Question

A dead body was found within a closed room of a house where the temperature was a constant 21° C. At the time of discovery the core temperature of the body was determined to be 27° C. One hour later a second measurement showed that the core temperature of the body was 25° C. Assume that the time of death corresponds to t = 0 and that the core temperature at that time was 37° C. Determine how many hours elapsed before the body was found. [Hint: Let t1 > 0 denote the time that the body was discovered.] (Round your answer to one decimal place.)

____ hr

Answer 1

$T^{\prime}=k\left(T-T_{m}\right) ~~~~\text{(Newton’s Law of Cooling)} \\[1 em] \therefore \int \frac{d T}{T-T_{m}}=\int k d t\\[1 em] \therefore \ln \left|T-T_{m}\right|=k t+C_{1} \rightarrow T(t)=T_{m}+c e^{k t} \\[1 em] \because T(\text{closed room})=21 \to \quad \quad T_{m}=21\\[1 em] \because T(\text{temperature at t=0 })=37 \to \quad \quad T(0)=37\\[1 em] \text{Substitute in }T(t)=T_{m}+c e^{k t}\\[1 em] \therefore 37=21+c e^{0} \rightarrow c=16 \\[1 em] \therefore T=21+16 e^{k t} ~~~~~(1)\\[1 em] \because \text{Where t is the time at discovery } \\[1 em] \because \text{The temperature was } ~ 27^{\circ} ~ \text{at the time of discovery} \\[1 em] \therefore 27=21+16 e^{k t}\rightarrow 16 e^{k t}=6 ~~~~~~(2) \\[1 em] \because \text{The temperature was } ~ 25^{\circ} ~ \text{after one hour(t+1)} \\[1 em] \therefore 25=21+16 e^{k (t+1)}\rightarrow 16 e^{k (t+1)}=4 ~~~~~~(3) \\[1 em] \because \text{Dividing (2) ~by (3)} \\[1 em] \frac{e^{k t}}{ e^{k (t+1)}}=\frac{6}{4}=\frac{3}{2}\\[1 em] e^{k t -kt-k}=\frac{3}{2}\rightarrow e^{-k}=\frac{3}{2} \rightarrow -k=\ln \frac{3}{2} \rightarrow k=-0.41 \\[1 em] \text{Substitute in (2) } \\[1 em] 16 ~e^{-0.41 t}=6\rightarrow e^{-0.41 t}=\frac{6}{16}=\frac{3}{8}\\[1 em] \therefore -0.41 ~t= \ln\frac{3}{8}\\[1 em] \therefore t= \ln\frac{3}{8}\div(-0.41)\\[1 em] \therefore t= 2.4 ~\text{hours }\\[1 em]$