Solution
Let us analyze the cooling process using Newton’s Law of Cooling. The rate of cooling is proportional to the temperature difference between the body and the surroundings.
$$ \frac{dT}{dt} = -k(T – T_s) $$
Analysis of the Methods:
We are given that the heat transfer from the milk to the cup is almost instantaneous. This implies that the time taken for the milk and cup to reach a common equilibrium temperature upon mixing is negligible.
Let $m$ be the mass of the milk, $M$ be the mass of the cup, $c$ be the specific heat of milk, and $C$ be the heat capacity of the cup. $T_1$ is the initial temperature of the milk, and $T_s$ is the surrounding temperature.
Method (a): Pour immediately. The milk and cup instantly reach a common temperature $T_{mix}$ via energy conservation. Then, the system cools as a single unit from $T_{mix}$ to $T_2$.
Method (b): Wait, then pour. The milk cools alone from $T_1$ to some intermediate temperature $T’$. Then it is poured, and the mixing causes the temperature to drop instantly to $T_2$.
Key Insight: Since the time required for internal heat redistribution (mixing) is zero, the total time depends only on the rate of heat loss to the surroundings. The rate of heat loss $dQ/dt$ is determined by the surface properties and the temperature difference $(T – T_s)$.
In both scenarios, to reach the final state $T_2$, the system must lose a specific total amount of thermal energy to the surroundings. Since the heat transfer coefficients and effective surface areas for cooling are comparable in the configuration where the cup is involved, and the driving potential is $\Delta T$, the total time elapsed to reach the final thermodynamic state is independent of the order of operations.
Conclusion: All three methods (a), (b), and (c) will take the same amount of time to reach the final temperature $T_2$.