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

Using Newton's law of cooling in the approximate form:

$$ \frac{\Delta T}{t} = k\left(\bar{T} - T_s\right) $$

where $$\bar{T}$$ is the average temperature and $$T_s$$ is the surrounding temperature.

First case: Cooling from 90° to 80° in time $$t$$:

$$\bar{T} = 85°$$, $$\Delta T = 10°$$

$$ \frac{10}{t} = k(85 - 20) = 65k \quad \cdots (1) $$

Second case: Cooling from 80° to 60° in time $$t'$$:

$$\bar{T} = 70°$$, $$\Delta T = 20°$$

$$ \frac{20}{t'} = k(70 - 20) = 50k \quad \cdots (2) $$

Dividing (2) by (1):

$$ \frac{20/t'}{10/t} = \frac{50k}{65k} $$

$$ \frac{2t}{t'} = \frac{10}{13} $$

$$ t' = \frac{2t \times 13}{10} = \frac{26t}{10} = \frac{13t}{5} $$

The correct answer is Option 4: $$\frac{13}{5}t$$.