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It is an everyday observation that hot and cold objects cool down or warm up to the temperature of their surroundings. If the temperature difference \(\Delta T\) between an object and its surroundings \(\left(\Delta T=T_{\mathrm{obj}}-T_{\mathrm{sur}}\right)\) is not too great, the rate of cooling or warming of the object is proportional, approximately, to this temperature difference; that is, $$ \frac{d(\Delta T)}{d t}=-A(\Delta T) $$ where \(A\) is a constant. (The minus sign appears because \(\Delta T\) decreases with time if \(\Delta T\) is positive and increases if \(\Delta T\) is negative.) This is known as Newton's law of cooling. (a) On what factors does \(A\) depend? What are its dimensions? (b) If at some instant \(t_{1}=0\) the temperature difference is \(\Delta T_{1}\), show that it is $$ \Delta T=\Delta T_{1} e^{-A t_{2}} $$ at a later time \(t_{2}\).

Step by step solution

01

Understanding the Proportionality Constant A

The factor A in Newton's law of cooling represents the rate at which the object exchanges heat with its surroundings. It depends on factors such as the nature of the object's surface (like material and texture), the surface area, and the properties of the surrounding medium (e.g., air or water). The more efficiently the object can exchange heat, the larger A will be.

02

Determining the Dimensions of A

To find the dimensions of A, analyze the given differential equation \ \( \frac{d( \Delta T )}{d t} = -A (\Delta T) \). The left-hand side has dimensions of temperature change per unit time, \( \left[ \frac{K}{T} \right] \) (where K is temperature and T is time). The right-hand side has dimensions of A multiplied by temperature difference, \( \left[ A \right] \left[ K \right] \). These must be equal, so \( \left[ A \right] = \left[ \frac{1}{T} \right] \). Therefore, A has the dimensions of inverse time, \( \left[ A \right] = T^{-1} \).

03

Setting Up the Differential Equation

We start with the differential equation: \[ \frac{d( \Delta T )}{d t} = -A (\Delta T ) \ \] We aim to solve this to find an expression for \( \Delta T \) as a function of time.

04

Separating Variables

Rearrange the equation to separate the variables: \[ \frac{1}{\Delta T} d( \Delta T ) = -A \, d t \ \] This allows us to integrate both sides with respect to their respective variables.

05

Integrating Both Sides

Integrate the left side with respect to \( \Delta T \) and the right side with respect to time t: \[ \int \frac{1}{\Delta T} d( \Delta T ) = \int -A \, d t \ \] The integrals yield: \[ \ln| \Delta T | = -A t + C \ \] where C is the constant of integration.

06

Solving for the Constant of Integration

To find C, use the initial condition that when t = 0, \( \Delta T = \Delta T_1 \): \[ \ln| \Delta T_1 | = C \ \] Therefore, \[ C = \ln| \Delta T_1 | \ \]

07

Substituting Back and Solving for \ \Delta T \

Substitute C back into the equation: \[ \ln| \Delta T | = -A t + \ln| \Delta T_1 | \ \] Exponentiate both sides to solve for \( \Delta T \): \[ \Delta T = \Delta T_1 e^{-A t} \ \]

08

Confirmation

Therefore, at a later time \( t_2 \), the temperature difference is given by: \[ \Delta T = \Delta T_1 e^{-A t_2} \ \]

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Key Concepts

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

Temperature Difference

In Newton's law of cooling, the temperature difference between an object and its surroundings is represented by \(\text{螖T = T}_{\text{obj}} - \text{T}_{\text{sur}}\).\
This difference is crucial because the rate at which an object cools or warms is directly proportional to how much hotter or colder it is compared to its environment.
For example, a hot cup of coffee in a cool room will cool down faster if the room is significantly cooler than the coffee.
The temperature difference (\text{螖T}) drives the heat exchange process, making it the foundation of Newton's law of cooling. The function shows how \( \frac{d( \text{螖T})}{ dt} = -A (\text{螖T} )\) involves \text{螖T} decreasing over time, indicating cooling, until equilibrium is reached.

Differential Equation

The core of Newton's law of cooling is the differential equation: \( \frac{d( \text{螖T})}{dt} = -A (\text{螖T} )\).
This equation states that the rate of change of the temperature difference (\text{螖T}) over time (t) is proportional to the current temperature difference.
In this case, 鈥楢鈥� is a proportionality constant which will have an effect on how fast the cooling or warming happens, with dimensions of inverse time (T^-1).
Differential equations like this one are powerful tools in describing dynamic processes in many fields, allowing us to predict how systems evolve over time.
Solving the differential equation helps to determine the temperature difference as a function of time, giving the solution: \( \text{螖T} = \text{螖T}_{1} e^{-At}\).

Heat Exchange

Heat exchange refers to the transfer of thermal energy between an object and its surroundings. In Newton's law of cooling, heat exchanges until thermal equilibrium is achieved.
The constant 鈥楢鈥� in our differential equation encompasses factors that affect this exchange, such as:

• The material's thermal conductivity (how well it conducts heat).
• The object's surface area (larger areas exchange more heat).
• The medium surrounding the object (air, water, etc.).

Efficient heat exchangers have larger values of 鈥楢鈥�, which lead to faster temperature changes.
Understanding these principles is essential for applications like climate control, engine cooling systems, and designing thermally efficient materials.鈥� } ] } # Breaking down the concept of heat exchange and heating mechanisms.} ] } } {