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How to solve Newton’s Law of Cooling Problems, such as Time and Temperature of a Dead Body
This is probably the nastiest kind of question you’ll ever find, it involves finding the time a body was killed using the temperature that the body was found at. We say it’s a “disgusting” kind of problem because most applications of math don’t have such kinds of applications.
Anyways, with these kinds of problems you’re given the ambient temperature, a temperature of the body and a time, and usually you’re asked to find out when the body was killed.
For these kinds of problems, you’ll want to use Newton’s Law of Cooling, and you want to memorize the simple ODE below:
\[\frac{{dT}}{{dt}} = – k\left( {T – {T_A}} \right)\]
Example problem:
- A corpse with temperature 27\(^OC\) is found at midnight, where the surrounding ambient temperature is constant at 17\(^OC\).
- The body is moved to a morgue with temperature 7\(^OC\).
- After one hour, the body temperature is 20\(^OC\).
- The body’s normal temperature is 36.8\(^OC\) at the time of death.
- How long before midnight was the time of death?
And you can solve the ODE without knowing the constant for now.
All of that extra information is really just an initial condition we can use to solve for the constant.
Remember k is just a number constant and is not dependent on t, so you can freely move it out of the integral.
\[\frac{{dT}}{{\left( {T – {T_A}} \right)}} = – kdt\]
\[\ln \left| {T – {T_A}} \right| = – kt + C\]
Then raise all terms to the power of e, to cancel out the ln. We have absolute symbols, but temperature here is above 0.
\[T – {T_A} = C{e^{ – kt}}\]
Reading the question statement, we actually have an initial condition. Set the time of discovery to be t = 0, then:
\[T(0) = {27^O}C\]
Substituting this in,
\[27 – 7 = C{e^{ – k(0)}}\]
Solving, C = 20
We’re also told that after 1 hour, the body temperature is 20\(^OC\).
Let’s define our function to represent time in hours, so T(1) = 20, and substitute it in, also knowing now that C = 20.
\[T – {T_A} = C{e^{ – kt}}\]
\[20 – 7 = 20{e^{ – k(1)}}\]
\[k = – \ln \left( {\frac{{13}}{{20}}} \right)\]
Good, this k value will always remain the same, BUT, C will not necessarily be the same! We use another variable, capital K to represent another C.
Substituting these in,
At time t = 0 which is the time of discovery, we’re told it’s body temperature is 27\(^OC\) and \({T_A} = {17^O}C\). This simplifies nicely because t = 0.
\[27 – 17 = K{e^0}\]
So, K = 10
Now we solve for the time that the body died. At the time of death the body was \({36.8^O}C\) and the surrounding temperature is \({17^O}C\). k always remains the same, if you want keep the exact value in your calculator, but it’s roughly 0.431.
\[36.8 – 17 = 10{e^{^{ – 0.431t}}}\]
Now just solve for t.
\[t = – \frac{1}{{0.431}}\ln \left( {\frac{{36.8 – 17}}{{10}}} \right)\]
t ends up being roughly negative 1.58. This means that the body died roughly 1.58 hours before it was discovered!
Try another ODE, the infamous “snowplow” problem you might see on an exam!