Question

A loaf of bread is removed from an oven and is left sitting in the kitchen to cool. Its temperature t minutes after being removed from the oven is given by the equation T(t) = 103e-0.0182t73. Find the temperature of the kitchen (assuming the temperature remains constant), the initial temperature of the bread when it is removed from the oven, and the temperature of the loaf of bread after 90 minutes.

Answer 1

Answer:

(a). 73° C.

(b). 176° C.

(c). 93.02°C.

Explanation:

In the question,

We have been given an equation,

$T(t) = 103e^{-0.0182t}+73.$

(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,

$T(\infty)=103e^{-0.0182*\infty}+73\\T(\infty)=103*0+73\\T(\infty)=0+73\\T(\infty)=73$

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,

$T(t) = 103e^{-0.0182t}+73.\\T(0)=103e^{-0.0182\times 0} + 73\\T(0)=103+73\\T(0)=176$

Therefore, the initial temperature of the bread is 176° C.

(c).

On putting the value of t = 90 minutes, we get,

$T(t) = 103e^{-0.0182t}+73.\\T(90)=103e^{-0.0182\times 90} + 73\\T(90)=20.0199+73\\T(90)=93.02$

Therefore, the temperature of the bread after 90 minutes is 93.02°C.