# 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 let 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 positive in a heating process, when T is less 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 differential equation is (dT)/(dt)+kT=30k . This is first-order linear differential equation. Integrating factor is I=e^(int kdt)=e^(kt) . After multiplying equation by 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 system of two equations:

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

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

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

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

Since T(0)=50 then 50=Ce^(-k*0)+150 or C=-100 .

Now equation has form T=-100e^(-kt)+150 ;

Since T(10)=75 then 75=-100e^(-10k)+150 or k=-(ln(0.75))/10 .

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

Now, we need tofind 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.