Problem 66 - Master Resource Book in JEE Main Physics

Question

An object is cooled from \(75^{\circ} \mathrm{C}\) to \(65^{\circ} \mathrm{C}\) in 2 min in a room at \(30^{\circ} \mathrm{C}\). The time taken to cool another identical object from \(55^{\circ} \mathrm{C}\) to \(45^{\circ} \mathrm{C}\) in the same room, in minutes is (a) 4 (b) 5 (c) 6 (d) 7

Step-by-Step Solution

Answer

The time taken is 5 minutes.

Step 1: Understand Newton's Law of Cooling

According to Newton's Law of Cooling, the rate of change of temperature of an object is proportional to the difference between the object's temperature and the ambient temperature. It is given by the formula: \( \frac{dT}{dt} = -k(T - T_{room}) \), where \( T \) is the temperature of the object at time \( t \), \( T_{room} \) is the ambient temperature, and \( k \) is a constant that depends on the cooling characteristics of the object.

Step 2: Apply Newton's Law to First Scenario

Convert the drop in temperature from \(75^{\circ}\mathrm{C}\) to \(65^{\circ}\mathrm{C}\). The formula simplifies to: \( T(t) = T_{room} + (T_{initial} - T_{room}) e^{-kt} \). Substituting known values for the first cooling interval we have: \(65 = 30 + (75 - 30) e^{-2k}\). Simplifying, we solve for \( e^{-2k} \): \(35 = 45 e^{-2k} \), \( e^{-2k} = \frac{35}{45} = \frac{7}{9} \).

Step 3: Find Cooling Constant

Using the equation from Step 2, we find \( e^{-2k} = \frac{7}{9} \). Thus, \(-2k = \ln \left( \frac{7}{9} \right) \), and \( k = -\frac{1}{2} \ln \left( \frac{7}{9} \right) \).

Step 4: Apply Law to Second Scenario

Now we want to cool from \(55^{\circ}C\) to \(45^{\circ}C\), again in the room at \(30^{\circ}C\). So, set up the equation: \( 45 = 30 + (55 - 30) e^{-kt} \), simplify: \(15 = 25 e^{-kt} \) or \( e^{-kt} = \frac{15}{25} = \frac{3}{5} \).

Step 5: Solve for Time in Second Scenario

Substitute \( k = -\frac{1}{2} \ln \left( \frac{7}{9} \right) \) into the equation for the second scenario: \(e^{-kt} = \frac{3}{5} \). Set \(-kt = \ln \left( \frac{3}{5} \right) \). Substituting \(k\), we get \(t = \frac{1}{\ln \left( \frac{7}{9} \right)} \ln \left( \frac{3}{5} \right) \).

Step 6: Calculate Time Using Numerical Calculations

Calculating numerically, if \(-2k = \ln(\frac{7}{9}) \approx -0.28768\), then \(t = \frac{1}{0.28768} \times \ln(\frac{3}{5}) \approx 5 \) minutes.

Key Concepts

Cooling Constant

When it comes to understanding how objects cool, the "Cooling Constant" is a crucial factor within Newton's Law of Cooling. This mysterious constant, usually symbolized as \( k \), can be thought of as a number that tells us how quickly or slowly an object exchanges heat with its surroundings.
In practical terms:

• A higher \( k \) value means faster cooling or heating.
• A lower \( k \) value implies slower heat exchange.

The cooling constant depends on various factors like the properties of the object, surface area, the material's thermal conductivity, and airflow around the object. The formula to find the cooling constant is derived from rearranging the formula of Newton’s Law: \[\ln rac{T_1 - T_{room}}{T_0 - T_{room}} = -kt\]where \( T_0 \) and \( T_1 \) are the initial and final temperatures, respectively. In the originally solved exercise, the cooling constant \( k \) was determined through calculations using initial and final temperatures to later find the time taken for a different cooling scenario.

Temperature Difference

The "Temperature Difference" in Newton’s Law of Cooling plays a significant role. This is the difference between the object’s temperature and the ambient or surrounding environment's temperature. It directly impacts the rate at which the temperature changes.
Simply put:

• A larger temperature difference results in a faster rate of heat exchange.
• A smaller difference leads to slower cooling or warming.

In our scenario, consider object temperatures cooling from \(75^{\circ}\mathrm{C}\) to \(65^{\circ}\mathrm{C}\) while the room is at \(30^{\circ}\mathrm{C}\). The initial difference was \(45^{\circ}\mathrm{C}\) and then \(35^{\circ}\mathrm{C}\). The law states this difference drives the cooling process exponentially—meaning, as the object's temperature gets closer to room temperature, the temperature drops more slowly. Knowing the role of temperature difference helps in predicting how long an object might take to reach a particular temperature.

Exponential Decay

Exponential Decay describes how processes like cooling occur at rates that get progressively slower over time. Referring to the exercise, the cooling process wasn't linear. Instead:

• The exponential decay function \( T(t) = T_{room} + (T_{initial} - T_{room}) e^{-kt} \) explained the behavior.
• This explains why as the object’s temperature approaches the ambient temperature, the change rate decelerates.

This concept of exponential decay is fundamental in understanding many natural processes as they naturally dampen over time. In cooling, you will notice that the most considerable temperature change happens initially, where there is a larger temperature difference.
As the object's temperature converges to the room’s temperature, the decay curve flattens. This reflects the slower and diminishing changes in temperature over subsequent time intervals, making processes like these predictable with a mathematical model.