Newton's Law of Cooling: A corpse was discovered at 2pm....

stars584

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In an investigation, a corpse was discovered at 2pm at body temp 35 degrees Celsius. 2 hrs later the body was 30 degrees C. if room temp is 20 degrees c. what time did the murder occur? ( normal body temp for a living person is 37 degrees c. )

Newton's law of cooling = (dT/(T-Tm)) = kdt


T(2)= 35
T(4)=30
T(?)= 37(is this correct)
Tm = 20

dT/T-20 =kdt

ln T-20 = Kt + C

e^ln(T-20) = e^kt+c

T-20=C1 e^kt

T(t)=20 +C1e^kt

T(2) = 35=20+C1e^k(2)
=15=C1e^k2

T(4)=30=20+C2e^k(4)
=10=C2e^k(4)

C1e^2k/e^k2 =15/e^2k

C1 = 15e^k(-2)

C2e^4k/e^4k =15/e^4k

C2 = 10e^(-4)k

C1 = C2

15e^k(-2) = 10e^(-4)k

3/2 = e^(-4)k/e^(-2)k

e^(-2)k = 3/2


-2K = ln (3/2)

K = (-1/2)(ln 3/2)

k= (-.2028)

C1 = 15e^(-2)(-0.2028)
C1= 15e^(0.4056)
C1 = 22.5030

T(t)=20 + 22.5030e^(-0.2028)t

Am going in the right direction.
 
let 2pm be t = 0.

\(\displaystyle \L \frac{dT}{dt} = k(T - 20)\)

\(\displaystyle \L \frac{dT}{T-20} = k dt\)

\(\displaystyle \L \ln(T-20) = kt+C_1\)

\(\displaystyle \L T = C_2e^{kt} + 20\)

at t = 0, T = 35 ...

\(\displaystyle \L 35 = C_2e^0 + 20\) ... \(\displaystyle \L C_2 = 15\)

\(\displaystyle \L T = 15e^{kt} + 20\)

at t = 2, T = 30 ...

\(\displaystyle \L 30 = 15e^{2k} + 20\)

\(\displaystyle \L 15e^{2k} = 10\) ... \(\displaystyle \L k = \frac{\ln{(\frac{2}{3})}}{2}\)

\(\displaystyle \L T = 15e^{kt} + 20\)

at the time of death, T = 37 ...

\(\displaystyle \L 37 = 15e^{kt} + 20\)

\(\displaystyle \L 17 = 15e^{kt}\)

\(\displaystyle \L kt = \ln{(\frac{17}{15})}\)

\(\displaystyle \L t = \frac{2}{\ln{(\frac{2}{3})}} \cdot \ln{(\frac{17}{15})} \approx -.617 \, hrs\)

... which works out to be about 1:23 pm
 
Re: Newton's Law of cooling

Hello, stars584!

I got the same function (basically) . . .


In an investigation, a corpse was discovered at 2pm at body temp 35°C.
Two hours later the body was 30°C.
If room temp is 20°C, what time did the murder occur?
(Normal body temp for a living person is 37°C.)

Newton's law of cooling: \(\displaystyle \frac{dT}{T\,-\,T_m}\:= \:k\,dt\)

T(2) = 35
T(4) =30
T(?) = 37 (is this correct?) yes!
Tm = 20

\(\displaystyle \frac{dT}{T\,-\,20}\:=\:k\,dt\)

\(\displaystyle \ln(T\,-\,20)\:=\:Kt\,+\,c\)

\(\displaystyle e^{\ln(T-20)}\:= \:e^{kt+c}\)

\(\displaystyle T\,-\,20\:=\:Ce^{kt}\)

\(\displaystyle T(t)\:=\:20\,+\,Ce^{kt}\;\) . . . Right!

Here's where I did it differently . . .

\(\displaystyle T(2):\;35\:=\:20\,+\,Ce^{2k}\)
. . . . \(\displaystyle Ce^{2k}\:=\:15\;\) [1]

\(\displaystyle T(4):\;30\:=\:20\,+\,Ce^{4k}\)
. . . . \(\displaystyle Ce^{4k} \:=\:10\;\) [2]

Divide [2] by [1]: \(\displaystyle \L\:\frac{Ce^{4k}}{Ce^{2k}}\:=\:\frac{10}{15}\;\;\Rightarrow\;\;e^{2k}\:=\:\frac{2}{3}\;\;\Rightarrow\;\;2k\:=\:\ln\left(\frac{2}{3}\right)\)

. . \(\displaystyle k\:=\:\frac{1}{2}\cdot\ln\left(\frac{2}{3}\right) \:=\:-0.202732554\:\approx\:-0.2027\)

The function (so far) is: \(\displaystyle \L\:T(t)\;=\;20\,+\,Ce^{-0.2027t}\)


From \(\displaystyle T(2)\,=\,35\), we have: \(\displaystyle \:20\,+\,Ce^{-0.2027(2)} \:=\:35\)

Then: \(\displaystyle \:Ce^{-0.4054} \:=\:15\;\;\Rightarrow\;\;C\:=\:15e^{0.4054} \:=\:22.4983512\:\approx\:22.5\)


Hence, the function is: \(\displaystyle \L\:T(t)\;=\;20\,+\,22.5e^{-0.2027t}\)

. . So we agree!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Solving: \(\displaystyle \:20\,+\,22.5e^{0.2027t}\:=\:37\)

. . we get: \(\displaystyle \: t\:=\:1.382841466\text{ hours }\,\approx\:1\text{ hour, }23\text{ minutes}\)

Since we are using \(\displaystyle t\,=\,0\) to represent 12 noon,
. . the murder took place at \(\displaystyle \fbox{1:23\text{ pm.}}\)

And we agree with skeeter, too . . .

 
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