heat
Specific heat
capacity
capacity
in J/g(
in J/g(o
o
C)
C)
Water
Water 4.18
4.18
Copper
Copper 0.385
0.385
Iron
Iron 0.449
0.449

5. Equation:
Equation:
q = mC T
q = mC T
 q = heat transferred (in joules)
q = heat transferred (in joules)
 m = mass (in grams)
m = mass (in grams)
 C = specific heat capacity
C = specific heat capacity
 T = change in temperature
T = change in temperature
T = T
T = Tfinal
final - T
- Tinitial
initial

6. Practice Problem
Practice Problem
What is the
What is the specific heat
specific heat of
of
silver if the
silver if the temperature of a
temperature of a
15.4 g
15.4 g sample of silver is
sample of silver is
increased from
increased from 20.0
20.0o
o
C
C to
to
31.2
31.2o
o
C
C when
when 40.5 J
40.5 J of heat is
of heat is
added?
added?

7. Givens:
Givens:
m = 15.4 g
m = 15.4 g
T
Ti
i = 20.0
= 20.0o
o
C
C
T
Tf
f = 31.2
= 31.2o
o
C
C
q = 40.5 J
q = 40.5 J
q = mC
q = mC T
∆T
∆
40.5=15.4(C)(31.2-20.0)
40.5=15.4(C)(31.2-20.0)
40.5=15.4(C)(11.2)
40.5=15.4(C)(11.2)
40.5=172.48(C)
40.5=172.48(C)
C = 0.235 J/g(
C = 0.235 J/g(o
o
C)
C)

8. Practice Problem
Practice Problem
What is the final temp of
What is the final temp of
silver if the temperature
silver if the temperature
of a 5.8 g sample of silver
of a 5.8 g sample of silver
starts out at 30.0
starts out at 30.0o
o
C and
C and
40.5 J of heat is added?
40.5 J of heat is added?
The specific heat of silver
The specific heat of silver
is .235 J/g(
is .235 J/g(o
o
C).
C).

9. Givens:
Givens:
m = 5.8 g
m = 5.8 g
T
Ti
i = 30.0
= 30.0o
o
C
C
q = 40.5 J
q = 40.5 J
C = 0.235
C = 0.235
T
Tf
f = ???
= ???
q = mC
q = mC T
∆T
∆
40.5=5.8(0.235)(T
40.5=5.8(0.235)(Tf
f -30.0)
-30.0)
40.5=1.363(T
40.5=1.363(Tf
f-30.0)
-30.0)
40.5=1.363T
40.5=1.363Tf
f – 40.89
– 40.89
81.39=1.363Tf
81.39=1.363Tf
T
Tf
f = 59.7139
= 59.7139
T
Tf
f = 60.
= 60.o
o
C
C

10. What quantity of heat is
What quantity of heat is
required to raise the
required to raise the
temperature of 100 mL of
temperature of 100 mL of
water from 45.6
water from 45.6
C to 52.8
C to 52.8
C?
C?
The specific heat of water is
The specific heat of water is
4.184 J/g(
4.184 J/g(
C) and water has a
C) and water has a
density of 1.00 grams/mL.
density of 1.00 grams/mL.
Practice Problem
Practice Problem

11. Givens:
Givens:
q = ?
q = ?
V = 100. mL
V = 100. mL
T
Ti
i = 45.6
= 45.6o
o
C
C
T
Tf
f = 52.8
= 52.8o
o
C
C
C = 4.184
C = 4.184
d
dwater
water=1.00
=1.00
g/mL
g/mL
100 mL = 100 g
100 mL = 100 g
q = mC
q = mC T
∆T
∆
q=100(4.184)(52.8-45.6)
q=100(4.184)(52.8-45.6)
q=3012.48
q=3012.48
q=3010 J
q=3010 J

12. Heat of Fusion &
Heat of Fusion &
Heat of Vaporization
Heat of Vaporization
 Specific heat works great for the part of
Specific heat works great for the part of
the heating and cooling process where
the heating and cooling process where
temperature is changing but what do
temperature is changing but what do
we do to calculate Q when we are in the
we do to calculate Q when we are in the
process of a phase change?
process of a phase change?
 Answer: We use heat of fusion or heat
Answer: We use heat of fusion or heat
of vaporization!
of vaporization!

13. Heat of Vaporization
Heat of Vaporization
 The amount of energy needed to
The amount of energy needed to
vaporize (l
vaporize (l 
 g) a specific amount
g) a specific amount
of a liquid at constant temperature
of a liquid at constant temperature
 Unit is usually J/g
Unit is usually J/g
Q = m(H
Q = m(Hv
v)
)

14. Example Problem
Example Problem
 A 115 g sample of liquid is boiled over a
A 115 g sample of liquid is boiled over a
10-minute period. As the liquid boils,
10-minute period. As the liquid boils,
the temperature remains constant with
the temperature remains constant with
2.27 x 10
2.27 x 105
5
J of heat is absorbed. At the
J of heat is absorbed. At the
end of the 10-minute period, all of the
end of the 10-minute period, all of the
liquid has been boiled away. What was
liquid has been boiled away. What was
the heat of vaporization of the liquid?
the heat of vaporization of the liquid?

15. Givens
Givens
Q = 2.27 x 10
Q = 2.27 x 105
5
J
J
m = 115 g
m = 115 g
Q = m(H
Q = m(Hv
v)
)
2.27 x 10
2.27 x 105
5
= 115(H
= 115(Hv
v)
)
H
Hv
v = 1973.91 J/g =
= 1973.91 J/g = 1970 J/g
1970 J/g

16. Heat of Fusion
Heat of Fusion
The amount of heat needed to
The amount of heat needed to
change a specific amount of a
change a specific amount of a
solid to a liquid at constant
solid to a liquid at constant
temperature.
temperature.
Unit is usually J/g
Unit is usually J/g
Q = m(H
Q = m(Hf
f)
)

17. Example Problem
Example Problem
The heat of fusion of ice at 0
The heat of fusion of ice at 0o
o
C
C
is 3.4 x 10
is 3.4 x 102
2
J/g. How much heat
J/g. How much heat
is needed to change 75g of ice
is needed to change 75g of ice
at 0
at 0o
o
C to liquid water at the
C to liquid water at the
same temperature?
same temperature?

18. Givens
Givens
H
Hf
f = 3.4 x 10
= 3.4 x 102
2
J/g
J/g
m = 75 g
m = 75 g
Q = m(H
Q = m(Hf
f)
)
Q = 75(3.4 x 10
Q = 75(3.4 x 102
2
)
)
Q = 2.55 x 10
Q = 2.55 x 104
4
J = 2.6 x 10
J = 2.6 x 104
4
J
J

19. Q
QTotal
Total
 When using multiple equations to solve
When using multiple equations to solve
for the energy change of a system, Q
for the energy change of a system, QT
T is
is
used.
used.
Q
QT
T = Q
= Q1
1 + Q
+ Q2
2 + etc.
+ etc.

20. Practice Problem
Practice Problem
 Water that has a mass of 283g is at
Water that has a mass of 283g is at
23.5
23.5o
o
C and is placed in a freezer that
C and is placed in a freezer that
has a temperature of -12.4
has a temperature of -12.4o
o
C. How
C. How
much energy must be removed from
much energy must be removed from
the water to reach this temperature?
the water to reach this temperature?