Question 1176057
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Hi
After 2 minutes, the temperature of the iron is 750 degrees.
T(t)=Ae^kt + Ts
t is time
•A is the difference between the initial temperature of the object and the surroundings (1100-350)
•k is a constant, the continuous rate of cooling of the object
T(2) = 750
750=750e^2k + 350
400 = 750e^2k
ln(400/750)/2 = k = -.314

(b) What will the temperature of the iron be after 10 minutes? 
T=750e^-3.14 + 350 = 382degrees

(c) How long will it take for the iron to reach 400 degrees?
400=750e^(-.314)t + 350
 ln(50/750)/-.314 = t = 8.6 min 

Wish You the Best in your Studies.
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