a cup of coffee has a temperature of 95°c when it is poured and allowed to cool in a room with a temperature…

Explanation:

Step1: Use Newton's Law of Cooling formula

Newton's Law of Cooling is given by (T(t)=T_{s}+(T_{0}-T_{s})e^{-kt}), where (T_{s}) is the surrounding temperature, (T_{0}) is the initial temperature of the object. Here, (T_{s} = 30^{\circ}C), (T_{0}=95^{\circ}C), so (T(t)=30+(95 - 30)e^{-kt}=30 + 65e^{-kt}).

Step2: Find the value of (k)

We know that when (t = 1), (T(1)=90^{\circ}C). Substitute into the formula: [ \begin{align*} 90&=30+65e^{-k\times1}\ 90 - 30&=65e^{-k}\ 60&=65e^{-k}\ e^{-k}&=\frac{60}{65}=\frac{12}{13}\ -k&=\ln(\frac{12}{13})\ k&=-\ln(\frac{12}{13})=\ln(\frac{13}{12})\approx0.0770 \end{align*} ]

Step3: Write the temperature function (T(t))

Substitute (k\approx0.0770) into (T(t)=30 + 65e^{-kt}), we get (T(t)=30+65e^{- 0.0770t})

Step4: Find the time (t) when (T(t) = 35^{\circ}C)

Set (T(t)=35), then [ \begin{align*} 35&=30+65e^{-0.0770t}\ 35 - 30&=65e^{-0.0770t}\ 5&=65e^{-0.0770t}\ e^{-0.0770t}&=\frac{5}{65}=\frac{1}{13}\ -0.0770t&=\ln(\frac{1}{13})\ t&=\frac{\ln(\frac{1}{13})}{- 0.0770}=\frac{-\ln(13)}{-0.0770}\approx34.19 \end{align*} ]

Answer:

(T(t)=30 + 65e^{-0.0770t}), and you must wait approximately (34.19) minutes until the coffee is (35^{\circ}C)