Newton's Law of Cooling

Definition Suppose you have a hot object in a temperature controlled room. The hot object radiates heat (thus cooling itself) in an attempt to reach the equilibrium temperature of the room. Initially, the object’s temperature is dropping at a very rapid rate, but as time goes on the rate at which the object is decreasing in temperature decays. Then, the rate of change of the temperature is proportional to the temperature of the object. Newton’s Law of Cooling An object with initial temperature T0 is sitting in a room with constant temperature Ts . If T (t) is the temperature of the object at time t, then dT = −k(T − Ts ), T (0) = T0 dt where the constant k > 0 is called a constant of proportionality and depends on the heat conductivity of the object. Newton’s Law of Cooling, cont’d Note that Newton’s Law of Cooling is a separable equation (though it is also a first-order linear equation), so we can solve it like a separable equation. dT = −k(T − Ts ) dt 1 dT = −k T − Ts dt ln |T − Ts | = −kt + C T − Ts T = C1 e −kt = C1 e −kt + Ts T (t) = (T0 − Ts )e −kt + Ts Analysis of Newton’s Law of Cooling So, now we have T (t) = (T0 − Ts )e −kt + Ts . If T0 > Ts (i.e. the object is hotter than the stable environment), then (T0 − Ts ) > 0. Hence, (T0 − Ts )e −kt is decreasing to 0 as t → ∞. This means that T (t) is decreasing to Ts as t → ∞, which makes sense since the object is cooling off until it reaches the stable temperature. On the other hand, if T0 < Ts . Now, we have an object that is colder than the environment (like a cold drink). We have T0 − Ts < 0 so that (T0 − Ts )e −kt is increasing to 0 as t → ∞. Thus, T (t) is increasing to Ts as t → ∞. Again this makes sense since the object is heating up until it reaches the stable temperature. Example A pie is removed from a 350◦ F oven and placed to cool in a room with temperature 75◦ F. In 15 minutes, the pie has a temperature of 150◦ F. Determine the time required to cool the pie to a temperature of 80◦ F. Solution: T (t) = (350 − 75)e −kt + 75 T (t) = 275e −kt + 75 150 = 275e −15k + 75 75 = 275e −15k   3 −1 ln k = 15 11  t/15 3 T (t) = 275 · + 75 11 Example A pie is removed from a 350◦ F oven and placed to cool in a room with temperature 75◦ F. In 15 minutes, the pie has a temperature of 150◦ F. Determine the time required to cool the pie to a temperature of 80◦ F. Solution:  80 = 275 · 1 55  = 3 11 t = 15 · 3 11 t/15 + 75 t/15 − ln(55) ≈ 46.264 minutes ln(3) − ln(11) Variation on Newton’s Law of Cooling In Newton’s Law of Cooling, the environment was assumed to have a constant temperature due to AC or heating. This does not always have to be the case though. For example, let R(t) be the temperature outside, which is certainly changing throughout the day. Let T (t) be the temperature inside a temperature controlled building, and let U(t) be the contribution to the rate of change of the building’s temperature by AC/Heating. Then, our model looks more like: dT = −k(T − R) + U. dt Note that this is a first-order linear ODE. Variation on Newton’s Law of Cooling dT dt dT dt dT + kT dt = −k(T − R) + U = −kT + (U + kR) = U + kR Z −kt T (t) = e e kt · (U + kR) dt Example Suppose the temperature outside varies according to the function  πt  R(t) = 80 − 10 · sin , 12 where t is measured in hours after midnight and R(t) is in degrees Fahrenheit. Also, suppose the AC is running constantly and cooling at U(t) = 1◦ /hour. At midnight the temperature in the building is 65◦ and at noon the temperature is 72◦ . Find T (t). Solution: Real life examples are crazy! Don’t actually solve this by hand.  πt   πt  − 0.718 sin( + 66.265 + 1.316e 0.073t T (t) = −2.581 cos 12 12