A bottle of juice initially has a temperature of 70°F. It is left to cool in a refrigerator that has a temperature of 45°F. After 10 minutes, the temperature of the juice is 55°F. Use Newton's Law of Cooling, T=C+(T, - C)e, to find a model for the temperature of the juice, T, after t minutes.
A bottle of juice initially has a temperature of 70°F. It is left to cool in a refrigerator that has a temperature of 45°F. After 10 minutes, the temperature of the juice is 55°F. Use Newton's Law of Cooling, T=C+(T, - C)e, to find a model for the temperature of the juice, T, after t minutes.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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![Section 3.5
ective #3: Use Newton's Law of Cooling.
em #3
Pencil Problem #3
It is left to cool in a
30°C. After 5
= object is 80°C.
3.
A bottle of juice initially has a temperature of
70°F. It is left to cool in a refrigerator that has a
temperature of 45°F. After 10 minutes, the
temperature of the juice is 55°F.
T = C+ (To –
3a. Use Newton's Law of Cooling,
mperature of the
T=C+(T, - C)e, to find a model for the
temperature of the juice, T, after t minutes.
= 100.
is 30°C: C= 30.
object is 80°C:
s into
isolating the
of each](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F47cd433c-af7e-4add-b878-a95e599142fb%2Fa132a078-c579-431a-94ef-653b4788a64f%2Fb8amyus_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Section 3.5
ective #3: Use Newton's Law of Cooling.
em #3
Pencil Problem #3
It is left to cool in a
30°C. After 5
= object is 80°C.
3.
A bottle of juice initially has a temperature of
70°F. It is left to cool in a refrigerator that has a
temperature of 45°F. After 10 minutes, the
temperature of the juice is 55°F.
T = C+ (To –
3a. Use Newton's Law of Cooling,
mperature of the
T=C+(T, - C)e, to find a model for the
temperature of the juice, T, after t minutes.
= 100.
is 30°C: C= 30.
object is 80°C:
s into
isolating the
of each
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