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 cooling process.
The cooling of hot water follows Newton's Law of Cooling, which states that the rate of change of the temperature of an object is directly proportional to the difference between its temperature and the ambient temperature. The equation governing this law is:
\[ \frac{dT}{dt} = -k (T - T_{\text{ambient}}), \]
where:
- \( T \) is the temperature of the object,
- \( T_{\text{ambient}} \) is the ambient temperature (room temperature),
- \( k \) is a constant that depends on the nature of the object.

Step 2: Time required for cooling.
The time \( t \) for the temperature to change from \( T_1 \) to \( T_2 \) is given by:
\[ \ln \left( \frac{T_1 - T_{\text{ambient}}}{T_2 - T_{\text{ambient}}} \right) = k \cdot t. \]
We can use this equation to calculate the time required for the cooling process.

Step 3: Time for cooling from 80°C to 60°C.
We are given that the temperature of the water decreases from 80°C to 60°C in 1 minute. Using the equation above:
\[ \ln \left( \frac{80 - 30}{60 - 30} \right) = k \times 60 \, \text{s}. \]
Simplifying: \[ \ln \left( \frac{50}{30} \right) = k \times 60. \]
\[ \ln \left( \frac{5}{3} \right) = k \times 60. \]
\[ k = \frac{\ln \left( \frac{5}{3} \right)}{60}. \]

Step 4: Time for cooling from 60°C to 50°C.
Now, we calculate the time it will take for the water to cool from 60°C to 50°C. Using the same equation:
\[ \ln \left( \frac{60 - 30}{50 - 30} \right) = k \cdot t. \]
Simplifying:
\[ \ln \left( \frac{30}{20} \right) = k \cdot t. \]
\[ \ln \left( \frac{3}{2} \right) = k \cdot t. \]
Now substitute the value of \( k \) from
Step 3:
\[ t = \frac{\ln \left( \frac{3}{2} \right)}{k}. \]
Substitute \( k \) and solve for \( t \).
Final Answer:
The time taken for the water to cool from 60°C to 50°C is:
\[ \boxed{48 \, \text{s}}. \]