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Two friends sit down to talk and enjoy a cup of coffee. One friend added a teaspoon of cream to his coffee immediately when the coffee was served. The other friend added the cream to his coffee after 5 minutes. We have to determine that whose coffee is hotter.

Let the initial temperature of coffee at time, $\mathrm{t}=0$be $T_0$.

Let the constant outside temperature be M and the temperature of the coffee be T at time t. So, by Newton鈥檚 law of cooling,

$$\begin{gathered} \frac{dT}{dt}鈭�\left(M-T\right) \\ \frac{dT}{dt}=-位\left(M-T\right) \end{gathered}$$

where, $位$is the constant of proportionality.

So, the initial value problem is,

$\frac{dT}{dt}=-位\left(M-T\right),\mathrm{T}\left(0\right)={\mathrm{T}}_0$

Now, integrating both sides,

$$\begin{gathered} 鈭�\frac{1}{M-T}dT=-鈭�位dt \\ ln\left(M-T\right)=-位t+C....................\left(2\right) \end{gathered}$$

Where, C is an arbitrary constant.

When $\mathrm{t}=0$,$\mathrm{T}={\mathrm{T}}_0$

Therefore, from (1)

$C=ln\left(M-T_0\right)$

Substituting this value of C in equation (1),

$$\begin{gathered} ln\left(M-T\right)=-位t+ln\left(M-T_0\right) \\ ln\left(M-T\right)-ln\left(M-T_0\right)=-位t \\ ln\left(\frac{M-T}{M-T_0}\right)=-位t \\ \frac{M-T}{M-T_0}=e^{-位t} \\ T=M-\left(M-T_0\right)e^{-位t} \end{gathered}$$

Thus, the temperature of coffee at time t is role="math" localid="1664178540900" $\mathit{T}\left(t\right)\mathbf{=}\mathit{M}\mathbf{-}\left(M-T_0\right){\mathit{e}}^{\mathbf{-}\mathbf{位}\mathbf{t}}$.

The temperature of coffee at time t is,

$T\left(t\right)=M-\left(M-T_0\right)e^{-位t}$ (From step2)

So, the temperature of coffee after 5minutes is,

$T\left(5\right)=M-\left(M-T_0\right)e^{-5位}$

Therefore, the cream is cooler than the air, so the relaxed friend who waits for 5 minutes before adding the cream will have the hottest coffee.