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review for exam 3 over chapters 6 and 11 score: 17.25/26 answered: 20/2…

Question

review for exam 3 over chapters 6 and 11
score: 17.25/26 answered: 20/26
question 18
an object with initial temperature 170°f is submerged in large tank of water whose temperature is 60°f.
find a formula for f(t), the temperature of the object after t minutes, if the cooling constant is k = 1.4
f(t)=
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Explanation:

Step1: Recall Newton's Law of Cooling formula

Newton's Law of Cooling is given by \(F(t)=T_s+(F_0 - T_s)e^{-kt}\), where \(T_s\) is the surrounding temperature, \(F_0\) is the initial temperature of the object, \(k\) is the cooling constant, and \(t\) is the time.
Here, \(T_s = 60\), \(F_0=170\), and \(k = 1.4\).

Step2: Substitute the values into the formula

Substitute \(T_s = 60\), \(F_0 = 170\), and \(k=1.4\) into the formula \(F(t)=T_s+(F_0 - T_s)e^{-kt}\).
We get \(F(t)=60+(170 - 60)e^{-1.4t}\).
Simplify the expression inside the parentheses: \(F(t)=60 + 110e^{-1.4t}\).

Answer:

\(F(t)=60+110e^{-1.4t}\)