Question

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newtons law of cooling states that for a cooling substance with initial temperature $t_0$, the temperature $t(t)$ after $t$
minutes can be modeled by the equation $t(t)=t_s+(t_0 - t_s)e^{-kt}$, where $t_s$ is the surrounding temperature
and $k$ is the substances cooling rate.
a liquid substance is heated to $80^{circ}c$. upon being removed from the heat, it cools to $60^{circ}c$ in 12 min.
what is the substances cooling rate when the surrounding air temperature is $50^{circ}c$?
round the answer to four decimal places.
0.0687
0.0732
0.0813
0.0916

Explanation:

Step1: Substitute the given values into the formula

Given \(T_0 = 80\), \(T_s=50\), \(t = 12\), \(T(12)=60\).
Substitute into \(T(t)=T_s+(T_0 - T_s)e^{-kt}\), we get \(60 = 50+(80 - 50)e^{-12k}\).

Step2: Simplify the equation

First, simplify \(60 = 50+(80 - 50)e^{-12k}\) to \(60-50 = 30e^{-12k}\), so \(10=30e^{-12k}\).
Then divide both sides by 30: \(\frac{10}{30}=e^{-12k}\), which is \(\frac{1}{3}=e^{-12k}\).

Step3: Take the natural logarithm of both sides

Take \(\ln\) on both sides: \(\ln(\frac{1}{3})=\ln(e^{-12k})\).
Since \(\ln(e^{-12k})=- 12k\) and \(\ln(\frac{1}{3})=-\ln(3)\), we have \(-\ln(3)=-12k\).

Step4: Solve for \(k\)

Solve for \(k\): \(k=\frac{\ln(3)}{12}\).
Calculate \(\ln(3)\approx1.0986\), then \(k=\frac{1.0986}{12}\approx0.0916\).

Answer:

0.0916