Answer:
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
[tex]T(t) = T_s+(T_0-T_s)e^{-kt}[/tex]
is the temperature at time t.
Substitute this for given two information to find k and T0
[tex]32.5 = 20+(37-20)e^{-kt} \\30.3 = 20+(37-20)e^{-k(t+1)}[/tex]
[tex]12.5 = (17)e^{-kt} ...i\\10.3 =17e^{-k(t+1)}...ii[/tex]
Divide I equation by II to get
[tex]1.2136=e^k\\k = 0.1936[/tex]
Using this we find t at 1.30 p.m.
[tex]32.5 = 20+(37-20)e^{-0.1936t} \\\\0.7353=e^{-0.1936t} \\t= 1.588[/tex]
i.e. approximately 1.6 hours or 1 hour 36 minutes lapsed at 1.30
Time of murder is
11 hrs 54 minutes.
