Answer
Let the temperature of object at any instant of time is repersented by $T(t)$ , $t$ repersents the time interval for which object is placed in room.
Now, according to question $\frac{dT}{dt}=-40^0F/h$, here negative sign repersents the decrease in temperature with respect time.
Also, $T(0)=615^0F$
On integrating, we have
$T(t)=-40t+c_1$
Using the intial values,
$615=-40(0)+c_1$
$c_1=615$
On putting this value in equation we have
$T(t)=-40t+615$
At $4 p.m$, $T=135^0F$
$135^0F=-40(t)+615$
$t=12 h$
Hence the object was placed at $4$ $a.m$
Work Step by Step
At $4$ a.m
