Transcript Document 7829788

Orders of Magnitude
Order of Magnitude – the common logarithm of a positive
quantity.
Examples
Mercury is about 5.79 x 10 10 meters from the Sun
Pluto is about 5.9 x 1012 meters from the Sun
Find the common logarithm of both numbers, and then
calculate the difference between these values…
 Pluto’s distance from the Sun is 2 orders of
magnitude greater than Mercury’s
Orders of Magnitude
Order of Magnitude – the common logarithm of a positive
quantity.
Examples
Determine how many orders of magnitude the quantities differ.
1. A kilometer and a meter
3
2. A $1 dollar bill and a penny
2
3. A horse weighing 400 kg and a mouse weighing 40 g
4. NYC with 7 million people and Earmuff Junction with
a population of 7  6
4
Application: Decibels
The level of sound intensity in decibels (dB) is
 I 
  10log  
 I0 
where  (beta) is the number of decibels, I is the sound
intensity in W/m 2, and I0 = 10 –12 W/m 2 is the threshold of
human hearing (the quietest audible sound intensity).
Application: Decibels
 I 
  10log  
The sound of a subway train is 100 dB, and the sound  I 0 
The level of sound intensity in decibels (dB) is
of a soft whisper is 10 dB. By how many orders of
magnitude do these quantities differ?
I1
I2
1  10 log  100  2  10 log  10
I0
I0
We seek the logarithm of the ratio I I
1
2
Application: Decibels
I1
I2
1  10 log  100  2  10 log  10
I0
I0
I1
I2
1   2  10 log  10 log  100  10
I0
I0
I1
10 log  90
I2
The two sound intensities
I1
 log  9 differ by 9 orders of magnitude!!!
I2
Application: Richter Scale
The Richter Scale magnitude R of an earthquake is
a
R  log  B
T
where a is the amplitude in micrometers of the vertical ground
motion at the receiving station, T is the period of the associated
seismic wave in seconds, and B accounts for the weakening of
the seismic wave with increasing distance from the epicenter
of the earthquake.
Application: R  log a  B
Richter Scale
T
How many more times severe was the 2001 earthquake it
Gujarat, India (R 1 = 7.9) than the 1999 earthquake in Athens,
Greece (R 2 = 5.9)?
a2
a1
R1  log  B  7.9 R2  log  B  5.9
T
T
We seek the ratio of severities a1 a2
a1
a2

 

R1  R2   log  B    log  B 
T
T

 

Application: R  log a  B
Richter Scale
T
a1
a2

 

R1  R2   log  B    log  B 
T
T




a1
a2
log  log  7.9  5.9
T
T
A Richter scale difference of 2
a1
corresponds to an amplitude ratio
log  2 of 100  The Gujarat quake was 100
a2
times as severe as the Athens quake!!!
 a1 a2  10  100
2
Application:
Chemical Acidity
The measure of acidity is pH, the opposite of the common log
of the hydrogen-ion concentration of a solution:
pH   log  H 

Note: More acidic solutions have higher hydrogen-ion
concentrations, and therefore have lower pH values…
Application: pH   log  H 
Chemical Acidity

Some especially sour vinegar has pH of 2.4, and a box of
baking soda has a pH of 8.4.
(a) What are their hydrogen-ion concentrations?
Vinegar:


 log  H   2.4
 H   10

Baking Soda:
2.4
 3.98110
 log  H    8.4
 H    10 8.4


log  H   2.4
3 moles
per liter


log  H   8.4
 3.98110
9 moles
per liter
Application: pH   log  H 
Chemical Acidity

Some especially sour vinegar has pH of 2.4, and a box of
baking soda has a pH of 8.4.
(b) How many times greater is the hydrogen-ion concentration
of the vinegar than that of the baking soda?
 H 

 H  
of vinegar
of baking soda
2.4
10
6
 8.4  10
10
times
greater
Application: pH   log  H 
Chemical Acidity

Some especially sour vinegar has pH of 2.4, and a box of
baking soda has a pH of 8.4.
(c) By how many orders of magnitude do the concentrations
differ?
The hydrogen-ion concentration of the vinegar is 6 orders
of magnitude greater than that of the baking soda…
This is exactly the difference in their pH values!!!
Application:
Newton’s Law of Cooling
An object that has been heated will cool to the temperature of
the medium in which it is placed (such as the surrounding air
or water). The temperature T of the object at time t can be
modeled by Newton’s Law of Cooling:
T  t   Tm  T0  Tm  e
Temp. of the
surrounding
medium
Initial Temp.
of the object
 kt
A constant
Application: T t   Tm  T0  Tm  ekt
Newton’s Law of Cooling
A hard-boiled egg at temperature 96 C is placed in 16 C water
to cool. Four minutes later the temperature of the egg is 45 C.
Use Newton’s Law of Cooling to determine when the egg will
be 20 C.
Identify terms: T0  96 Tm  16
Plug into equation:
T  t   16  80e
 kt
Use the point (4, 45) to solve for k:
T  4  16  80e
4 k
 45
Application: T t   Tm  T0  Tm  ekt
Newton’s Law of Cooling
A hard-boiled egg at temperature 96 C is placed in 16 C water
to cool. Four minutes later the temperature of the egg is 45 C.
Use Newton’s Law of Cooling to determine when the egg will
be 20 C.
T  4  16  80e
e
4 k
4 k
 29 80
4k  ln  29 80 
 45
ln  29 80 
k
4
 0.253
Save this value in
your calculator!!!
Application: T t   Tm  T0  Tm  ekt
Newton’s Law of Cooling
A hard-boiled egg at temperature 96 C is placed in 16 C water
to cool. Four minutes later the temperature of the egg is 45 C.
Use Newton’s Law of Cooling to determine when the egg will
be 20 C.
Finally, solve this equation:
20  16  80e kt
e
 kt
 1 20
kt  ln 1 20
ln 1 20 
t
k
 11.809
The temp. of the egg will be
20 C after about 11.8 minutes