   Chapter 12.5, Problem 57E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Newton's law of cooling Newton’s law of cooling (and warming) states that the rate of change of temperature u = u ( t ) of an object is proportional to the temperature difference between the object and its surroundings, where T is the constant temperature of the surroundings. That is, d u d t = k ( u − T ) ( k  constant ) Suppose an object at 0°C is placed in a room where the temperature is 20°C . If the temperature of the object is 8°C after 1 hour, how long will it take the object to reach 18°C ?

To determine

To calculate: The time require by the object to reach the temperature of 18° C for which the rate of cooling of the object by Newton’s law of cooling is given as, dudt=k(uT) where T is the temperature of the surroundings.

Explanation

Given information:

The rate of cooling of the object by Newton’s law of cooling is given as, dudt=k(uT) where T is the temperature of the surroundings. The room temperature is 20° C and initial temperature of the object is 0° C.

Formula used:

The Newton’s law of cooling is given as,

dudt=k(uT)

Where T is the temperature of the surroundings u is the initial temperature, t is any instantaneous time.

The logarithmic rule of integrals, 1xdx=ln|x|+C where x0.

The natural logarithm property,

logab=yb=ay.

Calculation:

Consider the rate of cooling of the object, dudt=k(uT).

Rewrite the provided differential equation,

duuT=kdt

Integrate both side of the equation as:

duuT=kdt

Now, use the logarithmic rule of integrals to get,

duuT=kdtln|uT|=kt+CuT=ekt+C

Solve further,

uT=ekteCuT=Aektu=T+Aekt

Now, initial temperature of the object is 0° C and initial temperature of the room is 20° C

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