
Newton's Law of Cooling Problems
Authored by Charles Martinez
Mathematics
11th - 12th Grade
CCSS covered

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8 questions
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1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Given the equation T(t)=68e-0.0174t + 72 which represents the temperature after t minutes, what would the temperature be after 1.5 hours?
about 60o F
about 100o F
about 87o F
Wicked Cold
Tags
CCSS.HSF.LE.A.1
CCSS.HSF.LE.A.2
CCSS.HSF.IF.A.2
CCSS.HSF.LE.B.5
CCSS.HSN.Q.A.1
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Given the that a hot pizza cools according to the equation P(t)=350e-0.35t + 75 and Matt and Tyler like to eat their pizza when it is 110o F, how long will they have to wait to eat their pizza?
FOR-EV-ER
about 6.5 minutes
about 12.5 minutes
about 1 minute
Tags
CCSS.HSF.LE.A.1
CCSS.HSF.LE.A.4
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Which cooling equation correctly represents the following scenario:
Boiling soup is 100oF when it is taken off of the stove in a 69oF room. After 15 minutes the temperature of the soup is 95oF.
S(p)=31e-0.011726t + 69
S(p)=100e-0.011726t + 69
S(p)=31e-.05435t + 100
S(p)=31e-.05435t + 69
Tags
CCSS.HSF.LE.A.1
CCSS.HSF.LE.A.2
CCSS.HSF.BF.A.1
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Given the cooling equation S(t)=31e-0.011726t + 69 where t is the amount of minutes and S(t) represents the temperature the soup, how long would it take for the soup to get to 80oF?
about 100 minutes
about 46 minutes
about 57 minutes
about 88 minutes
Tags
CCSS.HSF.LE.A.1
CCSS.HSF.LE.A.2
CCSS.HSA.CED.A.1
CCSS.HSA.SSE.A.1
CCSS.HSF.BF.A.1
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Given the cooling equation S(p)=31e-0.011726t + 69 where t is the amount of minutes and S(p) represents the temperature the soup, what would the temperature of the soup be after 2 and a half hours?
about 64oF
about 74oF
about 84oF
about 94oF
Tags
CCSS.HSF.LE.A.1
CCSS.HSF.LE.A.2
CCSS.HSA.CED.A.1
6.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Which cooling equation correctly represents the following scenario:
A cooked turkey is 165oF when it is taken out of the oven in a 75oF room. After 30 minutes the temperature of the turkey is 145oF.
T(t)=90e-0.148377t + 75
T(t)=90e-0.008377t + 75
T(t)=75e-0.148377t + 90
T(t)=75e-0.008377t + 90
Tags
CCSS.HSF.LE.A.1
CCSS.HSF.LE.A.2
CCSS.HSA.CED.A.1
CCSS.HSF.BF.A.1
CCSS.HSA.CED.A.2
7.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
Given the cooling equation T(t)=90e-0.008377t + 75 where T(t) is the temperature of a turkey after t minutes, what would the temperature be after 50 minutes?
about 200oF
about 164oF
about 134oF
about 104oF
Tags
CCSS.HSF.LE.A.1
CCSS.HSA.SSE.A.1
CCSS.HSF.IF.B.5
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