# Heat Energy - Chemistry

```
Energy is transferred
◦ Exothermic – heat is released
 Heat exits
◦ Endothermic – heat is required or absorbed by
reaction

Measure energy in Joules or calories

Temperature – average energy of molecular
motion
◦ Size and type of particle do not matter

Heat – total energy of molecular motion
◦ Speed, amount and type of particle matter


The heat needed to raise the temperature of
1g of a substance by 1oC
Water has a high heat capacity of 4.184
Joules or 1 calorie
J
cw  4.184 o
g C
Q  m  c  T
Q = heat (gained or lost)
m = mass (in grams)
c = specific heat value
∆T = change in temperature

How much heat is lost when solid aluminum
ingot with a mass of 4110g cools from 660oC
to 25oC? The specific heat of aluminum is
0.903 J/(g x oC).
Q  m  c  T
m = 4110g
c = 0.903 J/(g x oC)
∆T = 660oC – 25oC = 635oC
Q=?

J 
o
Q  4110 g  0.903 o  635 C
g C 


Q = 2, 356, 695 J
Or. . .
2, 357 kJ




What happens to the change in temperature
or equal masses of copper and water when
equal amounts of heat energy are given?
c for Cu = 0.387 J/(g x oC)
Use mass = 1.0g
Q=10.0J
Cu

J 
10.0 J  1.0 g  0.387 o T 
g C 

H2O

J 
10.0 J  1.0 g  4.18 o T 
g C 

∆T = 25.8oC
∆T = 2.4oC

Calorimeter – a device used to measure the
energy given off or absorbed during a
chemical or physical change.

A piece of unknown metal with mass 23.8 g is
heated to 100.0oC and dropped into 50mL of
water at 24.0oC. The final temperature of the
system is 32.5oC. What is the specific heat of
the metal? The density of water is 1g/mL.
Metal
m = 23.8g
∆T = 100 – 32.5 = 67.5oC
density 
Water
m = 50.0g
∆T = 32.5 – 24 = 8.5oC
c = 4.18 J/(g x oC)
mass
1g
x


volume 1mL
50mL
X = 50.0g
For water D = 1g/mL
Heat gained by water = Heat lost by metal

J  o
50.0 g  4.18 o  8.5 C  23.8g c  67.5o C
g C 





c = 1.1 J/(g x oC)

The amount of energy needed to convert 1g
of a substance from liquid to gas or from gas
to liquid.
Q  m  H v


How much heat is required to vaporize 15g of
liquid water?
Hv of water = 2260 J/g

J
Q  15 g  2260 
g

Q = 33,900J or 33.9 kJ


The amount of energy needed to melt 1g of
solid substance to liquid.
The amount of energy released when 1g of a
liquid freezes or becomes solid.
Q  m  H f


How much energy is released when 25g of
liquid water freezes?
Hf of water = 334 J/g

J
Q  25 g  334 
g

Q = 8,350 J or 8.35 kJ



A 30g sample of water is heated from 75oC to
135oC. How much energy is needed?
Remember, you are crossing a phase change
so you must account for that!
Use both specific heat and heat of
vaporization.
Q  m  c  T
Q  m  H v

J  o
Q  30 g  4.18 o  25 C
g C 



Q = 3,135 J
This accounts for liquid portion of water.

J
Q  30 g  2260 
g

Q = 67,800 J
This accounts for the phase change (liquid to gas)

J  o
Q  30 g  2.02 o  35 C
g C 



Q = 2,121 J
Accounts for gas (steam) portion of water.
So, total energy is:
3,135J + 67,800J + 2,121J =
73,056J or 73 kJ
```
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