Question

Hot water cools from 80°C to 60°C in 1 minute. In cooling from 60°C to 50°C it will take (room temperature = 30°C)

• A. 48 s
• B. 42 s
• C. 50 s
• D. 45 s

Hint:

Newton’s Law of Cooling can be used to predict the time required for an object to cool, based on the temperature difference and ambient temperature.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This problem involves Newton's Law of Cooling, where the rate of cooling is proportional to the temperature difference between the object and the surroundings.
Step 2: Key Formula or Approach:
Average form of Newton's Law of Cooling:
\[ \frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right) \]
Step 3: Detailed Explanation:
Case 1: \( 80^\circ\text{C} \) to \( 60^\circ\text{C} \) in \( t_1 = 60\text{ s} \). Surroundings \( T_s = 30^\circ\text{C} \).
\[ \frac{80 - 60}{60} = K \left( \frac{80 + 60}{2} - 30 \right) \]
\[ \frac{20}{60} = K(70 - 30) \Rightarrow \frac{1}{3} = 40K \Rightarrow K = \frac{1}{120} \]
Case 2: \( 60^\circ\text{C} \) to \( 50^\circ\text{C} \) in time \( t_2 \).
\[ \frac{60 - 50}{t_2} = K \left( \frac{60 + 50}{2} - 30 \right) \]
\[ \frac{10}{t_2} = \frac{1}{120} (55 - 30) \]
\[ \frac{10}{t_2} = \frac{25}{120} \Rightarrow t_2 = \frac{10 \times 120}{25} = \frac{1200}{25} = 48\text{ s} \]
Step 4: Final Answer:
The time taken will be \( 48\text{ seconds} \).