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3) A hot pie is just taken out from an oven with a set temperature of 175°F. It is placed near a window with an ambient temperature of 68°F. After 3 minutes, the temperature of the pie is 150°F, determine how long it will take the pie have a temperature of 100°F. Determine also the time when the temperature of the pie is half a degree above the ambient temperature.

3) A hot pie is just taken out from an oven with a set temperature of 175°F. It is placed near a window with an ambient temperature of 68°F. After 3 minutes, the temperature of the pie is 150°F, determine how long it will take the pie have a temperature of 100°F. Determine also the time when the temperature of the pie is half a degree above the ambient temperature.

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter5: A Survey Of Other Common Functions
Section5.4: Combining And Decomposing Functions
Problem 8E
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Newton's Law of Cooling Newton's Law of Cooling states that the time rate of change in the temperature of the body is directly proportional to the temperature difference between the body and its surrounding medium where the temperature is held constant. Let u be the temperature of the body and T is the temperature of the surrounding medium where the body is immersed into then by definition of the Newton's law of cooling: du dt : = k(T – u) or can be written as du dt -k(u – T) where k > 0 and is called the thermometer constant Finding the solution to this equation du -k(и — Т) dt du (-k) dt и — Т In(u – T) = -kt + c и — Т%3D се-kt then u = T + ce-kt Another form of this equation from other references is T = T, + (T, – T,)e-kt where T – temperature of the body at any time, t T; – surrounding or ambient temperature To – initial temperature of the body, or temperature at t = 0 k – thermometer constant; k > 0
Transcribed Image Text:Newton's Law of Cooling Newton's Law of Cooling states that the time rate of change in the temperature of the body is directly proportional to the temperature difference between the body and its surrounding medium where the temperature is held constant. Let u be the temperature of the body and T is the temperature of the surrounding medium where the body is immersed into then by definition of the Newton's law of cooling: du dt : = k(T – u) or can be written as du dt -k(u – T) where k > 0 and is called the thermometer constant Finding the solution to this equation du -k(и — Т) dt du (-k) dt и — Т In(u – T) = -kt + c и — Т%3D се-kt then u = T + ce-kt Another form of this equation from other references is T = T, + (T, – T,)e-kt where T – temperature of the body at any time, t T; – surrounding or ambient temperature To – initial temperature of the body, or temperature at t = 0 k – thermometer constant; k > 0
3) A hot pie is just taken out from an oven with a set temperature of 175°F. It is placed near a window with an ambient temperature of 68°F. After 3 minutes, the temperature of the pie is 150°F, determine how long it will take the pie have a temperature of 100°F. Determine also the time when the temperature of the pie is half a degree above the ambient temperature.
Transcribed Image Text:3) A hot pie is just taken out from an oven with a set temperature of 175°F. It is placed near a window with an ambient temperature of 68°F. After 3 minutes, the temperature of the pie is 150°F, determine how long it will take the pie have a temperature of 100°F. Determine also the time when the temperature of the pie is half a degree above the ambient temperature.
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