Two friends sit down to talk and enjoy a cup of coffee. When the coffee is served, the impatient friend immediately adds a teaspoon of cream to his coffee. The relaxed friend waits 5 min before adding a teaspoon of cream (which has been kept at a constant temperature). The two now begin to drink their coffee. Who has the hotter coffee? Assume that the cream is cooler than the air and has the same heat capacity per unit volume as the coffee, and that Newton’s law of cooling governs the heat transfer.

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’s law of cooling,

$$\begin{gathered} \frac{dT}{dt}\propto \left(M-T\right) \\ \frac{dT}{dt}=-\lambda \left(M-T\right) \end{gathered}$$

where, $\lambda$is the constant of proportionality.

So, the initial value problem is,

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

Now, integrating both sides,

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

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),

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

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

The temperature of coffee at time t is,

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

So, the temperature of coffee after 5minutes is,

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

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.