Question 1078775
Newton’s Law of Cooling Formula is expressed by

T(t) = Ts + (T0 – Ts) e-Kt

Where,

t = time,

T(t) = temperature of the given body at time t,

Ts = surrounding temperature,

To = initial temperature of the body,

k = constant.
 
So:
T(7)=68+(350-68)e^-kt
299.8=68+282e^-7k
e^-7k=0.82198581560283687943262411347518
ln e^-7k=ln 0.82198581560283687943262411347518
-7k=-0.1960321400324622534737708260656
k=0.02800459143320889335339583229509

Then:
T(12)=68+(350-68)e^-12(0.02800459143320889335339583229509)
T(12)=68+(282)0.71458373316960990812286116751929
T(12)=269.512 degrees 12 minutes out of the oven. ☺☺☺☺