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Chapter 12: Problem 10

An object is cooled from \(75^{\circ} \mathrm{C}\) to \(65^{\circ} \mathrm{C}\) in 2 minules in a room at \(30^{\circ} \mathrm{C}\). The time taken to cool the same object from \(55^{\circ} \mathrm{C}\) to \(45^{\circ} \mathrm{C}\) in the same room is (a) 5 minutes (b) 3 minutes (c) 4 minutes (d) 2 minutes

Short Answer

Expert verified
The time taken is approximately 4 minutes (option c).

Step by step solution

01

Understanding the Cooling Process

The situation described involves Newton's Law of Cooling, which states that the rate of cooling is proportional to the temperature difference between the object and the ambient temperature.
02

Newton's Law of Cooling Formula

Newton's Law of Cooling is given by the formula: \( T(t) = T_{ ext{ambient}} + (T_{ ext{initial}} - T_{ ext{ambient}}) imes e^{-kt}\), where \(T(t)\) is the temperature at time \(t\), \(T_{\text{ambient}}\) is the room temperature, \(T_{\text{initial}}\) is the initial temperature, and \(k\) is a constant.
03

Calculate First Cooling Phase Constant

Using the information that the object cools from \(75^{\circ}\text{C}\) to \(65^{\circ}\text{C}\), we can write the equation: \(65 = 30 + (75 - 30) \times e^{-2k}\). Simplifying this gives \(35 = 45 \times e^{-2k}\). Solving for \(k\) yields: \(e^{-2k} = \frac{35}{45}\), hence \(k = -\frac{1}{2}\ln\left(\frac{35}{45}\right)\).
04

Calculate Second Cooling Phase Time

Now, use the same formula for the second phase where the object cools from \(55^{\circ}\text{C}\) to \(45^{\circ}\text{C}\): \(45 = 30 + (55 - 30) \times e^{-kt}\). Simplifying gives \(15 = 25 \times e^{-kt}\). Solving for \(t\) yields: \(e^{-kt} = \frac{15}{25}\).
05

Solve for Time \(t\)

Since we know \(e^{-2k} = \frac{7}{9}\) from Step 3, substitute \(k\) from \(e^{-kt} = \frac{15}{25}\) to find \(t\). Express \(\left(\frac{3}{5}\right) = \left(\frac{7}{9}\right)^{t/2}\). Solve for \(t\): \(t = 2 \times \frac{\ln(3/5)}{\ln(7/9)}\). Evaluating this expression gives \(t \approx 4\) minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cooling Process
Cooling is a fascinating aspect of physics that occurs when an object loses heat to its surrounding environment. In the context of Newton's Law of Cooling, this is specifically described as the process where the rate at which an object cools is directly related to the difference in temperature between the object itself and the surrounding environment.
Understanding this fundamental principle helps explain why a hot object, such as a cup of coffee, cools down more quickly when there is a larger difference between its temperature and that of the room. Conversely, it cools more slowly as the temperatures approach each other.
  • It highlights the dynamic relationship between an object and its environment.
  • It forms the basis of numerous applications, like climate control and refrigeration.
Temperature Difference
The key factor driving the cooling process is the temperature difference between the object and its surrounding environment. This difference, often denoted as \( \Delta T \) in equations, is essential in determining the rate at which an object gains or loses heat. For instance, if an object is at a high temperature of 75掳C and the surrounding is 30掳C, there is a 45掳C temperature difference.
This substantial difference accelerates the cooling, whereas a smaller temperature gap, such as from 55掳C to 45掳C, results in a slower heat loss. Understanding this helps to safely and efficiently manage temperature changes across different materials and processes.
  • A large temperature difference means faster cooling.
  • As temperatures get closer, the difference decreases and so does the cooling rate.
Exponential Decay
The cooling process as per Newton's Law is characterized by exponential decay. This means that the temperature does not drop linearly but rather decreases at a rate that gets progressively slower over time. Mathematically, this is expressed through the formula \( T(t) = T_{\text{ambient}} + (T_{\text{initial}} - T_{\text{ambient}}) \cdot e^{-kt} \).
Here, \( e^{-kt} \) represents the exponential factor, where \( k \) is a constant dependent on the object's material properties. Exponential decay is notable in thermal physics as it explains why objects cool rapidly at first and then more gradually.
  • Easily predictable using the decay constant \( k \).
  • Enables calculating temperature at any given time \( t \).
Thermal Physics
Thermal physics is an extensive field that deals with the dynamics of heat flow, temperature changes, and energy transformations. As part of this, Newton's Law of Cooling provides insight into how objects reach thermal equilibrium with their environment.
It highlights how fundamental concepts like temperature difference, material properties, and exponential decay come together to describe real-world thermal processes. It helps in designing systems like heating, ventilation, and air conditioning (HVAC) by predicting how changes in one parameter affect overall thermal dynamics.
  • Connects microscopic interactions with macroscopic temperature changes.
  • Essential for energy management and machinery efficiency.

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Most popular questions from this chapter

When the tempcrature of a black body increases by \(0.1 \%\), the wavelength corresponding to maximum emission changes by \(0.13 \mu \mathrm{m}\). The initial wavelength corresponding to maximum emission is (a) \(0.26 \mu \mathrm{m}\) (b) \(0.13 \mathrm{um}\) (c) \(0.13 \mathrm{~mm}\) (d) \(0.26 \mathrm{~mm}\)

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