Suppose the temperature of a body when discovered is 85°F.Two hours laters, the temperature is 74°F and room temperature is 68°F.Find the time when the body was discovered after death(assume the body temperature to be 98.6°F at time of death.)?

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Answer 1 · 2018-07-04T21:25:12

$approx 1 " hour " 8 " mins"$

Using Newton's Law of cooling :

There are 3 conditions:

${(T(0) = T_o = 98.6),(T(tau) = 85 qquad larr" body discovered "),(T(tau + 2)= 74):}$

$square$ separates for integration:

$(dT)/(T - T_a) = -k \ dt$

$int_(T_o)^(T(t)) (dT)/(T - T_a) = -k t$

Because $T gt T_o$ :

$[ln(T - T_a)]_(T_o)^(T(t)) = -k t$

$(T - T_a)/(T_o - T_a) = e^(-k t)$

$:. bb( T = T_a + (T_o - T_a)e^(-k t) )$

The 3 conditions translate as:

${(T_o = 98.6),(85 = 68 + (98.6 - 68)e^(-k tau) qquad qquad qquad bbb((A))),(74 = 68 + (98.6 - 68)e^(-k (tau+ 2))qquad bbb((B))):}$

$bbb((B))/bbb((A)) implies e^(- 2k) = (74 - 68)/(85 - 68)$

$implies k = 1/2 ln(17/6)$

From $bbb((A))$ :

$- k tau = ln((85 - 68)/(98.6-68))$

$tau = (ln((98.6-68)/(85 - 68)))/(1/2 ln(17/6)) approx 1.129 " hrs"$