Newton’s Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the surrounding temperature.
Let the temperature of the object be T(t) at time t.
Let T0 denote the initial temperature of the object at time 0.
Let Ts denote the temperature of the surrounding medium.
Newton's Law of Cooling states dT/dt = -k (T - Ts), where k is a cooling constant and where the negative sign is for the case where the body is getting cooler. The solution of the differential equation is:
Newton published his Law of Cooling in 1701. Modern physicists believe Newton's Law holds up well and approximates the result in some but not all situations. Large temperature differences between the object and its surroundings, or non-constant temperatures, will cause difficulties. I leave the details to the physicists.
Problem 1: The dead body problem
Example: A dead body was 80 degrees Fahrenheit when it was discovered. Two hours later it had cooled to 75 degrees. The room is 60 degrees. What time did the body die?
Assume T0 at time of death = 98.6
Assume Ts the temperature of surrounding room = 60
Let k be the unknown cooling constant. k depends on the characteristics of the object and its environment and is often between 0.1 and 0.2.
T(t) = Ts + (T0 − Ts) * e^(−kt)
First equation, at time of discovery:
Let t = t_0 be time of discovery. 80 = 60 + (98.6 - 60) * e^(-k*t_0)
Second equation, two hours later:
At t = t_0 + 2, 75 = 60 + (98.6 - 60) * e^(-k*(t_0 + 2))
Solve first equation for k:
80 = 60 + (98.6 - 60) * e^(-k*t_0)
20 = 38.6 * e^(-k*t_0)
e^(-k*t_0) = 20 / 38.6 = .518
Solve second equation for k:
75 = 60 + (98.6 - 60) * e^(-k*(t_0 + 2))
15 = 38.6 * e^(-k*(t_0 + 2))
e^(-k*(t_0 + 2)) = 15 / 38.6 = .389
Solve for k by taking the ratio of the two equations, noting t_0 drops out:
[e^(-k*(t_0 + 2))] / [e^(-k*t_0) ] = .389 /518
e^(-2k) = .751
-2k = LN(.751)
k = - LN(.751) / 2
k = .143
Solve for time since death from first equation
80 = 60 + (98.6 - 60) * e^(-.143*t_0)
20 = 38.6 * e^(-.143*t_0)
e^(-.143*t_0) = 20 / 38.6 = .518
-.143*t_0 = LN(.518)
t_0 = - LN(.518) / .143
t_o = 4.8 hours
The body had been cooling for 4.8 hours before it was discovered.
Problem 2: The coffee with milk problem
Here is a different application of Newton's Law of Cooling.
Example: Who drinks the hotter coffee after five minutes: The person who puts milk in right away and then waits five minutes, or the person who waits five minutes and then puts in milk?
Assume initial temperature of coffee, T0 = 200 F
Assume temperature of the room, Ts = 72 F
Assume initial temperature of the milk, Tm = 40 F
Assume the mixture is 80% coffee and 20% milk
Assume cooling constant k = 0.1.
Case 1: Add cold milk immediately
mixture T(0) = .80*200 + .20*40 = 168
mixture T(5) = 72 + (168 − 72) * e^(−0.1*5) = 130.23
Case 2: Wait 5 minutes, then add room temperature milk
coffee T(5) = 72 + (200 - 72)*e^(-0.1*5) = 149.64
milk T(5) = 72 + (40 - 72)*e^(-0.1*5) = 52.59
mixture T(5) = .80*149.64 + .20*52.59 = 130.23
The conclusion is that cases 1 and 2 give the same temperature! However, there is also a case 3.
Case 3: Wait 5 minutes, then add cold milk
coffee T(5) = 72 + (200 - 72)*e^(-0.1*5) = 149.64
mixture T(5) = .80*149.64 + .20*40 = 127.71
For simplicity, I ignored a few things that probably have a minor effect on the final temperature: I assume the mixing process is instantaneous. I assume the cup does not absorb heat from the coffee. I assume stirring the mixture does accelerate heat transfer. Perhaps there are others. I can not measure the effect of these.
Thank you, Isaac Newton (who as a Brit, perhaps drank tea?).
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