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:
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.