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.

College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter6: Exponential And Logarithmic Functions
Section6.7: Exponential And Logarithmic Models
Problem 4SE: Define Newton’s Law of Cooling. Then name at least three real-world situations where Newton’s Law of...
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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
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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