The temperature of a roast varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the water temperature, A is the room temperature, and k is a positive constant. If a room temperature roast cools from 68°F to 25°F in 5 hours at freezer temperature of 20°F, how long (to the nearest hour) will it take the roast to cool to 21°F?

Explanation:

Given newton law of cooling as,

dT/dt=-k(T-A)

Where

T is water temperature

And A is room temperature

Solving this differential equations

Using variable separation

1/(T-A) dT=-k dt

Integrate both side

∫ 1/(T-A) dT= -∫k dt

In(T-A)=-kt+ c. c is constant

Take exponential of both side

T-A=exp(-kt+c)

T-A=exp(-kt)exp(c)

exp(c) is another constant let say C

T-A=Cexp(-kt)

T=A+Cexp(-kt)

Using the initial conditions given

If a room temperature roast cools from 68°F to 25°F in 5 hours

Given that freezer temperature of 20°F, then A=20°F

This shows that

At t=0 T=68°F

At t=5 T=25°F

T=A+Cexp(-kt)

68=A+Cexp(0)

Therefore,

A+C=68.

C=68-A=68-20

C=48

Therefore,

T=20+48exp(-kt)

At t=5 T=25°F

25=20+48exp(-5k)

48exp(-5k)=25-20

exp(-5k)=5/48

Take In of both sides

-5k=In(5/48)

-5k=-2.262

k=-2.262/-5

k=0.4524.

Therefore

T=20+48exp(-0.4524t)

How long will it take for T=21°F

21=20+48exp(-0.4524t)

21-20=48exp(-0.4524t)

1=48exp(-0.4524t)

exp(-0.4524t)=1/48

exp(-0.4524t)=0.02083

Take In of both sides

-0.4524t=In0.02083

-0.4524t=-3.8712

t=8.56hours

Nearest whole number is 9 hours

Which is approximately 9hours

electrical power

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this