Flag Thermal Physics> It is an everyday observation that hot an...
question mark

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 ∆T between an object and its surrounding (∆T = Tobj - Tsur) is not too great, the rate of cooling or warming of the object is proportional, approximately, to this temperature difference; that is,
src=data:image/png;base64,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
Where A is a constant. The minus sign appears because ∆T decrease with time if ∆T is positive and increase if ∆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=0 the temperature difference is ∆t0 show that it is ∆T= ∆T0e –at a time t later.

Shane Macguire , 10 Years ago
Grade upto college level
anser 1 Answers
Askiitians Tutor Team

Newton's law of cooling provides a fascinating insight into how objects exchange heat with their environment. Let's break down your questions regarding the factors that influence the constant A and the mathematical expression for temperature change over time.

Understanding the Constant A

The constant A in Newton's law of cooling is crucial as it determines the rate at which an object cools or warms. Several factors influence this constant:

  • Surface Area: A larger surface area allows for more heat exchange with the surroundings, leading to a higher value of A.
  • Material Properties: Different materials have varying thermal conductivities. For instance, metals typically have higher thermal conductivities than wood, affecting the rate of heat transfer.
  • Nature of Surroundings: The medium surrounding the object (air, water, vacuum) plays a significant role. For example, heat transfer in water is generally more efficient than in air due to its higher heat capacity.
  • Temperature Gradient: The greater the temperature difference between the object and its surroundings, the more pronounced the heat transfer, which can also affect the value of A.

Dimensions of A

The dimensions of the constant A can be derived from the equation of Newton's law of cooling. The law can be expressed as:

\(\frac{d(∆T)}{dt} = -A(∆T)\)

Here, ∆T is the temperature difference, which has dimensions of temperature (Θ). The rate of change of ∆T with respect to time (d(∆T)/dt) has dimensions of temperature per time (Θ/T). Therefore, we can deduce the dimensions of A:

Dimensions of A = Dimensions of (d(∆T)/dt) / Dimensions of (∆T) = (Θ/T) / (Θ) = 1/T

Thus, A has dimensions of inverse time, typically expressed in units such as s-1.

Deriving the Temperature Change Over Time

Now, let's delve into the mathematical aspect of your question regarding the temperature difference over time. Starting from the initial condition at time t=0, where the temperature difference is ∆T0, we can express the change in temperature difference as follows:

Given the differential equation:

\(\frac{d(∆T)}{dt} = -A(∆T)\)

This is a separable differential equation. We can rearrange it to isolate the variables:

\(\frac{d(∆T)}{∆T} = -A dt\)

Next, we integrate both sides:

\(\int \frac{1}{∆T} d(∆T) = -A \int dt\)

This yields:

\(\ln |∆T| = -At + C\)

Where C is the constant of integration. To solve for C, we apply the initial condition at t=0:

At t=0, ∆T = ∆T0, so:

\(\ln |∆T_0| = C\)

Substituting back into our equation gives:

\(\ln |∆T| = -At + \ln |∆T_0|\)

Exponentiating both sides results in:

|∆T| = |∆T_0| e-At

Thus, we can express the temperature difference at any time t as:

∆T = ∆T0 e-At

This equation illustrates how the temperature difference decreases exponentially over time, confirming that as time progresses, the temperature of the object approaches that of its surroundings.

Conclusion

In summary, the constant A is influenced by various factors such as surface area, material properties, and the surrounding medium, with dimensions of inverse time. The relationship between temperature difference and time is elegantly captured in the exponential decay formula derived from Newton's law of cooling. This understanding not only deepens our grasp of thermal dynamics but also has practical implications in fields ranging from engineering to environmental science.

ApprovedApproved
Last Activity: 7 Months ago
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments