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Step-by-Step Solution
Step 1: State Newton’s Law of Cooling
According to Newton's law of cooling, the rate of change of temperature of a body is proportional to the difference between the body's temperature and the ambient (or surrounding) temperature. Mathematically:
$ \frac{dT}{dt} = K \bigl(T - T_0\bigr) $
where:
$ T $ is the temperature of the object (coffee in this case).
$ T_0 $ is the ambient (room) temperature.
$ K $ is a constant of proportionality.
Step 2: Identify the Temperature Drops and Time Intervals
From the problem statement:
The coffee cools from $90^\circ \!C$ to $80^\circ \!C$ in $t$ minutes.
The room temperature is $20^\circ \!C$.
We wish to find the time, $t'$, it takes to cool from $80^\circ \!C$ to $60^\circ \!C$ under the same ambient conditions.
Step 3: Use Average Temperature for Approximation
For convenience in typical Newton's law of cooling problems, we take the "average" temperature of the coffee during the cooling from an initial temperature $T_1$ to a final temperature $T_2$. This is often used to simplify the analysis when the exact integral form is not used. Therefore, during the first interval:
Initial temperature $T_1 = 90^\circ \!C$
Final temperature $T_2 = 80^\circ \!C$
Average temperature $T_{av1} = \frac{90 + 80}{2} = 85^\circ \!C$
Step 4: Relate the Cooling Rate for the First Interval
Over the time $t$ for the temperature to drop by $10^\circ \!C$ (from $90^\circ \!C$ to $80^\circ \!C$), we write:
$ \frac{10}{t} = K \bigl( T_{av1} - T_0 \bigr) = K \bigl(85 - 20 \bigr). $
So,
$ \frac{10}{t} = K \times 65. $
Step 5: Relate the Cooling Rate for the Second Interval
Next, for the coffee to cool from $80^\circ \!C$ to $60^\circ \!C$, the total drop is $20^\circ \!C$. The average temperature in this interval is:
$ T_{av2} = \frac{80 + 60}{2} = 70^\circ \!C. $
Hence, for the time taken $t'$ to cool by $20^\circ \!C,$ we have:
$ \frac{20}{t'} = K \bigl(70 - 20 \bigr) = K \times 50. $
Step 6: Eliminate the Constant K
We have two equations involving $K$:
$ \frac{10}{t} = 65K \quad \text{and} \quad \frac{20}{t'} = 50K. $
From the first equation, we get $ K = \frac{10}{65 t} $. Substitute this into the second equation:
$ \frac{20}{t'} = 50 \left( \frac{10}{65 t} \right). $
So:
$ \frac{20}{t'} = \frac{500}{65 \, t}. $
Step 7: Solve for t'
Rearrange to find $ t' $:
$ t' = \frac{20 \times 65 \, t}{500} = \frac{1300}{500} \, t = \frac{13}{5} \, t. $
Step 8: State the Final Answer
Thus, the time taken to cool from $80^\circ \!C$ to $60^\circ \!C$ is:
$ t' = \frac{13}{5} \, t. $
