Question

A bottle of

milkmilk

initially has a temperature of

7575degrees°F.

It is left to cool in a refrigerator that has a temperature of

4545degrees°F.

After 10 minutes the temperature of the

milkmilk

is

5757degrees°F.

a. Use​ Newton's Law of​ Cooling,

Upper T equals Upper C plus left parenthesis Upper T 0 minus Upper C right parenthesis e Superscript ktT=C+T0−Cekt​,

to find a model for the temperature of the

milkmilk​,

T​,

after t minutes.

T	​=	Upper C plus left parenthesis Upper T 0 minus Upper C right parenthesis e Superscript ktC+T0−Cekt
T	​=	45 plus left parenthesis 30 right parenthesis e Superscript nothing t45+(30)e t	left arrow←	Solve for k and enter the answer
		​(Round to four decimal​ places.)

b. What is the temperature of the

milkmilk

after 15​ minutes?

Tequals=nothingdegrees°F

​(Type an integer. Round to nearest​ degree.)

c. When will the temperature of the

milkmilk

be

5252degrees°​F?

tequals=nothing

minutes

Answer 1

Newton’s law of cooling formula is expressed by,

T(t) = Ts + (To – Ts) ekt

Where,

t = time,

T(t) = temperature of the given body at time t,

Ts = surrounding temperature,

To = initial temperature of the body,

k = constant.

Here,

surrounding temperature, Ts = 45 F.

The initial temperature of the body, To = 75 F.

Hence, the equation after putting the values is.

T(t) = 45 + (75 – 45) ekt

==> T(t) = 45 + 30 ekt

(a)

Given, after 10 minutes, temperature of the milk, T10 = 57 F

Putting in above expression, we get

57 = 45 + 30 e10k

==> e10k = 0.4

==> k=-0.0916 min-1

Hence, expression for temperature of milk with time is given by

T(t) = 45 + 30 e-0.0916t ----(1)

(b)

Temperature of milk after 15 minutes is given by

T(15) = 45 + 30 e-0.0916*15

T(15) = 52.59 F = 53 F

(c)

To find time at which the temperature of the milk will be 52 F, put T(t)=52 in equation (1)

52 = 45 + 30 e-0.0916t

==> e-0.0916t = 0.233

==> t = 15.8874 minutes