Question

you place a cup of 205°f coffee on a table in a room that is 72°f, and 10 minutes later, it is 195°f. approximately how long will it be before the coffee is 180°f? use newtons law of cooling: ( t(t)=t_{a}+(t_{0}-t_{a})e^{-kt} ) 1 hour 15 minutes 45 minutes 25 minutes

Explanation:

Step1: Find the value of \(k\)

Given \(T_0 = 205\), \(T_A=72\), \(t = 10\), \(T(10)=195\)
Substitute into \(T(t)=T_A+(T_0 - T_A)e^{-kt}\)
\(195=72+(205 - 72)e^{-10k}\)
\(195 - 72=133e^{-10k}\)
\(123 = 133e^{-10k}\)
\(e^{-10k}=\frac{123}{133}\)
Take the natural logarithm of both sides: \(- 10k=\ln(\frac{123}{133})\)
\(k=-\frac{1}{10}\ln(\frac{123}{133})\approx-\frac{1}{10}\times(- 0.078)=0.0078\)

Step2: Find the value of \(t\) when \(T(t) = 180\)

Substitute \(T_0 = 205\), \(T_A = 72\), \(k = 0.0078\) and \(T(t)=180\) into \(T(t)=T_A+(T_0 - T_A)e^{-kt}\)
\(180=72+(205 - 72)e^{-0.0078t}\)
\(180 - 72=133e^{-0.0078t}\)
\(108 = 133e^{-0.0078t}\)
\(e^{-0.0078t}=\frac{108}{133}\)
Take the natural logarithm of both sides: \(-0.0078t=\ln(\frac{108}{133})\)
\(t=-\frac{\ln(\frac{108}{133})}{0.0078}\approx-\frac{- 0.209}{0.0078}\approx27\) (closest to 25 minutes)

Answer:

25 minutes