[Math] Newtons Law of Cooling: In a Murder Investigation, the Temperature of the Corpse was 32.5 Degrees Celsius at 1.30 PM and 30.3 Degree Celsius

calculusintegrationordinary differential equationsphysics

In a Murder Investigation, the temperature of the corpse was $32.5^\circ C$ at $1:30$ PM and $30.3^\circ C$ an hour later. Normal body temperature is $37^\circ C$ and the temperature of the surroundings was $20^\circ C$. When did the murder take place?

Newton's Law of Cooling $$k(T-T_s)$$

$$k(37-20)$$

$$k=17$$

Where do I go from here?

Best Answer

You made several mistakes already. We can set up this problem like this:

$$\frac{dT}{dt}=-k(T-T_s)=-k(T-20),T(0)=32.5,T(1)=30.3$$

Here $0$ corresponds to the first measurement and $1$ corresponds to the second measurement an hour later*. You need to solve this equation (which will take some calculus, not just algebra as you did), and then use the two given temperatures to compute $k$. Once you know $k$, then you fully know $T(t)$, and you can solve the equation $T(t)=37$ for $t$.

* There are other ways to choose the time coordinates; for instance, you could make $0$ be the time of the murder if you wanted. This slightly changes the rest of the problem.

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