At 10:00 am, Rick arrived at the location of a gromflomite who had died. The temperature of thelocation was 64°F, and at the time of death, the body temperature of the gromflomite was 85.3°F.Rick took the body’s temperature at 10:30 am, at which time it was 76.3°F, and then againat 11:00 am, at which the temperature was 72.7°F. At what time did the gromflomite die?[Round answer to the nearest minute.]Answer: _____________________

If we suppose that a mass of constant heat capacity is cooled at a rate that is proportional to the temperature difference between the body and the environment then we can use an exponencial model for the body temperature as

T - T env = a*e^(-t/b)

then

if we use the reference t=0 as 10:30 am , then at t=0 , T= 76.3°F and

T - T env = a*e^(-t/b)

76.3°F-64°F = a*e^(-0/b)

a = 76.3°F-64°F = 12.3 °F

then for t= 0.5 hours (11:00 am from 10:30 am) ,T= 72.7°F and

T - T env = a*e^(-t/b)

72.7°F - 64°F = 12.3 °F*e^(-0.5 h /b)

b = -0.5 h/ln [( 72.7°F - 64°F) / 12.3 °F ] = 1.44 h

finally for Td=85.3°F the time of death td= is

Td - T env = a*e^(-td/b)

td= -b* ln[(Td - T env)/a]

td= - 1.44 h * ln[(85.3 °F - 64°F )/12.3 °F ] ]= - 0.79 h from 10:30 am

therefore

td= - 0.79 h*60 min/hour= -47.4 min from 10:30 am = 9:42 am

Note :

we applied the following physical model to end with the exponencial function

Q= m*c * (T-To) → dQ/dt= m*c*dT/dt

and -dQ/dT= U*A*(T- T env)

then

m*c*dT/dt = -U*A*(T- T env)

-dT/dt = U*A/(m*C) *(T - T env)

ln ( T- T env) = U*A/(m*C) (-t) + C

T- T env = e^(U*A/(m*C) t + C) = e^[U*A/(m*C)*(-t)] *e^C

T- T env = a*e^(-t/b)