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

it would be kevlar ^^

the ima (higher than 1m) will be greater than the ima with the height of 1m.