Newton's Law of Cooling. The initial temperature of an object is 37.8 degree C. If it took 10 minutes for the object to cool off to 32.2 degree C at a room where it is 15.6 degree C, then how many minutes more will it take for the object for it to be at 26.7 degree C? Use this formula: U=Uo + Ce^(kt) Where: U= temperature, Uo= air temperature, k= constant of proportionality

Newton's Law of Cooling. The initial temperature of an object is 37.8 degree C. If it took 10 minutes for the object to cool off to 32.2 degree C at a room where it is 15.6 degree C, then how many minutes more will it take for the object for it to be at 26.7 degree C? Use this formula: U=Uo + Ce^(kt) Where: U= temperature, Uo= air temperature, k= constant of proportionality

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter6: The Trigonometric Functions

Section6.7: Applied Problems

Problem 70E

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A driver of a car stopped at a gas station to fill up his gas tank. He looked at his watch, and the time read exactly 3:40 p.m. At this time, he started pumping gas into the tank. At exactly 3:44, the tank was full and he noticed that he had pumped 10.7 gallons. What is the average rate of flow of the gasoline into the gas tank?
A bottle of soda with a temperature of 71 Fahrenheit was taken off a shelf and placed ina refrigerator with an internal temperature of 35 .After ten minutes, the internal temperature of thesoda was 63F . Use Newton’s Law of cooling towrite a formula that models this situation. To thenearest degree, what will the temperature of thesoda be after one hour?
Define Newton’s Law of Cooling. Then name at least three real-world situations where Newton’s Law of Cooling would be applied.

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