Question

an body cools down from 80°C to 60°C in 5 minutes The temperature pf the surrounding is 20°C. The time taken to cool from 60°C to 40°C is

• A. \(\frac{25}{3}s\)
• B. 500 s
• C. 450 s
• D. 420 s

Answer 1

The Correct Option is B

To solve this problem, we need to apply Newton's Law of Cooling, which states that the rate of heat loss of a body is directly proportional to the difference in temperature between the body and its surroundings, provided the temperature difference is small.

The formula for Newton's Law of Cooling can be expressed as:

\(\frac{d\theta}{dt} = -k(\theta - \theta_{\text{surrounding}})\) 

Where:

• \(\theta\) is the temperature of the body at time \(t\).
• \(\theta_{\text{surrounding}}\) is the surrounding temperature.
• \(k\) is the cooling constant.

Let's solve it step by step:

1. The initial temperature \(T_1\) is 80°C and the final temperature \(T_2\) is 60°C for the first condition with surrounding temperature \(T_s\) as 20°C.
2. According to Newton's formula, the difference in temperature over time is given by:
3. In the first case:
   • Initial temperature difference \(\Delta T_1 = 80 - 20 = 60\)°C,
   • Final temperature difference \(\Delta T_2 = 60 - 20 = 40\)°C.
   • The time taken for this change is 5 minutes or 300 seconds.
4. Using this information, we calculate the factor of cooling constant \(\frac{\Delta T_2}{\Delta T_1} = \exp(-kt_1)\).
5. Rearranging gives:
6. Now, apply the same formula for the time taken to cool from 60°C to 40°C, where the new initial temperature difference is 40°C, and final temperature difference is 20°C:
7. Solving for \(t_2\), we get this equation:
8. Calculate the value:
9. This simplifies to:

Therefore, the time taken to cool from 60°C to 40°C is 500 seconds. Hence, the correct answer is 500 s.