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 a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is :

The correct option is (13/5)t.

To find the time taken for the coffee to cool, we can use the average form of Newton's Law of Cooling, which is expressed as:
$\frac{T_1-T_2}{t}=K\left(\frac{T_1+T_2}{2}-T_s\right)$
where:
• $T_1$ is the initial temperature of the body.
• $T_2$ is the final temperature of the body.
• $T_s$ is the surrounding room temperature.
• $t$ is the time taken to cool.
• $K$ is a positive constant representing cooling efficiency.

Step 1: Apply the formula for the first case
In the first scenario, the coffee cools from $90°C$ to $80°C$ in time $t$ minutes when the room temperature is $20°C$.
Substituting these values into the cooling formula:
$\frac{90-80}{t}=K\left(\frac{90+80}{2}-20\right)$
$\frac{10}{t}=K\left(85-20\right)$
$\frac{10}{t}=65K$ — (Equation 1)

Step 2: Apply the formula for the second case
In the second scenario, the coffee cools from $80°C$ to $60°C$ in an unknown time $t\text{'}$ at the same room temperature of $20°C$.
Substituting these values:
$\frac{80-60}{t\text{'}}=K\left(\frac{80+60}{2}-20\right)$
$\frac{20}{t\text{'}}=K\left(70-20\right)$
$\frac{20}{t\text{'}}=50K$ — (Equation 2)

Step 3: Solve for the unknown time t'
Divide Equation 1 by Equation 2:
$\frac{10/t}{20/t\text{'}}=\frac{65K}{50K}$
Simplifying the fractions:
$\frac{t\text{'}}{2t}=\frac{13}{10}$
Isolate $t\text{'}$:
$t\text{'}=2t\times \frac{13}{10}$
$t\text{'}=\frac{13}{5}t$

Thus, the time taken by the similar cup of coffee to cool from $80°C$ to $60°C$ is (13/5)t.