Newton's Law of Cooling Explained

Newton's Law of Cooling describes how an object loses heat to its surroundings. It states that the rate of heat loss is directly proportional to the difference in temperature between the object and the environment. This means that an object cools faster when it is much hotter than its surroundings, and the cooling process slows down as the object's temperature gets closer to the surroundings' temperature.

Cooling Rate Formulation

The mathematical representation of Newton's Law of Cooling is given by the differential equation:

Where:

• $ T $ is the temperature of the object at time $ t $.
• $ T_s $ is the constant temperature of the surroundings.
• $ \frac{dT}{dt} $ is the rate of change of temperature with time.
• $ k $ is a positive constant representing the rate of cooling, which depends on the object's properties and the surrounding conditions.

Integrating this equation gives us the time ($ \Delta t $) it takes for an object to cool from an initial temperature ($ T_1 $) to a final temperature ($ T_2 $):

Here, $ \ln $ represents the natural logarithm.

Cooling Process Analysis

We are given two cooling scenarios under the same external conditions:

First Cooling Interval (90°C to 80°C)

• Initial Temperature ($ T_1 $): $ 90^\circ C $
• Final Temperature ($ T_2 $): $ 80^\circ C $
• Time Taken ($ \Delta t_1 $): $ 5 $ minutes

Applying Newton's Law of Cooling formula:

Second Cooling Interval (80°C to 70°C)

• Initial Temperature ($ T_1 $): $ 80^\circ C $
• Final Temperature ($ T_2 $): $ 70^\circ C $
• Time Taken ($ \Delta t_2 $): $ ?$ (This is what we need to find)

Applying the formula for the second interval:

Comparing Cooling Times

To determine $ \Delta t_2 $, we compare the temperature differences in both intervals. Let's assume the surrounding temperature $ T_s $ is less than $ 70^\circ C $.

• The temperature difference in the first interval ranges from $ (90^\circ C - T_s) $ down to $ (80^\circ C - T_s) $.
• The temperature difference in the second interval ranges from $ (80^\circ C - T_s) $ down to $ (70^\circ C - T_s) $.

Crucially, the temperature difference is larger throughout the first interval ($ 90^\circ C $ to $ 80^\circ C $) than it is throughout the second interval ($ 80^\circ C $ to $ 70^\circ C $). Since the rate of cooling is directly proportional to this temperature difference, the object cools faster when the temperature difference is larger.

Therefore:

• The cooling rate from $ 90^\circ C $ to $ 80^\circ C $ is higher.
• The cooling rate from $ 80^\circ C $ to $ 70^\circ C $ is lower.

To achieve the same $ 10^\circ C $ drop in temperature, it will take longer when the cooling rate is slower. This implies that $ \Delta t_2 $ must be greater than $ \Delta t_1 $.

Physics MCQ Conclusion

Since the cooling rate decreases as the object's temperature approaches the surroundings' temperature, the time required to cool through a $ 10^\circ C $ interval will increase as the temperature drops. Cooling from $ 80^\circ C $ to $ 70^\circ C $ involves a smaller average temperature difference compared to cooling from $ 90^\circ C $ to $ 80^\circ C $. Thus, it will take longer than the initial 5 minutes.

The correct answer is that the body will take more than 5 minutes.