# Temperature Problems

Newton's law of cooling, which is equally applicable to heating, states that the time rate of change of the temperature of a body is proportional to the temperature difference between the body and its surrounding medium. Let T denote the temperature of the body and T_m denote the temperature of the surrounding medium. Then, the time rate of change of the temperature of the body is (dT)/(dt), and Newton's law of cooling can be formulated as (dT)/(dt)=-k(T-T_m), or as (dT)/(dt)+kT=kT_m, where k is a positive constant of proportionality. Once k is chosen positive, the minus sign is required in Newton's law to make (dT)/(dt) negative in a cooling process, when T is greater than T_m, and to make it positive in a heating process, when T is smaller than T_m.

Example 1. A body at an unknown temperature is placed in a room which is held at a constant temperature of 30° F. If after 10 minutes the temperature of the body is 0° F and after 20 minutes the temperature of the body is 15° F, find the unknown initial temperature.

We have that T_m=30; so, the differential equation is (dT)/(dt)+kT=30k. This is a first-order linear differential equation. The integrating factor is I=e^(int kdt)=e^(kt). After multiplying the equation by the integrating factor, we obtain that e^(kt)(dT)/(dt)+kTe^(kt)=30ke^(kt), or (d(Te^(kt)))/(dt)=30ke^(kt).

Integrating both sides gives Te^(kt)=30e^(kt)+C, or T=Ce^(-kt)+30.

We are given that T(10)=0, or 0=Ce^(-10k)+30. Also, T(20)=15, or 15=Ce^(-20t)+30.

Thus, we have a system of two equations:

{(Ce^(-10k)=-30),(Ce^(-20k)=-15):}

Dividing the first equation by the second gives e^(10k)=2. Now, from the first equation, we have that C=-30e^(10k)=-30*2=-60.

Finally, T_0=T(0)=Ce^(-k*0)+30=C+30=-60+30=-30.

Let's take a look at another interesting example.

Example 2. A body at a temperature of 50° F is placed in an oven whose temperature is kept at 150° F. If after 10 minutes the temperature of the body is 75° F, find the time required for the body to reach a temperature of 100° F.

We have that T(0)=50, T(10)=75, T_m=150.

So, the differential equation is (dT)/(dt)+kT=150k. Again, as in example 1, this is a linear first-order differential equation. Its solution is T=Ce^(-kt)+150.

Since T(0)=50, we have that 50=Ce^(-k*0)+150, or C=-100.

Now, the equation has the form T=-100e^(-kt)+150.

Since T(10)=75, we have that 75=-100e^(-10k)+150, or k=-(ln(0.75))/10.

Finally, the equation has the following form: T=-100e^(ln(0.75)/10t)+150.

Now, we need to find such t that T(t)=100:

100=-100e^((ln(0.75))/10t)+150, or t=10 ln(0.5)/(ln(0.75))~~24.0942 minutes.