#### Answer

The murder took place at 11:54 am

#### Work Step by Step

Let the initial temperature be $32.5^{\circ}C$ at 1:30 pm
We can find $k$:
$T(t) = 20+12.5~e^{kt}$
$T(1) = 20+12.5~e^{k} = 30.3$
$12.5~e^{k} = 10.3$
$e^{k} = \frac{10.3}{12.5}$
$k = ln(\frac{10.3}{12.5})$
$k = -0.193584749$
Now we can let the initial temperature be $37.0^{\circ}C$.
Then:
$T(t) = 20+17~e^{-0.193584749~t}$
We can find the time $t$ when the temperature is $32.5^{\circ}C$:
$T(t) = 20+17~e^{-0.193584749~t}$
$20+17~e^{-0.193584749~t} = 32.5$
$e^{-0.193584749~t} = \frac{12.5}{17}$
$-0.193584749~t = ln(\frac{12.5}{17})$
$t = \frac{ln(\frac{12.5}{17})}{-0.193584749}$
$t = 1.6$
The murder took place 1.6 hours (i.e. 1 hour and 36 minutes) before 1:30 pm. Therefore, the murder took place at 11:54 am.