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Q 3.3-1E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 107
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

A cup of hot coffee initially at 95°C cools to 80°C in 5 min while sitting in a room of temperature 21°C. Using just Newton’s law of cooling, determine when the temperature of the coffee will be a nice 50°C.

The temperature of coffee will reach 50°C after 20.7 minutes.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step1: Analyzing the given statement

The initial temperature of a cup of hot coffee is 95°C and its temperature after sitting in a room of temperature 21°C for 5 min, becomes 80°C. By using Newton’s law of cooling, we have to determine the time after which the temperature of the coffee will reach 50°C.

Newton’s Law of Cooling is,

T(t)=M0+(T0-M0)e-kt······(1)

Here, we will take the values as,

Initial temperature of coffee,T0=95oC

Temperature of the room,M0=21oC

Temperature after 5 min,T(5)=80oC

Step 2: To find the value of k in the formula of Newton’s Law of cooling

Using the given values in equation (1), to find the value of k,

T5=21+95-21e-5k 80=21+74e-5k80-21=74e-5k 59=74e-5k e5k=7459 5k=ln1.254 k=ln1.2545 k=0.0453

One will use this value of k in next step to find the time after which the temperature of coffee will reach 50°C.

Step 3: To determine the time after which the temperature of the coffee will reach 50°C

Substituting T(t)=50oCin equation (1),

50=21+95-21e-(0.0453)t 50-21=74e-(0.0453)te(0.0453)t=74290.0453t=ln2.552 t=ln2.5520.0453 t=20.7min

Hence, the temperature of coffee will reach 50°C after 20.7 minutes.

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