Question

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.)?

Answer 1

`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"`