Question

Newton's law of cooling indicates that the temperature of a warn object will decrease exponentially with time and will approach the temperature of the

Exponential models
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asked 2021-02-09
Newton's law of cooling indicates that the temperature of a warn object will decrease exponentially with time and will approach the temperature of the surrounding air. The temperature T(t) is modeled by \(\displaystyle{T}{\left({t}\right)}={T}_{{a}}+{\left({T}_{{0}}-{T}_{{a}}\right)}{e}^{{-{k}{t}}}\). In this model, \(\displaystyle{T}_{{a}}\) represents the temperature of the surrounding air, \(\displaystyle{T}_{{0}}\) represents the initial temperature of the object and t is the time after the object starts cooling. The value of k is the cooling rate and is a constant related to the physical properties the object.
A cake comes out of the oven at 335F and is placed on a cooling rack in a 70F kitchen. After checking the temperature several minutes later, it is determined that the cooling rate k is 0.050. Write a function that models the temperature T(t) of the cake t minutes after being removed from the oven.

Answers (1)

2021-02-10

Given:
The initial temperature (T0 ) of the cake is \(335^{\circ}F\).
The tempearture (Ta) of the cake when place in cooling tray is \(70^{\circ}F\).
The cooling rate (k) is 0.050.
Known fact:
By the Newton's law of cooling , the temperature is modeled as
\(\displaystyle{T}{\left({t}\right)}={T}_{{a}}+{\left({T}_{{0}}-{T}_{{a}}\right)}{e}^{{-{k}{t}}}\)
where, \(\displaystyle{T}_{{a}}\) is the temperature surronded by air.
\(\displaystyle{T}_{{0}}\) is the initial temperature
t is the time after the object starts cooling,
k is the cooling rate.
Calculation:
The function that models the temperature of the cake t minutes after being removed from the oven is computed as follows.
Substitute
\(\displaystyle{T}_{{0}}={335}{F},{T}_{{a}}={70}{F}\) and \(k=0.050 \in\) \(\displaystyle{T}{\left({t}\right)}={T}_{{a}}+{\left({T}_{{0}}+{T}_{{a}}\right)}{e}^{{-{k}{t}}}\)
\(\displaystyle{T}{\left({t}\right)}={70}+{\left({335}-{70}\right)}{e}^{{-{\left({0.050}\right)}{t}}}\)
\(\displaystyle={70}+{265}{e}^{{-{\left({0.050}\right)}{t}}}\)
Answer:
The function that models the temperature of the cake t minutes after it is removed from the oven is \(\displaystyle{T}{\left({t}\right)}={70}+{265}{e}^{{-{\left({0.050}\right)}{t}}}\)

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