Question

water whose temperature is at 100°c is left to cool in a room where the temperature is 60°c. after 3 minutes, the water temperature is 90°. if the water temperature t is a function of time t given by t = 60 + 40e^{kt}, find the time for the water temperature to reach 65°c. round to the nearest hundredth of a minute. minutes question help: ebook

Explanation:

Step1: Find the value of \(k\)

Given \(T = 60+40e^{kt}\), when \(t = 3\), \(T=90\).
Substitute into the equation: \(90=60 + 40e^{3k}\).
First, simplify: \(90-60=40e^{3k}\), so \(30 = 40e^{3k}\).
Then \(e^{3k}=\frac{30}{40}=\frac{3}{4}\).
Take the natural logarithm of both sides: \(\ln(e^{3k})=\ln(\frac{3}{4})\).
Using the property \(\ln(e^{x})=x\), we get \(3k=\ln(\frac{3}{4})\).
So \(k=\frac{1}{3}\ln(\frac{3}{4})\approx\frac{1}{3}(- 0.2877)\approx - 0.0959\).

Step2: Find the time \(t\) when \(T = 65\)

Substitute \(T = 65\) and \(k\approx - 0.0959\) into \(T = 60+40e^{kt}\).
We have \(65=60 + 40e^{-0.0959t}\).
Simplify: \(65 - 60=40e^{-0.0959t}\), so \(5 = 40e^{-0.0959t}\).
Then \(e^{-0.0959t}=\frac{5}{40}=\frac{1}{8}\).
Take the natural logarithm of both sides: \(\ln(e^{-0.0959t})=\ln(\frac{1}{8})\).
Using \(\ln(e^{x})=x\), we get \(-0.0959t=\ln(\frac{1}{8})\).
Since \(\ln(\frac{1}{8})=- \ln(8)\approx - 2.0794\), then \(t=\frac{-2.0794}{- 0.0959}\approx21.68\).

Answer:

\(21.68\)