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JEE Newton’s Cooling Science Thermodynamics

Newton’s Cooling

In this article we will derive the equation of Newton’s Cooling

In this article, we will derive the equation of Newton’s Cooling. So, let’s get started.

Newton’s Cooling

Thermal Conductivity

Consider a rod of length l and area A. Let the left end be in contact with a large object that is at a temperature TA. Let the right end be in contact with a large object that is at a temperature TB. What is the rate H at which energy is transfered through the rod?
The units of H are energy/time. What does H depend on?

Area: The larger the area A, the faster energy can be transfered, so H ∝ A.
Length: The longer the length l that the rod is, the slower is the energy transfer, so we will guess that H ∝ 1/l.

Temperature difference: The larger that TA − TB is, the faster the energy will be transfered, so H ∝ (TA − TB).
Putting these ideas together we have

H ∝(TA − TB)A/l (1)

Changing the proportionality by an equal sign brings in a constant k:

H = k(TA − TB)A/l (2)

where k is called the thermal conductivity.

It is interesting to include both the properties of heat capacity and thermal conductivity in one application. Suppose the object at the left end of the rod has a mass
m and a specific heat capacity of cv. Let the temperature on the right, TB, be held constant, but let the temperature of the object on the left decrease as it loses energy.

Let T be the temperature of the object on the left at time t. Then we have:

H = k(T − TB)A/l (3)

where H is the energy loss per second of the left object. The amount of energy, Q, the object loses in a time ∆t is

Q = H∆t = k(T − TB)A/l.∆t (4)

Since Q = mcv∆T, we have

mcv.∆T/∆t= −k(T − TB)A/l (5)

Taking the limit as ∆t goes to zero results in a simple differential equation:

dT/dt = −k(T − TB)A/mcvl (6)

Since TB is a constant, dT/dt = d(T−TB)/dt. If we define the temperature difference between the left and right sides, (T − TB), as δT, we have:

d(δT)/dt = −kA/mcvl.(δT) (7)

This is a simple differential equation whose solution is an exponentially decaying function:

δT = (δT)oe^− kA/mcvl.t (8)

where (δT)o is the initial temperature difference between the left and right side. Thus the temperature decreases exponentially to its final value. This type of cooling is called Newton’s cooling.

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