Download Resistive Heating - Physics 420 UBC Physics Demonstrations

April 13th, 2010
Mark
4th Year Physics student at UBC,
Graduated at Riverside Secondary in
Port Coquitlam
Basketball, Soccer, Music, and Sailing
Burns Bread
Creates Light
Transmits Electricity
Hot Dog…..
Hot Dog…..
Tastes Delicious 

Toaster…

Light Bulb…

Power Lines…

Hot Dog…

In this presentation, I
plan to explain the
concept that connects
each one of these four
objects

What is Resistive Heating?
Who discovered Resistive Heating?
What causes Resistive Heating?

A little Experiment







Introduction
Experimentation
Calculations
Conclusion
Conclusion

“The process by which a current is passed
through a conductor and heat is released”
Wikipedia

“The process by which a current is passed
through a conductor and heat is released”
Wikipedia

Also Known as:




Resistive Heating
Joule Heating
Ohmic Heating
Electric Heating

“The process by which a current is passed
through a conductor and heat is released”
Wikipedia

Also Known as:




Resistive Heating
Joule Heating
Ohmic Heating
Electric Heating
All names mean the same
thing but we will be using
this name, ‘Joule Heating’
from here on, named after
its discoverer.

First experimented with in 1841

First experimented with in 1841

James Prescott Joule

Hence the name: ‘Joule Heating’

First experimented with in 1841

James Prescott Joule


Hence the name: ‘Joule Heating’
What did he do?



Passed a current through a wire of fixed length,
Immersed in a bath of water of fixed volume/mass
Observed the temperature in the water varried with
current, length of wire, and time.

What did he do?




Passed a current through a wire of fixed length,
Immersed in a bath of water of fixed volume/mass
Observed the temperature in the water varried with
current, length of wire, and time.
He came up with the relation:
Q  I  R t
2
Where:
Q = amount of Energy produced (in the form
of heat) UNITS: joules
1 joule = kg m /s2
I = Current UNITS: amperes
R = Resistance UNITS: ohms
t = Time UNITS: seconds
Q  I  R t
2


In a complete circuit, electrons move around
the circuit.
Often these electrons ‘collide’ with ions, and
the electrons share kinetic energy with the ions



Ion: a charged atom, ie: NaCl+ (Salt)
The kinetic energy excites the ions, and makes
them vibrate.
This increase in kinetic energy displays itself as
heat and a temperature increase occurs.


In a complete circuit, electrons move around
the circuit.
Often these electrons ‘collide’ with ions, and
the electrons share kinetic energy with the ions




Ion: a charged atom, ie: NaCl+ (Salt)
The kinetic energy excites the ions, and makes
them vibrate.
This increase in kinetic energy displays itself as
heat and a temperature increase occurs.
SUMMARY: Energy is transfered from the
electrical power supply to the conductor and
any materials with which it is in thermal
contact.

Show “Battery-Resistor Circuit”

What do the blue spheres represent?

What do the green spheres represent?

Why do the green and blue spheres interact?

What changes when the voltage is increased?
 WHY?

So what does this have to do with:

Burnt Toast

Light Bulbs

Power transmission lines

Hot Dogs???

Burnt Toast?

Inside the toaster are small wires called ‘filaments’,
 The elements heat up when a current is passed through
them and so it cooks/burns your toast.

Burnt Toast?

Inside the toaster are small wires called ‘filaments’,
 The elements heat up when a current is passed through
them and so it cooks/burns your toast.

Light Bulbs?

Like the toaster, the light bulb also has a filament.
 The filament is made of such a material that it can
withstand extreme temperatures and glow brightly.
 This material is usually ‘Tungsten’ which has a melting
temperature of 3880 C

Power Transmission Wires?

The wires are required to transmit large
amounts of electricity over long distances
The amount of heat lost due to Joule Heating
must be reduced as much as possible to
minimize energy lost to heat

We know:

V=IR



Power Transmission Wires?
So if we maximize V, then we can minimize I, and
so minimize Q
Transformers are used to step up the voltage and
reduce the current, while keeping the resistance
constant
Vlarge = Ismall R
If ‘I’ is reduced by ½, then energy lost can be
reduced by a 4!!!
And so: Q = Ismall2 x R x t



Power Transmission Wires?
The power lines used to transport power large
distances have VERY large voltages….
110 kV – 230 kV

Could you use the concepts of joule heating to
cook a hot dog?


Could you use the concepts of joule heating to
cook a hot dog?
OF COURSE!!!...... Why?


Could you use the concepts of joule heating to
cook a hot dog?
Hot Dogs are delicious, and so they contain
salt:

Remember: Salt = NaCl+ which is an ion!


We will place a Hot Dog in between two
electrodes and apply a voltage to cook the hot
dog.
We perform 3 quick experiments:



1) one hot dog
2) two hot dogs, connected in a parallel circuit
3) two hot dogs, connected in a series circuit

We will be using “Schneiders All Beef Wieners”

110 calories, 9g fat, 4g protein, and 380 mg SODIUM 
“Taste the difference quality makes!”

Lets get started: first: cook one hot dog at a
time.

We will be timing the procedure, from the moment
the voltage is applied, till roughly when the meat
stops cooking.
Time
Trial 1 Current
Trial 2 Current
Avg. Current
Avg. Resistance
0
1.5
1.5
1.5
80
30
1.94
2.03
1.985
60.4534005
40
2.27
2.57
2.42
49.58677686
50
2.62
3.09
2.855
42.03152364
60
3.04
3.3
3.17
37.85488959
70
3.3
3.45
3.375
35.55555556
80
3.41
3.36
3.385
35.45051699
90
3.32
3.18
3.25
36.92307692
100
3.04
2.98
3.01
39.86710963
110
2.89
2.87
2.88
41.66666667
120
2.8
2.77
2.785
43.08797127
130
2.68
2.66
2.67
44.94382022
140
2.6
2.57
2.585
46.42166344
145
2.56
2.52
2.54
47.24409449
150
2.51
2.47
2.49
48.19277108
155
2.46
2.44
2.45
48.97959184
160
2.41
2.41
2.41
49.79253112
165
2.33
2.37
2.35
51.06382979
170
2.27
2.32
2.295
52.2875817
175
2.18
2.27
2.225
53.93258427
180
2.05
2.21
2.13
56.33802817
185
1.95
2.13
2.04
58.82352941
190
1.8
2.05
1.925
62.33766234
195
0.9
1.93
1.415
84.80565371
200
0.17
1.67
0.92
130.4347826
205
0.11
1.02
0.565
212.3893805
210
0.09
0.2
0.145
827.5862069
215
0.08
0.13
0.105
1142.857143
220
0.08
0.11
0.095
1263.157895
225
0.07
0.1
0.085
1411.764706
230
0.07
0.09
0.08
1500
235
0.07
0.08
0.075
1600
Time
Trial 1 Current
Trial 2 Current
Avg. Current
Avg. Resistance
0
1.5
1.5
1.5
80
30
1.94
2.03
1.985
60.4534005
40
2.27
2.57
2.42
49.58677686
50
2.62
3.09
2.855
42.03152364
60
3.04
3.3
3.17
37.85488959
70
3.3
3.45
3.375
35.55555556
80
3.41
3.36
3.385
35.45051699
90
3.32
3.18
3.25
36.92307692
100
3.04
2.98
3.01
39.86710963
110
2.89
2.87
2.88
41.66666667
120
2.8
2.77
2.785
43.08797127
130
2.68
2.66
2.67
44.94382022
140
2.6
2.57
2.585
46.42166344
145
2.56
2.52
2.54
47.24409449
150
2.51
2.47
2.49
48.19277108
155
2.46
2.44
2.45
48.97959184
160
2.41
2.41
2.41
49.79253112
165
2.33
2.37
2.35
51.06382979
170
2.27
2.32
2.295
52.2875817
175
2.18
2.27
2.225
53.93258427
180
2.05
2.21
2.13
56.33802817
185
1.95
2.13
2.04
58.82352941
190
1.8
2.05
1.925
62.33766234
195
0.9
1.93
1.415
84.80565371
200
0.17
1.67
0.92
130.4347826
205
0.11
1.02
0.565
212.3893805
210
0.09
0.2
0.145
827.5862069
215
0.08
0.13
0.105
1142.857143
220
0.08
0.11
0.095
1263.157895
225
0.07
0.1
0.085
1411.764706
230
0.07
0.09
0.08
1500
235
0.07
0.08
0.075
1600
QHD
1
The meat stops cooking
because the connection at
the electrodes gets burned
out!

RTotal
RTotal
1
1


R HD R HD
1
 R HD 
2
V  I RTotal
 weiner contains  V 
The moisture in the
2



the
Ions,
NaCl,
and when the ITotal  2
V
2
 ITotal Rmoisture
R HD and evapourates, the RHD 
boils
HD  
 V 2RHD 
hotdog stops
 cooking
Q


RHD 
HD

V=I R, thus using only the measured current
and the known voltage, we can find the
resistance
Iave = 2.505mAmps, 0.002505 Amps

V = 120 Volts

V/I = 120V/0.002505 Amps = 47904 Ohms

= 47.9 kOhms
Thus: The average resistance of a hotdog is ~50kOhms.
V  I R
V
I 
R HD


2


V
2
QHD  I R HD t  
 R HD t
R HD 

 V 2 

t  QHD
RHD 
For one wiener


The average resistance of a hot dog is 50
kOhms
What does this number depend on?


The average resistance of a hot dog is 50
kOhms
What does this number depend on?
Size of hotdog
 Moisture content
 Temperature
 Fat content (other brands, varieties)
 Meat type (Chicken, Beef, Turkey…)


Parallel Circuits:

1
Total resistance is calculated:

RTotal
1
1


R HD R HD

Parallel Circuits:

1
Total resistance is calculated:
RTotal
1
1


R HD R HD
1
RTotal

1
 R HD
2
Rtotal  25kOhm


Parallel Circuits:

Rtotal  25kOhm
Total Current is calculated:
V  I RTotal

ITotal

 V  120V
 

RHD  25kOhm
ITotal  0.0048Amps



Parallel Circuits:

Total Current is calculated:
ITotal  0.0048Amps


This is a different current than when cooking only one
weiner.
 Does this make sense???
Ione weiner = 2.505mAmps
 Itwo weiners, parallel = 4.8mAmps

Parallel Circuits:
 Does this make sense???
Ione weiner = 0.002505Amps
ITotal  0.0048Amps
We know by the current law for parallel circuits
that:
Itotal = I1 + I2

So really: each hot dog is receiving the same
current as it was before (when there was only one
weiner cooking)

Parallel Circuits:
Itotal = I1 + I2
Thus: the current law justifies that the cooking
time is the same as the previous experiment….
But what happens when the hotdogs are connected
in series?

Series Circuits:

Total resistance is calculated:
RTotal  RHD  RHD  2RHD


Series Circuits:

Total resistance is calculated:
RTotal  RHD  RHD  2RHD
V  I RTotal
V  I 2RHD



V
I
2R HD

Series Circuits:

Total current is calculated:
V
I
2R HD
120V
 0.0012Amps
2(50kOhm)



Series Circuits:
1
Itotal  I1
2
Thus: the resistance law for series circuits justifies
that the cooking time is at least double what it was
in the previous
experiment….

Why more than double????

Series Circuits:
1
Itotal  I1
2
Thus: the resistance law for series circuits justifies
that the cooking time is at least double what it was
in the previous experiment….

Remember Joule’s equation for heat:
Q  I  R t
2

Series Circuits:
1
Itotal  I1
2
Remember Joule’s equation for heat:
Q  I  R t
2
 in the new Itotal:
If we plug
1 
Q   I1   R t
2 
2


Series Circuits:
1 
Q   I1   R t
2 
2
Which is the same as:
1 2
Q  I1  R t
4
the heat absorbed is actually ¼ the amount of
So
heat as in the first case, so it will take ~4 times as
long to cook the hot dog.

We Cooked hot dogs:
three times:
One at a time
 Two in parallel
 Two in series


We achieved 2
different results:

One at a time
 ~3min, 15 sec to cook

Two in parallel:
 ~3min, 15 sec to cook

Two in series:
 ~13 mins to cook

The concept of joule heating is used often:

In everyday kitchen appliances,
 Toaster, Iron, Electric Coffee Maker…

The same idea is used to Transport electricity, and
minimize loss by maximizing voltage

Also can be used to cook hotdogs!
But only in parallel circuitry, as series is a waste of
time!

How much $$$ did it cost us to cook one hot
dog?

ASSUME:
 Voltage used: 120V
 Average current: 0.002505 Amps
 Time to cook: 3 minutes, 15 seconds
 Average cost of electricity in BC: $0.06724 / kWh

ASSUME:
 Voltage used: 120V
 Average current: 0.002505 Amps
 Time to cook: 3 minutes, 15 seconds
 Average cost of electricity in BC: $0.06724 / kWh
 1 Watt = 1 Volt x 1 Amp

ASSUME:
 Voltage used: 120V
 Average current: 0.002505 Amps
 Time to cook: 3 minutes, 15 seconds
 Average cost of electricity in BC: $0.06724 / kWh
 1 Watt = 1 Volt x 1 Amp
 1 kilo-Watt-hour = the amount of energy in kilo Watts
consumed per hour

ASSUME:
 Voltage used: 120V
 Average current: 0.002505 Amps
 Time to cook: 3 minutes, 15 seconds
 Average cost of electricity in BC: $0.06724 / kWh
 1 Watt = 1 Volt x 1 Amp
 1 kilo-Watt-hour = the amount of energy in kilo Watts consumed per hour
3 minutes, 15 seconds = 3.25 minutes
3.25 minutes / 60 minutes = 0.0542 Hours

ASSUME:
 Voltage used: 120V
 Average current: 0.002505 Amps
 Time to cook: 3 minutes, 15 seconds
 Average cost of electricity in BC: $0.06724 / kWh
 1 Watt = 1 Volt x 1 Amp
 1 kilo-Watt-hour = the amount of energy in kilo Watts consumed per hour
 3.25 minutes / 60 minutes = 0.0542 Hours
TOTAL COST = Voltage x Current x Time x Cost per
hour / 1000
 Voltage used: 120V
 Average current: 0.002505 Amps
 Time to cook: 3 minutes, 15 seconds
 Average cost of electricity in BC: $0.06724 / kWh
 1 Watt = 1 Volt x 1 Amp
 1 kilo-Watt-hour = the amount of energy in kilo Watts consumed per hour
 3.25 minutes / 60 minutes = 0.0542 Hours
TOTAL COST = Voltage x Current x Time x Cost per
hour / 1000
= 120V x 0.002505Amps x 0.0542 hours x $0.06724
/ 1000
 Voltage used: 120V
 Average current: 0.002505 Amps
 Time to cook: 3 minutes, 15 seconds
 Average cost of electricity in BC: $0.06724 / kWh
 1 Watt = 1 Volt x 1 Amp
 1 kilo-Watt-hour = the amount of energy in kilo Watts consumed per hour
 3.25 minutes / 60 minutes = 0.0542 Hours
TOTAL COST = Voltage x Current x Time x Cost per
hour / 1000
= 120V x 0.002505Amps x 0.0542 hours x $0.06724
/ 1000
=$0.000001096 or 0.0001096 cents…… DIRT CHEAP
=$0.000001096 or 0.0001096 cents…… DIRT CHEAP
Because of BC’s abundant hydroelectricity generation.
Ontario’s power costs are double ours, and california’s
are more than TRIPPLE!

We found that the resistance of the hot dog is
~50kOhms
Why?
What factors contribute to the resistance?

We found that the resistance of the hot dog is
~50kOhms
Why?
What factors contribute to the resistance?
-Cross sectional area
-Length
-Density
-Ionic Content

What would happen if the hot dog were a
different shape?
 PhET simulation #2

What would happen if the hot dog were a
different shape?

What would be a better conductor?
 A hot dog or a copper wire?
Why?

What would happen if the hot dog were a
different shape?

What would be a better conductor?
 A hot dog or a copper wire?
Why?
The Wire: It has more free ions, and so conducts
better

What would conduct better?

A hot dog, or a LED (Light Emitting Diode)

What would conduct better?


A hot dog or an LED?
How can we test this?

If we plug both the hot dog and the LED into the
same circuit, the current should take the path of least
resistance

What would conduct better?


A hot dog or an LED?
How can we test this?

If we plug both the hot dog and the LED into the
same circuit, the current should take the path of least
resistance
TIME TO TEST THIS:

Place 1 LED in hotdog, electrodes ~2mm apart.

Place 1 LED in hotdog, electrodes ~2mm apart.

Place 1 LED in hotdog, electrodes ~2mm apart.

Move LED electrodes to ~3 cm apart.

Will anything be different?

Move LED electrodes to ~3 cm apart.
Hotdog electrodes are ~70mm apart, hot dog is
~50kOhms of resistance, 0.002505 Amps Current, 120
Volts.
 When LED electrodes are 2mm apart:
 2mm/70mm = 0.0286
 0.0286 x 120V = 3.44V

LED’s are usually rated at 5 V, so anything wider
than ……..
WHAT ELECTRODE DISTANCE WILL DAMAGE
THE LED?
Hotdog electrodes are ~70mm apart, hot dog is
~50kOhms of resistance, 0.002505 Amps Current, 120
Volts.
 When LED electrodes are 2mm apart:
 2mm/70mm = 0.0286
 0.0286 x 120V = 3.44V

LED’s are usually rated at 5 V, so anything wider
than a 2.917 mm electrode seperation will damage
the LED.
5.0V/120V x 70mm = 2.917mm
SUMMARY:
Joule heating is the process that does basic things
like:
- cook toast,
- heats an electric kettle
- controls loss of long power transmissions
- lights lightbulbs
- COOKS HOTDOGS!
SUMMARY:
Joule heating is the process that does basic things
like:
- COOKS HOTDOGS!
It works by exciting the charged molecules inside of a
resistor and increasing their kinetic energy. This
energy is released in the form of heat.