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
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Expert's answer
2020-06-23T14:23:03-0400
"T^{\\prime}=k\\left(T-T_{m}\\right) ~~~~\\text{(Newton\u2019s Law of Cooling)} \\\\[1 em] \n\\therefore \\int \\frac{d T}{T-T_{m}}=\\int k d t\\\\[1 em] \n\\therefore \\ln \\left|T-T_{m}\\right|=k t+C_{1} \\rightarrow T(t)=T_{m}+c e^{k t} \\\\[1 em] \n \\because T(\\text{closed room})=21 \\to \\quad \\quad T_{m}=21\\\\[1 em]\n \\because T(\\text{temperature at t=0 })=37 \\to \\quad \\quad T(0)=37\\\\[1 em]\n\\text{Substitute in }T(t)=T_{m}+c e^{k t}\\\\[1 em]\n\\therefore 37=21+c e^{0} \\rightarrow c=16 \\\\[1 em] \n\\therefore T=21+16 e^{k t} ~~~~~(1)\\\\[1 em]\n\\because \\text{Where t is the time at discovery } \\\\[1 em] \n\\because \\text{The temperature was } ~ 27^{\\circ} ~ \\text{at the time of discovery} \\\\[1 em] \n\\therefore 27=21+16 e^{k t}\\rightarrow 16 e^{k t}=6 ~~~~~~(2) \\\\[1 em] \n\\because \\text{The temperature was } ~ 25^{\\circ} ~ \\text{after one hour(t+1)} \\\\[1 em] \n\\therefore 25=21+16 e^{k (t+1)}\\rightarrow 16 e^{k (t+1)}=4 ~~~~~~(3) \\\\[1 em]\n \\because \\text{Dividing (2) ~by (3)} \\\\[1 em] \n \\frac{e^{k t}}{ e^{k (t+1)}}=\\frac{6}{4}=\\frac{3}{2}\\\\[1 em] \ne^{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] \n\n \\text{Substitute in (2) } \\\\[1 em]\n \n16 ~e^{-0.41 t}=6\\rightarrow e^{-0.41 t}=\\frac{6}{16}=\\frac{3}{8}\\\\[1 em] \n \\therefore -0.41 ~t= \\ln\\frac{3}{8}\\\\[1 em] \n \\therefore t= \\ln\\frac{3}{8}\\div(-0.41)\\\\[1 em] \n \\therefore t= 2.4 ~\\text{hours }\\\\[1 em]"
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