Physics 2B: Lecture 3 Secs. 11.7-11.9

heat pumps

entropy

Example 3.1Suppose a single cycle of the heat engine below takes .3 seconds, how much work is done after 10 minutes?

Example 3.1Suppose a single cycle of the heat engine below takes .3 seconds, how much work is done after 10 minutes?

10 min = 10 min⇣60 sec

1 min

⌘= 600 sec

Step 1. convert total time from mins to secs.

Example 3.1Suppose a single cycle of the heat engine below takes .3 seconds, how much work is done after 10 minutes?

10 min = 10 min⇣60 sec

1 min

⌘= 600 sec

Step 1. convert total time from mins to secs.

Step 2. determine how many cycles in 10 mins

Ncycles = 600/.3 = 2000

Example 3.1Suppose a single cycle of the heat engine below takes .3 seconds, how much work is done after 10 minutes?

10 min = 10 min⇣60 sec

1 min

⌘= 600 sec

Step 1. convert total time from mins to secs.

Step 2. determine how many cycles in 10 mins

Ncycles = 600/.3 = 2000

Step 3. find work done in 10 mins

Wdone in 10 mins

= Ncycles

Wout

= (2000)(50 J) = 10, 000 J

Example 3.1Suppose a single cycle of the heat engine below takes .3 seconds, how much work is done after 10 minutes?

10 min = 10 min⇣60 sec

1 min

⌘= 600 sec

Step 1. convert total time from mins to secs.

Step 2. determine how many cycles in 10 mins

Ncycles = 600/.3 = 2000

Step 3. find work done in 10 mins

Wdone in 10 mins

= Ncycles

Wout

= (2000)(50 J) = 10, 000 J

What’s the power (output) of this process?

Example 3.1Suppose a single cycle of the heat engine below takes .3 seconds, how much work is done after 10 minutes?

10 min = 10 min⇣60 sec

1 min

⌘= 600 sec

Step 1. convert total time from mins to secs.

Step 2. determine how many cycles in 10 mins

Ncycles = 600/.3 = 2000

Step 3. find work done in 10 mins

Wdone in 10 mins

= Ncycles

Wout

= (2000)(50 J) = 10, 000 J

What’s the power (output) of this process?

Answer: P = (50J)/(.3s) = 166.7 J/s

heat engines vs. heat pumps

heat engine heat pump

we simply reversed the arrows: work is done on the system to transport heat as opposed to heat being used for work

QH

QC

Wout

QH

QC

Win

refrigerators and heat pumps

a heat engine used heat to perform work

a heat pump uses work to transfer heat

in a single cycle, it looks like heat is going from cold to hot; indeed it is. this can occur because work is being done on the system

QH

QC

Win

there is no violation of the 2nd law of thermodynamics

refrigeratorsa fridge removes heat from the cold environment (the inside of your fridge) and exhausts heat into the hot environment (your apartment)

QH

QC

is the amount of work that must be done to transfer heat

instead of the efficiency, we characterize a fridge by its coefficient of performance

unlike the efficiency, the fridge COP can be any non-negative number

0 COPfridge < 1

TH

TC

QH

QC

Win

Win = QH �QC

Win

COPfridge =QC

Win=

QC

QH �QC

refrigerators

just like there was a maximally efficient heat engine — the Carnot heat engine — the Carnot fridge has the maximal fridge COP

COPfridge(Carnot) =TC

TH � TC

TH

TC

QH

QC

Win

Example 3.2Suppose we want to keep the Carnot freezer in your 20 C apartment at -2 C, what’s its fridge COP?

COPfridge(Carnot) =TC

TH � TC=

271 K

293 K � 271 K= 12.3

TH

TC

QH

QC

Win

heater

TH

TC

QH

QC

Win

a heater is a heat pump

in contrast to the fridge, we’re interested in how much heat is transported to the hot reservoir, rather than how much heat is removed from the cold one

COPheater =QH

Win=

QH

QH �QC

1 < COPheater < 1

TH

TC

QH

QC

Win

Carnot heater

a Carnot heater has the maximal heater COP

COPheater =TH

TH � TC

Example 3.3

QH

QC

W

TH

TC

TH

TC

QH

QC

W

Suppose we reversed the arrows in the heat transfer diagram of our heater to make a heat engine. How do the heater COP and efficiency compare? Take and . |QH | = 13 J |W | = 10 J

Example 3.3

QH

QC

W

TH

TC

TH

TC

QH

QC

W

Suppose we reversed the arrows in the heat transfer diagram of our heater to make a heat engine. How do the heater COP and efficiency compare? Take and . |QH | = 13 J |W | = 10 J

COPheater =QH

W=

1

1�QC/QH=

1

eStep 1:

Example 3.3

QH

QC

W

TH

TC

TH

TC

QH

QC

W

Suppose we reversed the arrows in the heat transfer diagram of our heater to make a heat engine. How do the heater COP and efficiency compare? Take and . |QH | = 13 J |W | = 10 J

COPheater =QH

W=

1

1�QC/QH=

1

eStep 1:

Step 2: evaluate COPheater =13 J

10 J= 1.3 e =

10 J

13 J= .8

entropy

QH

QC

W

TH

TC

let’s return to the heat engine (an equivalent discussion applies to the heat pump)

for a Carnot or ideal heat engine, we have two expressions for the efficiency that we can identify

1� QC

QH= e = 1� TC

TH

which we can rearrange to

QC

TC=

QH

TH

entropy S (quantification of the level of

“disorder” in a system)

QH

QC

W

TH

TC

QC

TC=

QH

TH

S =Q

Tentropy is conserved by a Carnot heat engine

heat energy is not conserved, the excess is used for work

Wout

= QH �QC = S(TH � TC)

Rudolf Clausius

T1 < T2

T1 T2

entropy and the 2nd of thermodynamics

we have learned that heat “always” goes from hot to cold (if no work is done)

What’s the change in entropy that quantifies this?

�S =Q

T1� Q

T2> 0

2nd law (third form): entropy change of the TOTAL macroscopic system is non-negative

entropy provides a quantitative measure of irreversibility

entropy and the 2nd of thermodynamics

T1 < T2

T1 T2

2nd law: entropy change of total system is non-negative

heat does NOT go from hot to cold because it would violate the 2nd law

violation of 2nd law �S = � Q

T1+

Q

T2< 0

entropy and the 2nd of thermodynamics

T1 < T2

T1 T2

2nd law: entropy change of total system is non-negative

heat does NOT go from hot to cold because it would violate the 2nd law

violation of 2nd law �S = � Q

T1+

Q

T2< 0

Example 3.4A block slides down an inclined plane at temperature T = 300 K. Without friction, its kinetic energy at the bottom is 10 J; with friction, the kinetic energy is 8 J. How does the entropy of the block and plane change when there is friction?

Answer: it increases (2nd law). How much?

Example 3.4A block slides down an inclined plane at temperature T = 300 K. Without friction, its kinetic energy at the bottom is 10 J; with friction, the kinetic energy is 8 J. How does the entropy of the block and plane change when there is friction?

Step 1: assume no change to the temperature

Answer: it increases (2nd law). How much?

Example 3.4A block slides down an inclined plane at temperature T = 300 K. Without friction, its kinetic energy at the bottom is 10 J; with friction, the kinetic energy is 8 J. How does the entropy of the block and plane change when there is friction?

Step 1: assume no change to the temperature

Step 2: 10 J - 8 J = 2 J is the energy lost to friction; we assume this all goes into heat Q

Answer: it increases (2nd law). How much?

Example 3.4A block slides down an inclined plane at temperature T = 300 K. Without friction, its kinetic energy at the bottom is 10 J; with friction, the kinetic energy is 8 J. How does the entropy of the block and plane change when there is friction?

Step 1: assume no change to the temperature

Step 2: 10 J - 8 J = 2 J is the energy lost to friction; we assume this all goes into heat Q

2nd law: entropy change of total system is non-negative

Step 3:

�S =Q

T=

2 J

300 K= .007 J/K > 0

Answer: it increases (2nd law). How much?

summary

QH

QC

W

TH

TC

TH

TC

QH

QC

Win

equivalent forms of the 2nd law:

heat engine heat pump

(i) entropy change of total system is non-negative

(ii) systems tend to become disordered

(iii) entropy of a system is non-negative

next lecture: Wednesday Jan 18

more on entropy and Sections 12.1 - 12.3

Monday Jan 16 is MLK Day — NO CLASS

Note: lots of definitions and new notation in the new sections