A cup of coffee cools from $90°C$to $80°C$ in t minutes when the room temperature is$20°C$. The time taken by the similar cup of coffee to cool from $80°C$ to $60°C$ at the same room temperature is :

Solving using the Average Temperature Method of Newton’s Law of Cooling

Newton’s Law of Cooling states:

$$\frac{dT}{dt} = -k (T - T_{\text{room}})$$

where:

$T$ is the temperature of the object (coffee),

$T_{\text{room}}$ is the ambient room temperature,

$k$ is a proportionality constant,

$t$ is time.

Using the average temperature method, we approximate the cooling rate using the average temperature difference from the surroundings:

$$\text{Rate of cooling} \propto \text{Average excess temperature}$$

 $$t \propto \frac{\Delta T}{\text{Average temperature excess}}$$

where: $\Delta T = T_{\text{initial}} - T_{\text{final}}$,

Average temperature excess is:

$$T_{\text{avg}} - T_{\text{room}} = \frac{T_{\text{initial}} + T_{\text{final}}}{2} - T_{\text{room}}$$

Step 1: Compute Time for Cooling from 90°C to 80°C

Given:

$T_{\text{initial}} = 90^\circ C$,

$T_{\text{final}} = 80^\circ C$,

$T_{\text{room}} = 20^\circ C$.

The average temperature excess:

$$T_{\text{avg}} = \frac{90 + 80}{2} = 85^\circ C$$

 $$T_{\text{avg}} - T_{\text{room}} = 85 - 20 = 65^\circ C$$

Time taken is proportional to:

$$t \propto \frac{(90 - 80)}{65} = \frac{10}{65}$$

Let this time be $t$.

Step 2: Compute Time for Cooling from 80°C to 60°C

Given:

$T_{\text{initial}} = 80^\circ C$,

$T_{\text{final}} = 60^\circ C$.

The average temperature excess:

$$T_{\text{avg}} = \frac{80 + 60}{2} = 70^\circ C$$

 $$T_{\text{avg}} - T_{\text{room}} = 70 - 20 = 50^\circ C$$

Time taken is proportional to:

$$t' \propto \frac{(80 - 60)}{50} = \frac{20}{50}$$

Step 3: Compute Ratio of Times

$$\frac{t'}{t} = \frac{\frac{20}{50}}{\frac{10}{65}}$$$$= \frac{20}{50} \times \frac{65}{10}$$ $$= \frac{20 \times 65}{50 \times 10}$$ $=\frac{1300}{500}=\frac{13}{5}$

$t^{\text{'}}=\frac{13}{5}t$ 

Final Answer:

The time taken to cool from 80°C to 60°C is $\frac{13}{5}$ times the time taken to cool from 90°C to 80°C.

Thus, $t^{\text{'}}=\frac{13}{5}t$