. . . Newton’s Law of cooling/warming Consider pouring a ° C cup of coffee and kept it on the table in an ° C room.
What happens to the temperature of the coffee? We observe that the cup of coffee will cool off until it reaches the room temperature. Now consider taking a ° C glass of cold water from the refrigerator and kept it on the table in a ° C room. What happens to the temperature of the cold water?
Similarly, we can observe the water will warm up until it reaches room temperature. According to Newton’s law of cooling or warming , the rate at which the temperature of a body changes is proportional to the difference between the temperature of the body and the temperature of the surrounding medium the so-called ambient temperature . If T t ( ) represents the temperature of a body at time t , T m the temperature of the surrounding medium, and dT dt the rate at which the temperature of the body changes, then Newton’s law of cooling(or warming) is dT dt T m ∝ or dT dt k T T m ) , where k is constant of proportionality. In either case, cooling or warming, if T m is constant, it stands to reason that k < .
Example . In a murder investigation, a corpse was found by a detective at exactly p.m. Being alert, the detective also measured the body temperature and found it to be o F. Two hours later, the detective measured the body temperature again and found it to be o F.
If the room temperature is o F, and assuming that the body temperature of the person before death was . o F, at what time did the murder occur? ; 88789 69315 ) = ) = − Let T be the temperature of the body at any time t and with time taken to be p.m. By Newton’s law of cooling, dT dt k T or dT kdt Integrating on both sides, we get log − kt C or − Ce kt .
When t , and so C = − When t , we have − = − e k . Thus, k = Ordinary Differential Equations Hence, the solution is = − t log or T Now, we would like to find the value of t , for which T t ( ) = , and t = ≈− It appears that the person was murdered at about . p.m.