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A body cools from \(50^{\circ} \mathrm{C}\) to \(49^{\circ} \mathrm{C}\) in \(5 \mathrm{~s}\). How long will it take to cool from \(40^{\circ} \mathrm{C}\) to \(39^{\circ} \mathrm{C}\) ? Assume temperature of surroundings to be \(30^{\circ} \mathrm{C}\) and Newton's law of cooling is valid [BVP Engg- 2008] (a) \(2.5 \mathrm{~s}\) (b) \(10 \mathrm{~s}\) (c) \(20 \mathrm{~s}\) (d) \(5 \mathrm{~s}\)

Short Answer

Expert verified
The body will take 10 seconds to cool from 40°C to 39°C. (Answer: b)

Step by step solution

01

Understand Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference in temperature between the object and its surroundings. Mathematically, it is represented as \( \frac{d\vartheta}{dt} = -k(\vartheta - \vartheta_s) \), where \( \vartheta \) is the temperature of the body at time \( t \), \( \vartheta_s \) is the temperature of the surroundings, and \( k \) is the cooling constant.
02

Set Up the Equation for the First Cooling

For the initial scenario, the body cools from 50°C to 49°C, with surroundings at 30°C in 5 seconds. We use the formula derived from Newton's Law: \( t = \frac{1}{k} \ln \frac{\vartheta_0 - \vartheta_s}{\vartheta - \vartheta_s} \). Here, \( t = 5 \) seconds, \( \vartheta_0 = 50 \), \( \vartheta = 49 \), and \( \vartheta_s = 30 \). Substitute these values to find \( k \).
03

Solve for the Cooling Constant k

Substitute the values into the formula: \( 5 = \frac{1}{k} \ln \frac{50 - 30}{49 - 30} \). Simplifying gives \( 5 = \frac{1}{k} \ln \frac{20}{19} \). Solving for \( k \), we find that \( k = \frac{1}{5} \ln \frac{20}{19} \).
04

Set Up the Equation for the Second Cooling

Now, we need to find the time taken to cool from 40°C to 39°C. Using the same formula \( t = \frac{1}{k} \ln \frac{\vartheta_0 - \vartheta_s}{\vartheta - \vartheta_s} \), substitute \( \vartheta_0 = 40 \), \( \vartheta = 39 \), and \( \vartheta_s = 30 \).
05

Calculate the Time for the Second Cooling

Substitute the known values into the formula: \( t = \frac{1}{\left(\frac{1}{5} \ln \frac{20}{19}\right)} \ln \frac{40 - 30}{39 - 30} \). This simplifies to \( t = 5 \ln \frac{10}{9} / \ln \frac{20}{19} \). Calculate this value.
06

Final Calculation and Result

Upon calculation, \( \ln \frac{10}{9} = 0.105 \) and \( \ln \frac{20}{19} = 0.051 \). Thus, \( t = 5 \cdot \frac{0.105}{0.051} \approx 10 \) seconds.

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Key Concepts

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

Rate of Change of Temperature
Newton's Law of Cooling describes how an object's temperature changes over time when exposed to a surrounding environment with a different temperature. The "rate of change of temperature" represents how quickly this change happens. Mathematically, the rate of change is expressed as \( \frac{d\vartheta}{dt} = -k(\vartheta - \vartheta_s) \). Here, \( \vartheta \) is the object's temperature at a given time, \( \vartheta_s \) is the ambient temperature, and \( k \) is the cooling constant.
In Newton's Law, the rate of temperature change is proportional to the difference between the object's temperature and its surrounding temperature. This means the greater the temperature difference, the faster the object cools. Imagine a hot cup of coffee placed in a cool room. Initially, the coffee will cool quickly, but as the temperature difference decreases, the cooling rate also decreases.
  • The larger the initial temperature difference, the quicker the rate of change.
  • Smaller differences result in slower temperature changes over time.
Understanding how the rate of change affects temperature is crucial when applying Newton's Law to real-world scenarios such as predicting cooling times.
Cooling Constant
The cooling constant \( k \) is a crucial part of Newton's Law of Cooling. It's a measure of how quickly heat is transferred from an object to its environment. The cooling constant provides a link between the rate of temperature change and the difference in temperatures.
Mathematically, you can solve for \( k \) using experimental data. For instance, if a body cools from 50°C to 49°C in 5 seconds, with the surroundings at 30°C, you can plug values into the equation \( 5 = \frac{1}{k} \ln \frac{20}{19} \) to solve for \( k \).
  • A smaller value of \( k \) indicates slower heat exchange and cooling.
  • A larger \( k \) represents faster cooling rates.
Determining the cooling constant allows predictions of how different materials or substances will cool over time under various conditions. By knowing this constant, we can use Newton's Law to model cooling processes accurately.
Temperature Difference
The temperature difference in Newton's Law of Cooling is the driving force behind the cooling process. It is the difference between the temperature of the object \( \vartheta \) and the surrounding temperature \( \vartheta_s \). This difference determines how quickly the object will cool.
The equation \( \frac{d\vartheta}{dt} = -k(\vartheta - \vartheta_s) \) highlights this relationship. At a larger temperature difference, the rate of change of temperature is greater, leading to faster cooling. Conversely, as an object approaches the ambient temperature, the rate of cooling slows significantly.
  • Higher temperature differences can lead to more rapid temperature changes.
  • As temperatures balance, cooling slows and eventually stops.
Understanding how temperature difference impacts cooling is vital for practical applications like designing thermal systems, ensuring proper storage conditions for temperature-sensitive items, and more. Newton's Law provides valuable insights into these processes by focusing on temperature differences.

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Most popular questions from this chapter

The average kinetic energy of a gas molecule at \(27^{\circ} \mathrm{C}\) is \(6.21 \times 10^{-21} \mathrm{~J} .\) Its average kinetic energy at \(127^{\circ} \mathrm{C}\) will be (a) \(12.2 \times 10^{-21} \mathrm{~J}\) (b) \(8.28 \times 10^{-21} \mathrm{~J}\) (c) \(10.35 \times 10^{-21} \mathrm{~J}\) (d) \(11.35 \times 10^{-21} \mathrm{~J}\)

It is known that wax contracts on solidification. If molten wax is taken in a large vessel and it is allowed to cool slowly, then (a) it will start solidifying from the top to downward (b) it will starts solidifying from the bottom to upward (c) it will start solidifying from the middle, upward and downward at equal rates (d) the whole mass will solidify simultaneously

A steel scale measures the length of a copper wire as \(80.0 \mathrm{~cm}\), when both are at \(20^{\circ} \mathrm{C}\) (the calibration temperature for scale). What would be the scale read for the length of the wire when both are at \(40^{\circ} \mathrm{C}\) ? (Given \(\alpha_{\text {nteel }}=11 \times 10^{-6}\) per \(^{\circ} \mathrm{C}\) and \(\alpha_{\text {copper }}=17 \times 10^{-6}\) per \(^{\circ} \mathrm{C}\) ) (a) \(80.0096 \mathrm{~cm}\) (b) \(80.0272 \mathrm{~cm}\) (c) \(1 \mathrm{~cm}\) (d) \(25.2 \mathrm{~cm}\)

A bar of iron is \(10 \mathrm{~cm}\) at \(20^{\circ} \mathrm{C}\). At \(19^{\circ} \mathrm{C}\) it will be \((\alpha\) of iron \(\left.=11 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right)\) (a) \(11 \times 10^{-6} \mathrm{~cm}\) longer (b) \(11 \times 10^{-6} \mathrm{~cm}\) shorter (c) \(11 \times 10^{-5} \mathrm{~cm}\) shorter (d) \(11 \times 10^{-5} \mathrm{~cm}\) longer

A cylinder of radius \(r\) and thermal conductivity \(K_{1}\) is surrounded by a cylindrical shell of linear radius \(r\) and outer radius \(2 r\), whose thermal conductivity is \(K_{2}\). There is no loss of heat across cylindrical surfaces, when the ends of the combined system are maintained at temperatures \(T_{1}\) and \(T_{2}\). The effective thermal conductivity of the system, in the steady state is (a) \(\frac{K_{1} K_{1}}{K_{1}+K_{2}}\) (b) \(K_{1}+K_{2}\) (c) \(\frac{K_{1}+3 K_{2}}{4}\) (d) \(\frac{3 K_{1}+K_{2}}{4}\)

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