A cup of coffee cools from 908C to 808C in t minutes, when the room temperature is 208C. The time taken by a similar cup of coffee to cool from 808C to 608C at a room temperature same at 208C is :

The correct answer is 13/5 t.

Note: The temperatures in the question text appear as "908C", "808C", etc., due to an OCR error. These correctly represent 90°C, 80°C, 60°C, and 20°C.

According to the average form of Newton's Law of Cooling, the rate of change of temperature of an object is proportional to the difference between its average temperature and the ambient temperature (room temperature). The formula is given by:

$\frac{T_1-T_2}{t}=K(\frac{T_1+T_2}{2}-T_0)$

Where:

• $T_1$ is the initial temperature.
• $T_2$ is the final temperature.
• $t$ is the time taken to cool.
• $T_0$ is the room temperature.
• $K$ is a positive cooling constant.

Case 1: The coffee cools from 90°C to 80°C in time $t$, with a room temperature of 20°C.

Plugging these values into the formula:

$\frac{90-80}{t}=K(\frac{90+80}{2}-20)$

$\frac{10}{t}=K(85-20)$

$\frac{10}{t}=65K$     ... (Equation 1)

Case 2: The coffee cools from 80°C to 60°C in a new time $\mathrm{t\text{'}}$, with the same room temperature of 20°C.

Applying the formula again:

$\frac{80-60}{\mathrm{t\text{'}}}=K(\frac{80+60}{2}-20)$

$\frac{20}{\mathrm{t\text{'}}}=K(70-20)$

$\frac{20}{\mathrm{t\text{'}}}=50K$     ... (Equation 2)

Now, divide Equation 1 by Equation 2 to eliminate $K$:

$\frac{\left(\frac{10}{t}\right)}{\left(\frac{20}{\mathrm{t\text{'}}}\right)}=\frac{65K}{50K}$

$\frac{10}{t}\times \frac{\mathrm{t\text{'}}}{20}=\frac{65}{50}$

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

Multiplying both sides by $2t$ gives:

$\mathrm{t\text{'}}=2t\times \frac{13}{10}$

$\mathrm{t\text{'}}=\frac{26}{10}t=\frac{13}{5}t$

Therefore, the time taken for the coffee to cool from 80°C to 60°C is 13/5 t.