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...

Question

The temperature of a roast varies according to Newton's Law of Cooling:  , 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?

A	1
B	9
C	12
D	24

Answer 1

In this problem, T(t) will represent the temperature of the roast (in degrees Fahrenheit) after t hours. We can see from the given information that A = 20 and T(0) = 68. Next, we make the change of variable y(t) = T(t) - 20 (note that y(t) is nonnegative in the context of this problem and that y(t) is positive for finite values of t). Then we have that

y'(t) = \frac{d}{dt}[T(t) - 20] = T'(t)

\frac{dy}{dt} = -ky

y(0) = T(0) - 20 = 68 - 20 = 48

Use of the separation of variables method on this differential equation involving y gives us

\frac{1}{y}dy = -k dt (valid since y ≠ 0 for finite values of t)

⇒ ln|y| = -kt + C

⇒ |y| = e^{-kt + C}

⇒ y = e^{C}e^{-kt} (since y is positive for finite values of t)

⇒ y(t) = Ae^{-kt} (where A = e^{C})

The initial condition y(0) = 48 then gives us

y(0) = Ae^{-k*0}

= A*e^{0}

= A*1

= A

⇒ A = 48

⇒ y(t) = 48e^{-kt}

Since we are also given that T(5) = 25, we know that y(5) = T(5) - 20 = 25 - 20 = 5. Further, we can also conclude that

y(5) = 48e^{-k*5}

= 48e^{-5k}

= 5

⇒ 48e^{-5k} = 5

⇒ e^{-5k} = \frac{5}{48}

⇒ -5k = ln(\frac{5}{48})

⇒ k = \frac{1}{-5}ln(\frac{5}{48})

= \frac{1}{5}ln((\frac{5}{48})^{-1})

= \frac{1}{5}ln(\frac{48}{5})

≅ 0.4524

⇒ y(t) = 48e^{-[\frac{1}{5}ln(\frac{48}{5})]t}

≅ 48e^{-0.4524t}

⇒ T(t) - 20 = 48e^{-[\frac{1}{5}ln(\frac{48}{5})]t}

≅ 48e^{-0.4524t}

⇒ T(t) = 20 + 48e^{-[\frac{1}{5}ln(\frac{48}{5})]t}

≅ 20 + 48e^{-0.4524t}

Letting T(t) = 21, we finally have that

20 + 48e^{-0.4524t} = 21

⇒ 48e^{-0.4524t} = 21 - 20 = 1

⇒ e^{-0.4524t} = \frac{1}{48}

⇒ -0.4524t = ln(\frac{1}{48})

⇒ t = \frac{1}{-0.4524}ln(\frac{1}{48})

= \frac{1}{0.4524}ln((\frac{1}{48})^{-1})

= \frac{1}{0.4524}ln(48)

= 8.557031412

≅ 9

Thus, the roast will take 9 hours to cool down to 21 degrees Fahrenheit and the answer is B.

Answer 2

In this case, we are given T(0) = 68, T(5)=25, and A=20 (the freezer temperature which is also the room temperature since the roast is placed in the freezer). We want to find t when T(t)=21. Plugging the given information into the equation gives dT/dt = - k*(T-20). Hence dT/(T-20) = -k*dt. Integration of both sides yields

ln|T-20|=-kt +C1, so

|T-20|= e^(-kt+C1)

=e^(-kt) * e^(C1)

=C2*e^(-kt), where C2=e^(C1) is a constant.

So T-20=Ce^(-kt), where C is a constant absorbing the sign of |T-20|.

As a result, we have T(t)= 20 + Ce^(-kt). As 68=T(0)=20+C, then C=48 and so T(t)=20+48e^(-kt).

Since T(5) = 25, then 25=20+48e^(-5k) which yields e^(-5k)=5/48, so k=0.4524. Hence, we have

T(t)=20+48e^(-0.4524t).

Now 21=T(t)=20+48e^(-0.4524t) gives e^(-0.4524t)=1/48 and so t=8.557 and it can be rounded to be 9.

So B is the right choice.