In a murder investigation, the temperature of the

Ответ:

kayleebueno

20.01.2022

Step-by-step explanation:

Given that in a murder investigation, the temperature of the corpse was 32.5°C at 1:30 pm and 30.3°C an hour later.

Normal body temperature is 37.0°C and the temperature of the surroundings was 20.0°C.

As per Newton law of cooling we have

$T(t) = T_s+(T_0-T_s)e^{-kt}$

is the temperature at time t.

Substitute this for given two information to find k and T0

$32.5 = 20+(37-20)e^{-kt} \\30.3 = 20+(37-20)e^{-k(t+1)}$

$12.5 = (17)e^{-kt} ...i\\10.3 =17e^{-k(t+1)}...ii$

Divide I equation by II to get

$1.2136=e^k\\k = 0.1936$

Using this we find t at 1.30 p.m.

$32.5 = 20+(37-20)e^{-0.1936t} \\\\0.7353=e^{-0.1936t} \\t= 1.588$

i.e. approximately 1.6 hours or 1 hour 36 minutes lapsed at 1.30

Time of murder is

11 hrs 54 minutes.

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