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4. A cherry pie with a temperature of 120°C was placed on a table to cool in a room of constant temperature 25°C. After 5 minutes, the temperature of the pie had dropped to 89°C. How long, after the pie was placed, would the temperature be 65°C? Use Newton's Law of Cooling : = -k(T – T,). dt

4. A cherry pie with a temperature of 120°C was placed on a table to cool in a room of constant temperature 25°C. After 5 minutes, the temperature of the pie had dropped to 89°C. How long, after the pie was placed, would the temperature be 65°C? Use Newton's Law of Cooling : = -k(T – T,). dt

College Algebra
College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter3: Functions
Section3.3: Rates Of Change And Behavior Of Graphs
Problem 45SE: A driver of a car stopped at a gas station to fill up his gas tank. He looked at his watch, and the...
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4. A cherry pie with a temperature of 120°C was placed on a table to cool in a room of constant temperature 25°C. After 5 minutes, the temperature of the pie had dropped to 89°C. How long, after the pie was placed, would the temperature be 65°C? Use Newton's dT Law of Cooling : = -k(T – T,). dt
Transcribed Image Text:4. A cherry pie with a temperature of 120°C was placed on a table to cool in a room of constant temperature 25°C. After 5 minutes, the temperature of the pie had dropped to 89°C. How long, after the pie was placed, would the temperature be 65°C? Use Newton's dT Law of Cooling : = -k(T – T,). dt
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