(a). 73° C.
(b). 176° C.
(c). 93.02°C.
In the question,
We have been given an equation,
[tex]T(t) = 103e^{-0.0182t}+73.[/tex]
(a).
We need to find the temperature of the kitchen,
So,
Time, t → ∞
On putting the value of t as infinite in the equation we get,
[tex]T(\infty)=103e^{-0.0182*\infty}+73\\T(\infty)=103*0+73\\T(\infty)=0+73\\T(\infty)=73\\[/tex]
Therefore, the temperature of the Kitchen is 73° C.
(b).
The initial temperature of the bread when it is removed from the oven is at, t = 0 s
So,
On putting the value of t = 0, we get,
[tex]T(t) = 103e^{-0.0182t}+73.\\T(0)=103e^{-0.0182\times 0} + 73\\T(0)=103+73\\T(0)=176[/tex]
Therefore, the initial temperature of the bread is 176° C.
(c).
On putting the value of t = 90 minutes, we get,
[tex]T(t) = 103e^{-0.0182t}+73.\\T(90)=103e^{-0.0182\times 90} + 73\\T(90)=20.0199+73\\T(90)=93.02[/tex]
Therefore, the temperature of the bread after 90 minutes is 93.02°C.