Question

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:

• A. \( \frac{13}{5} t \)
• B. \( \frac{10}{13} t \)
• C. \( \frac{13}{10} t \)
• D. \( \frac{5}{13} t \)

Hint:

Use Newton's law of cooling to solve such problems. The temperature difference between the object and the environment is proportional to the rate of cooling.

Answer 1

The Correct Option is A

By using the average form of Newton's law of cooling: \[ \frac{90 - 80}{t} = k \left( \frac{90 + 80 - 20}{2} \right) \quad \cdots (i) \] \[ \frac{80 - 60}{t'} = k \left( \frac{80 + 60 - 20}{2} \right) \quad \cdots (ii) \] From equations (i) and (ii), we get: \[ 10 \times t' = 65 \times t \] \[ t' = \frac{65}{50} \times 2t = \frac{13}{5} t \] Thus, the time taken to cool from 80°C to 60°C is \( \frac{13}{5} t \), so the correct answer is option (1).