Newton's Law of Cooling - Time

The Newton's Law of Cooling Time calculator uses Newton's Law of Cooling to determine the time associated with a temperature based on the ambient temperature, the initial and final body temperatures and the time between the final and initial body temperature readings.

INSTRUCTIONS: Choose units and enter the following:

• (T1) Temperature One
• (T2) Temperature Two
• (td) Time between T1 and T2 Temperature Observations
• (amb) Ambient Temperature
• (T3) Temperature Three

Time of T3(To3): The calculator returns the time associated the third temperature in hours.  However, this can be automatically converted to compatible units via the pull-down menue

The Math / Science

The formula for Newtons Law of Cooling used in this calculator is:

`To3 = (-1/k)*log((T3-amb)/(T1-amb))`

with `k = (-1/(td))*log((T2-amb)/(T1 -amb))`

where:

• To3 = Time associated with temperature 3
• T1 = Initial Temperature
• T2 = Second Temperature
• td = Time difference between observations of T1 and T2
• amb = Ambient Temperature
• T3 = Third Temperature

Newton's Law of Cooling states that the rate of change of an object's temperature is proportional to the difference between its temperature and the ambient temperature. 

     `(dT)/(dt) =-k(T-T_(a))`

     `(dT)/(dt)= k (T-T_(a))`

There are two "versions" of Newton's Law of Cooling that you might find, due to whether or not the temperature of the object is less than or greater than the ambient temperature. Most textbooks use the first version, but it is really a matter of personal preference; you just have to note that the sign of the proportionality constant, k, will change based on which version you use and the constraints of the problem. 

In this vCalc Equation for Newton's Law of Cooling, we do the work for you if you are trying to evaluate a time where there was a particular temperature.
You do not need our equation to evaluate this, and we will provide an example for you where you can find the answer using pen and paper, rather than the equation above. 

Example Problem

The body of a murder victim is found at 10pm in a room with a maintained temperature of 55 degrees Fahrenheit. The body is determined to have a temperature of 85 degrees Fahrenheit. At 11pm, its temperature has dropped to 81 degrees. Use Newton’s Law of Cooling (that the transfer of heat from a hot object to a neighboring cold object is proportional to the temperature difference between them) to determine the time of the murder.

First we note that this is a first-order, separable differential equation.
We can separate and solve the equation for T first:
     `(dT)/(dt) = -k(T-T_(a))`

     `int_(T_(0))^(T)((dT)/(T-T_(a))) = int_(0)^(t) (- k*dt)`

     `ln(T-T_(a))/(T_0 -T_a)=-k*t+c`

     `T=T_(a) +(T_(0) - T_(a))(e^(-k*t))`
Note that in this step we absorbed `e^(c)` into the constant c. Also , in most problems we can assume c=1. 

Now we use the ambient temperature of 55 degrees, and the temperature difference of the body from 85 to 81 degrees in a span of one hour to find the proportionality constant, k.

     `(85-55)/(81-55)=e^(-k*1)`

     `-ln((85-55)/(81-55))=k`

     `0.1431008436=k`

Plugging k back into our equation for T=98.6, yields the following for t:

     `t= -1/(0.1431008436)*ln((98.6-55)/(85-55))`
     `t=-2.612561598 text( hours)`

See Also

http://www.sosmath.com/diffeq/first/application/newton/newton.html
https://www.khanacademy.org/math/differential-equations/first-order-differential-equations/modeling-with-differential-equations/v/newtons-law-of-cooling
https://en.wikipedia.org/wiki/Newton's_law_of_cooling