Newton's Law of Cooling

Question

after the coffee is heated to 200 degrees , it is set in a room at 50 degrees.It cools at a rate that is proportional to the difference between its current temperature and that of the surrounding atmosphere. Let C(t) be the difference between the temperature of the coffee and the temperature of the coffee and that of the room, where t is time measured minutes after the coffee starts cooling. When t=0, the rate of cooling 2 degrees/min. 

 

Write the equation for D(t). How long will it take for the mik to cool to 120 degrees. 

Answer 1

First you need to figure out the equation for D(t)

 

dT/dt is proportional to (T - Ta)

 

Because the coffee is cooling down the sign of the derivative will be negative:

 

dT/dt=-k(t-Ta) from dT=-ky

 

You can do the integration here or you may already know that the above equation can be written in this form:

 

C(t)=C₀e^-kt

 

In this problem:

 

 

 C(t) =T(t)-T_a  = Temperature difference between coffee and temperature in the room at time t.
 C_o = T(0) -T_a = T_o-T_a  = Initial temperature difference at time t=0

 

substituting we get:

 

T(t)-Ta=(To-Ta)e^-kt

 

T(t)=Ta+(To-Ta)e^-kt

 

Now we need to find the k value:

 

dT/dt=2^o/min.

 

dT/dt=-k(t-Ta)

 

2^o/min.=-k(200-50)

 

k= -.0133 so now we can substitute this value for k along with the other values in our derived equation:

 

(t)=50+(200-50)e^-.0133t

 

(t) = 120^o

 

120^o=50+(200-50)e^-.0133t

 

70^o=150^oe^-.0133t

 

70/150=e^-0133t

 

150/70=e^.0133

 

ln(150/70)=.0133t

 

t=57.3 minutes

 

Answer 2

When you solve Newton's law of cooling with your numbers, you get

T = 50 + 150 e^-kt

Since

dT/dt (t=0) = -k 150 = -2, we have k =1/75 min^-1, so that

T = 50 +150 e^-t/75

 

You can solve for t when T=120:

 

120 = 50 + 150 e^-t/75 ⇒ t= 75 ln(15/7) ≈ 57.2 min

 

Answer 3

T₂ = T₀ + (T₁ - T₀) * e^{(-k * Δt)}

 

T₂ = 120°

T₀ = 50°

T₁ = 200°

k = 2° min^-1

 

T₂ - T₀ = (T₁ - T₀)*e^{(-k * Δt)}

 

[(T₂ - T₀]/[T₁ - T₀)] = e^{(-k * Δt)}

 

(70°)/(150°) = e^{-[(2°min) * Δt]}

 

-2° min^-1 * Δt = ln (70°) - ln (150°) = -0.76214

 

Δt = 0.38 min.

 

Bella, this answer of mine seems ridiculously short, but I can't find a flaw in my calculation.

Answer 4

T= temperature of coffee at time t
t = time (any units, but to be consistent with the given information, choose minutes)
T_a = temperature of the surrounding atmosphere = 50 degrees

T₀ = initial temperature of coffee = 200 degrees

The temperature is proportional to the difference between its current temperature and it surrounding temperature. We can express this mathematically using this differential equation

dT/dt = -k(T-T_a)

Separate the variables .
dT/(T-T_a) = -kdt

Integrate.

ln (T-T_a) = -kt +C

Eliminate the natural log.

(T-T_a) = e^-kt+C

(T-T_a) = Ce^-kt

When t = 0, what is C? It's simply the initial difference in temperatures. C = (T₀ - T_a)

(T-T_a) = (T₀ - T_a)e^-kt

and the problem asks you to represent T-T_a with C(t).

C(t) = (T₀ - T_a)e^_-kt

For this specific problem, the rate of cooling is 2 degrees/min at t=0, which means

 

dT/dt = -k(T-T_a)

 

-2 deg/min = -k (200-50)

k = 2/150 = 1/75 ~ 0.01333

 

and of course T₀ - T_a  = 200 deg - 50 deg = 150 deg

Substituting into our expression for C(t), we finally have the expression for the difference of temperature of the coffee and the surrounding atmosphere as a function of time

 

C(t) = (150 deg)e^-0.01333t

If the milk cools to 120 degrees, then C(t) = 120 deg - 50 deg = 70 deg.

70 deg = (150 deg)e^-0.01333t

Solve for t.

70/150 = e^-0.01333t

 

ln (70/150) = -0.01333t

t = ln(70/150)/(-0.01333)

t = -0.7621/(-0.01333)

t = 57.16 min