Do Now

On your Daily Organizer, describe each scenario as
conduction, convection or radiation:
A.
B.
C.
D.
E.
A fire warming up someone’s body who is sitting near
the fire
A pot of water heating up on the stove
Putting an ice pack on a burn
Light from a lamp shines on you and warms you up.
Holding a mug filled with coffee to warm up your
hands
Today’s Agenda
5 min
• Do Now
5 min
• Important Dates
25 min
• First Law of Thermodynamics
20 min
• Second Law of Thermodynamics
5 min
• Exit Ticket
Today’s Agenda
5 min
• Do Now
5 min
• Important Dates
25 min
• First Law of Thermodynamics
20 min
• Second Law of Thermodynamics
5 min
• Exit Ticket
Important Dates & Reminders


Homework This Week
Progress Reports
 Missing

Last Unit Exam Next Thursday
 Study

Guide Tomorrow
February Monthly News Project
 Due

Lab Notebooks
Next Thursday Feb. 28th
Congrats to both Girls’ and Boys’ Basketball!
Today’s Agenda
5 min
• Do Now
5 min
• Important Dates
25 min
• First Law of Thermodynamics
20 min
• Second Law of Thermodynamics
5 min
• Exit Ticket
First Law of Thermodynamics
The First Law of Thermodynamics basically states
that in a closed system, energy can neither be
created nor destroyed, only transformed or
transferred. There is an energy balance in the
universe.
What does this sound a lot like???
Connection to Heat

Whenever heat is added to a system, it transforms
to an equal amount of some other form of energy.
 That
form of usable energy is WORK.
Conservation of Energy

Some visual examples of this principle
st
1
Law Equation
Q = DE – W
Q = (EF– EI) -W
Q = heat added TO THE SYSTEM
DE = D in internal energy
W = Work done ON THE SYSTEM
If the system does work, W is NEGATIVE
Example

A total of 135 J of work is done on a gaseous
refrigerant as it undergoes compression. If the
internal energy of the gas increases from 0 J
to 156 J during the process, what is the total
amount of energy removed from the gas by
heat?
Example Answer

A total of 135 J of work is done on a gaseous
refrigerant as it undergoes compression. If the
internal energy of the gas increases from 0 J
to 156 J during the process, what is the total
amount of energy removed from the gas by
heat?
Q
= ΔE – W
 Q = 156 J – 135 J
 Q = 21 J
Guided Practice

An object has an initial internal energy of 48 J. The
internal energy increases to 73 J when 38 J of work
is done. What is the amount of energy in the form
of heat that was gained by the system during this?
Guided Practice Answer

An object has an initial internal energy of 48 J. The
internal energy increases to 73 J when 38 J of work
is done. What is the amount of energy in the form
of heat that was gained by the system during this?
Q
= ΔE – W
Q
Q
= (EF – EI) – W
= (73 J – 48 J) – 38 J
 Q = -13 J
Step Up

The internal energy of a system is initially 27 J.
After 33 J of heat is added to the system, the
internal energy is measured to be 86 J. How much
work was done on the system?
Step Up Answer

The internal energy of a system is initially 27 J.
After 33 J of heat is added to the system, the
internal energy is measured to be 86 J. How much
work was done on the system?
Q
= ΔE – W
Q
 33
= (EF – EI) – W
J = (86 J – 27 J) – W
 33
J = 59 J – W
 W = 59 J – 33 J
W
= 26 J
Step Further

The internal energy of a system is initially 45 J. A
total of 28 J of energy is added to the system by
heat while the system does -31 J of work. What is
the system’s final internal energy?
Q
= ΔE – W
Q
= (EF – EI) – W
Step Further Answer

The internal energy of a system is initially 45 J. A
total of 28 J of energy is added to the system by
heat while the system does -31 J of work. What is
the system’s final internal energy?
Q
= ΔE – W
Q
 28
= (EF – EI) – W
J = (EF – 45 J) – (-31) J
 28
J = (EF – 45 J) + 31 J
 -3 J = EF – 45 J
 EF
= 42 J
Practice Problems
Complete the practice problems on the Daily
Organizer under First Law of Thermodynamics
Today’s Agenda
5 min
• Do Now
5 min
• Important Dates
25 min
• First Law of Thermodynamics
20 min
• Second Law of Thermodynamics
5 min
• Exit Ticket
Second Law of Thermodynamics
The entropy of the universe increases in all natural
processes and reactions.
Entropy is the measure of a system’s disorder.
Second Law of Thermodynamics


The entropy, or disorder, of the universe increases in
all natural processes.
After cleaning your room, it always has a tendency
to become messy again.
 This
is a result of the second law.
 To
understand disorder more, think about gas in a jar that is
suddenly released. What is going to happen?

As the disorder in the universe increases, the energy
is transformed into less usable forms
Calculating Entropy

When a body absorbs
an amount of heat Q
from a reservoir at
temperature T, the
body gains and the
surroundings lose an
amount of entropy
 Positive
S: Entropy
Increased
 Negative S: Entropy
Decreased
Example 1

A system has 91 J of heat transferred to it raising
its temperature 15 °C. What was the increase in
entropy?
Example 1 Answer

A system has 91 J of heat transferred to it raising
its temperature 15 °C. What was the increase in
entropy?
 ΔS
= Q/ΔT
 ΔS = 91 J/15 °C
 ΔS = 6.1 J/°C
Guided Practice

A system starts out at 56 °C. It loses 117 J of heat
causing its temperature to decrease to 11 °C. What
was the decrease in entropy?
Guided Practice Answer

A system starts out at 56 °C. It loses 117 J of heat
causing its temperature to decrease to 11 °C. What
was the decrease in entropy?
 ΔS
= Q/ΔT
 ΔS = 17 J/(11 °C – 56 °C)
 ΔS = -2.6 J/°C
Step Up

When a system increases its temperature from 14
°C to 29 °C, its entropy also increases by 9 J/°C.
How much heat energy was transferred into the
system?
Step Up Answer

When a system increases its temperature from 14
°C to 29 °C, its entropy also increases by 9 J/°C.
How much heat energy was transferred into the
system?
 ΔS
= Q/ΔT
 9 J/°C = Q/(29 °C – 14 °C)
 Q = 135 J
Step Further

When a system initially at 59 °C has 65 J of heat
transferred into it, its entropy increases by 14 J/°C.
What was the final temperature of the system?
Step Further Answer

When a system initially at 59 °C has 65 J of heat
transferred into it, its entropy increases by 14 J/°C.
What was the final temperature of the system?
 ΔS
= Q/ΔT
 14 J/°C = 65 J/(TF – 59 °C)
 TF
– 59 °C = 65 J/(14 J/°C)
 TF – 59 °C = 4.3 °C

 TF
Add 59 to both sides!
= 63.3 °C
Practice Problems
Complete the practice problems on the Daily
Organizer under Second Law of
Thermodynamics
Today’s Agenda
5 min
• Do Now
5 min
• Important Dates
25 min
• First Law of Thermodynamics
20 min
• Second Law of Thermodynamics
5 min
• Exit Ticket