Newton\'s law of cooling gives the temperature T(T) of an object at time t in te

Question

Newton's law of cooling gives the temperature T(T) of an object at time t in terms of T_0, is temperature at t = 0, and T_s, the temperature of the surroundings. T(t) = T_s + (T_0 - T_s)e^-kt A police officer arrives at a crime scene in a hotel room at 9:18 PM, where he finds a dead body. He immediately measures the body's temperature and find it to be 79.5 degree F. Exactly one hour later he measures the temperature again, and find it to be 78.0 degree F. Determine the time of death, assuming that victim body temperature was normal (98.6 degree F) prior to death, and that the room temperature was constant at 69 degree F.

Explanation / Answer

Here at t1=9:18 PM means at the time police arrived body temperature is T(t)=79.5 degreeF

And at t=t2 means when the death occured the temperature of body is T0=98.6 degreeF

Temperature of surroundings is Ts=63 degreeF

Now, T(t)=Ts+(T0-Ts)e-kt

wher tis time of death

now, k=1/t0

So, T(t)=Ts+(T0-Ts)e-t/t0

t0=9:18 PM =33480 sec

79.5=63 + (98.6 - 63)e-t/33480

16.5=35.6e-t/33480

0.4635=e-t/33480

so, -t/33480=ln 0.4635=-0.7689

So, t=26019.58 sec.

So time of death will be 7:14 PM.

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