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Pizza, Hot Chocolate and
Newton’s Law of Cooling
Jason Colombino
Salem High School
Master Teacher Project
Entry Ticket
a.
b.

Identify the initial amount and growth rate for each function.
Evaluate each function for t = 5

1.

f(t) = 2(1.05)t

2. f(t) = 123(1 – 0.9)t

3. f(t) = e2t

Turn and Talk: Compare and contrast your reasoning and answers with a
partner.
e stands for Euler
• e is a constant and is an irrational number.
• Called “Euler’s constant” named after Leohard
Euler
• Approximately equal to 2.718….
• e is the base of natural logarithms (ln)
• So ln(e) = 1 just like the log 10 = 1
Sir Issac Newton
Newton’s Law of Cooling
T(t) = Ts + (T0 – Ts)e-kt
T(t) = temperature at time t
Ts = temperature of surrounding
environment
T0 = initial temperature of object
k = constant (different for each object)
Example: Ouch, that’s hot!!!
Joe and Jessica make pizza at his house. When they take the pizza out of the
oven, it has an initial temperature of 400° F. The thermometer at Joe’s house
reads 70° F. After 1 minute, Joe takes a bite and burns his mouth (the pizza is
still 350° F). How long should Joe and Jessica wait to be able to eat the pizza
at a reasonable temperature of 120°?
Turn and Talk: Identify the known and unknown variables –

T(t) =
Ts =
T0 =
k=
Cooling Pizza
So, how do we find the value of k?
Adding a constant
Turn and Talk: What do you think will happen if
we adjust the constant Ts (for example, what
happens to the time it takes the pizza to cool
if Joe’s house was warmer or cooler)?
Group Work: Changing the Surrounding
Temperature
• How long will it take for the pizza to cool to 120°
if . . . .
a. It is placed in the refrigerator (40°) and
b. It is placed in the freezer (0°)
Summing it Up: Describe the effect of changing the
constant surrounding temperature on the cooling
of
the pizza. Was your initial hypothesis correct?
Elaborate.
Group Work: Cold Pizza
•

Joe’s friend Sam really likes frozen pizza for some reason. The problem is
the pizza just came out of the oven and is a scorching 400°.

•

Joe and Same decide to immediately put the pizza in the freezer, which is
set at a frigid 0° F (assume k is the same constant we found in the first
pizza example).

1. How long do Joe and Sam have to wait until the pizza is a nice and cold
45°F?
2. How long would they have to wait if they put the pizza in a refrigerator set
at 40°?
3. Sketch a graph for all three scenarios (class example, and numbers 1 and
2 from this example). Describe the effect of changing the surrounding
temperature on the pizza.
Exit Ticket: Hot Chocolate
• Ted is at the football game and gets a cup of hot
chocolate that is initially 190°F. It is a cold night
for football, as the scoreboard indicates that it is
15°. After 1 minute, the hot chocolate is now
140°.
• Approximately how long (to the nearest minute)
does Ted have to drink the hot chocolate before
it gets too cold (Ted won’t drink the hot
chocolate once it gets to 65°F)?

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Pizza, Hot Chocolate and Newton's Law of Cooling

  • 1. Pizza, Hot Chocolate and Newton’s Law of Cooling Jason Colombino Salem High School Master Teacher Project
  • 2. Entry Ticket a. b. Identify the initial amount and growth rate for each function. Evaluate each function for t = 5 1. f(t) = 2(1.05)t 2. f(t) = 123(1 – 0.9)t 3. f(t) = e2t Turn and Talk: Compare and contrast your reasoning and answers with a partner.
  • 3. e stands for Euler • e is a constant and is an irrational number. • Called “Euler’s constant” named after Leohard Euler • Approximately equal to 2.718…. • e is the base of natural logarithms (ln) • So ln(e) = 1 just like the log 10 = 1
  • 5. Newton’s Law of Cooling T(t) = Ts + (T0 – Ts)e-kt T(t) = temperature at time t Ts = temperature of surrounding environment T0 = initial temperature of object k = constant (different for each object)
  • 6. Example: Ouch, that’s hot!!! Joe and Jessica make pizza at his house. When they take the pizza out of the oven, it has an initial temperature of 400° F. The thermometer at Joe’s house reads 70° F. After 1 minute, Joe takes a bite and burns his mouth (the pizza is still 350° F). How long should Joe and Jessica wait to be able to eat the pizza at a reasonable temperature of 120°? Turn and Talk: Identify the known and unknown variables – T(t) = Ts = T0 = k=
  • 7. Cooling Pizza So, how do we find the value of k?
  • 8. Adding a constant Turn and Talk: What do you think will happen if we adjust the constant Ts (for example, what happens to the time it takes the pizza to cool if Joe’s house was warmer or cooler)?
  • 9. Group Work: Changing the Surrounding Temperature • How long will it take for the pizza to cool to 120° if . . . . a. It is placed in the refrigerator (40°) and b. It is placed in the freezer (0°) Summing it Up: Describe the effect of changing the constant surrounding temperature on the cooling of the pizza. Was your initial hypothesis correct? Elaborate.
  • 10. Group Work: Cold Pizza • Joe’s friend Sam really likes frozen pizza for some reason. The problem is the pizza just came out of the oven and is a scorching 400°. • Joe and Same decide to immediately put the pizza in the freezer, which is set at a frigid 0° F (assume k is the same constant we found in the first pizza example). 1. How long do Joe and Sam have to wait until the pizza is a nice and cold 45°F? 2. How long would they have to wait if they put the pizza in a refrigerator set at 40°? 3. Sketch a graph for all three scenarios (class example, and numbers 1 and 2 from this example). Describe the effect of changing the surrounding temperature on the pizza.
  • 11. Exit Ticket: Hot Chocolate • Ted is at the football game and gets a cup of hot chocolate that is initially 190°F. It is a cold night for football, as the scoreboard indicates that it is 15°. After 1 minute, the hot chocolate is now 140°. • Approximately how long (to the nearest minute) does Ted have to drink the hot chocolate before it gets too cold (Ted won’t drink the hot chocolate once it gets to 65°F)?

Notas del editor

  1. Show 2 short videos: first on Newton biography (optional) AND 2. overview of Newton’s Law of Cooling.
  2. k in this example is approximately equal to -0.16
  3. Assume same value for k = -.16
  4. K equals approximately 0.3365