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.)?

1 Answer
Jul 4, 2018

Answer:

#approx 1 " hour " 8 " mins"#

Explanation:

Using Newton's Law of cooling :

  • #dot T = -k (T - T_a) \ square qquad "where " T_a = 68 " is ambient" #

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"#