In this article, we are going to study to solve numerical problems based on Newton’s law of cooling.

Newton’s Law of Cooling:

The rate of loss of heat by a body is directly proportional to its excess temperature over that of the surroundings provided that this excess is small.

Let θ and θ_o, be the temperature of a body and its surroundings respectively. Let dQ / dt be the rate of loss of heat, So from Newton’s Law of Cooling,

Where k is a constant. Sometimes constant is denoted by C

Example – 01:

A metal sphere, when suspended in a constant temperature enclosure, cools from 80 °C to 70 °C in 5 minutes and to 62^oC in the next five minutes. Calculate the temperature of the enclosure.

Solution:

Let θ_o be the temperature of the surroundings.

Consider a cooling from 80 °C to 70 °C:

Initial temperature = θ₁ = 80 °C, Final temperature = θ₂ = 70 °C, Time taken t = 5 min

By Newton’s Law of Cooling

Consider a cooling from 70 °C to 62°C:

Initial temperature = θ₁ = 70 °C, Final temperature = θ₂ = 62 °C, Time taken t = 5 min

Dividing equation (1) by (2)

132 – 2 θ_o = 120 – 1.6θ_o

12 = 0.4 θ_o

θ_o = 12/0.4 = 30 °C

Ans: Surrounding temperature is 30 °C.

Example – 02:

A metal sphere cools at the rate of 3 °C per minute when its temperature is 50 °C. Find its rate of cooling at 40 °C if the temperature of the surroundings is 25 °C.

Solution:

Consider the cooling when the temperature was 50 °C:

Rate of cooling (dθ/dt)₁= 3 °C per minute, temperature of body = θ₁ = 50°C, temperature of surroundings = θ_o = 25 °C

Newton’s Law of Cooling

Consider the cooling when the temperature was 40 °C:

Temperature of body = θ₁ = 40 °C, temperature of surroundings = θ_o = 25 °C, Rate of cooling (dθ/dt)₂= ?

Ans: The rate of cooling at 40 °C is 1.8 °C per minute,

Example – 03:

A body cools at the rate of 0.5 °C/s when it is 500C above the surroundings. What is the rate of cooling when it is at 30 °C above the same surroundings?

Solution:

Consider the cooling when the temperature is 50 °C above the surroundings:

Rate of cooling (dθ/dt)₁= 0.5 °C per second, the temperature of the body above surroundings = (θ₁ – θ_o)= 50 °C,

By Newton’s law of cooling

Consider the cooling when the temperature is 30 °C above the surroundings:

Temperature of the body above surroundings = (θ₁ – θ_o)= 30 °C, Rate of cooling (dθ/dt)₂= ?

Ans: The rate of cooling at 30 °C above the  surroundings is1.8 °C per minute,

Example – 04:

A metal sphere cools at the rate of 0.6 °C per minute when its temperature is 30 °C above the surroundings. At what rate will it cool when its temperature is 20 °C above surroundings, other conditions remaining constant?

Solution:

Consider the cooling when the temperature is 30 °C above the surroundings:

Rate of cooling (dθ/dt)₁= 0.6 °C per minute, the temperature of the body above surroundings = (θ₁ – θ_o)= 30 °C,

By Newton’s Law of Cooling

Consider the cooling when the temperature is 20 °C above the surroundings:

Temperature of the body above surroundings = (θ₁ – θ_o)= 20 °C,

Rate of cooling (dθ/dt)₂= ?

Ans: The rate of cooling at 20 °C above the  surroundings is 0.4 °C per minute,

Example – 05:

A body at 50 °C cools in surroundings at 30 °C. At what temperature will its rate of cooling be half that at the beginning?

Solution:

Temperature of surroundings = θ_o = 30 °C

Consider the cooling when the temperature was θ₁= 50 °C

By Newton’s Law of Cooling

Consider the cooling when the temperature was θ₁ °C:

Rate of cooling (dθ/dt)₂= ½ (dθ/dt)₁

Ans: at 40 °C the rate of cooling be half that at the beginning

Example – 06:

A body cools from 75 °C to 55 °C in ten minutes when the surrounding temperature is 31°C. At what average temperature will its rate of cooling be ¼ th that at the start?

Solution:

Let θ_o be the temperature of the surroundings.

Consider a cooling from 75 °C to 55 °C :

Initial temperature = θ₁ = 75 °C, Final temperature = θ₂ = 55 °C, Time taken t = 10 min

Consider the cooling when the temperature was θ₁ °C:

Rate of cooling (dθ/dt)₂= 1/4 (dθ/dt)₁

Ans: At temperature 39.5 °C the rate of cooling be ¼ th that at the start

Example – 07:

A body cools from 60 °C to 50 °C in 5 minutes. How much time will it take to cool from 50 °C to 44 °C if the surrounding temperature is 32 °C?

Solution:

Let θ_o = 44 °C  be the temperature of surroundings.

Consider a cooling from 60 °C to 50 °C :

Initial temperature = θ₁ = 60 °C, Final temperature = θ₂ = 50 °C, Time taken t = 5 min

By Newton’s law of cooling

Consider a cooling from 50 °C to 44 °C :

Initial temperature = θ₁ = 50 °C, Final temperature = θ₂ = 44 °C

Ans: Time taken to cool from 50 °C to 44 °C is 4.6 min

Example – 08:

A body cools from 72 °C to 60 °C in 10 minutes. How much time will it take to cool from 60 °C to 52 °C if the temperature of the surroundings is 36 °C?

Solution:

Surrounding temperature = θ_o = 36 °C

Consider a cooling from 72 °C to 60 °C:

Initial temperature = θ₁ = 72 °C, Final temperature = θ₂ = 60 °C, Time taken t = 10 min

By Newton’s law of cooling

Consider a cooling from 60 °C to 52 °C :

Initial temperature = θ₁ = 60 °C, Final temperature = θ₂ = 52 °C

Ans: The time taken to cool from 60 °C to 52 °C is 10 min

Example – 09:

A body cools from 750C to 70 °C in 2 minutes. What will additional time it take to cool to 60 °C if the room temperature is 30 °C?

Solution:

Surrounding temperature = θ_o = 30 °C

Consider a cooling from 75 °C to 70 °C :

Initial temperature = θ₁ = 75 °C, Final temperature = θ₂ = 70 °C, Time taken t = 2 min

Consider a cooling from 70 °C to 60 °C :

Initial temperature = θ₁ = 70 °C, Final temperature = θ₂ = 60 °C

Ans: The time taken to cool from 70 °C to 60 °C is 34/7 min

Example – 10:

A heated metal ball is placed in cooler surroundings. Its rate of cooling is 2 °C per minute when its temperature is 60 °C and 1.2 °C per minute when its temperature is 52 °C. Find the temperature of the surroundings and the rate of cooling when the temperature of the ball is 48 °C.

Solution:

Let θ_o be the temperature of the surroundings.

Consider a cooling at 60 °C:

Temperature = θ₁ = 60 °C, Rate of cooling = 2 °C per minute

Consider a cooling at 52 °C:

Temperature = θ₂ = 52 °C, Rate of cooling = 1.2 °C per minute

Dividing equation (1) by (2)

∴   52 – θ_o = 36 – 0.6 θ_o

∴   16 = 0.4 θ_o

∴   θ_o = 40 °C

substituting in equation (1)

2 = C (60 – 40)

∴   C = 1/10 min^-1

Consider a cooling at 48 °C:

Temperature = θ₃ = 48 °C,

Ans: Temperature of surrounding is 40 °C and rate of cooling at 48 °C is 0.8 °C per min

Example – 11:

A copper ball cools from 62 °C to 50 °C in 10 minutes and to 42 °C in the next 10 minutes. Calculate the temperature at the end of next 10 minutes.

Solution:

Let θ_o be the temperature of the surroundings.

Consider a cooling from 62 °C to 50 °C:

Initial temperature = θ₁ = 62 °C, Final temperature = θ₂ = 50 °C, Time taken t = 10 min

By Newton’s law of cooling

Consider a cooling from 50 °C to 42 °C:

Initial temperature = θ₁ = 50 °C, Final temperature = θ₂ = 42 °C, Time taken t = 10 min

Dividing equation (2) by (1)

∴    138 – 3θ_o = 112 – 2θ_o

∴   θ_o  =26 ^oC

Surrounding temperature is 26 ^oC

substituting in equation (1)

1.2 = C( 56 -26)

∴    C = 1.2/30 = 1/25 min^-1

Consider further cooling from 42 ^oC to θ₂ ^oC:

Initial temperature = θ₁ = 50 ^oC, Final temperature = θ₂, Time taken t = 10 min

Ans: Temperature after next 10 minutes is 36.7 ^oC

Example – 12:

A body cools from 60 ^oC to 52 ^oC in 5 minutes and from 52 ^oC to 44 ^oC in next 7.5 minutes. Determine its temperature in the next 10 minutes.

Solution:

Let θ_o be the temperature of the surroundings.

Consider a cooling from 60 ^oC to 52 ^oC :

temperature = θ₁ = 60 ^oC, Final temperature = θ₂ = 52 ^oC, Time taken t = 5 min

By Newton’s law of cooling

Consider a cooling from 52 ^oC to 44 ^oC :

Initial temperature = θ₁ = 52 ^oC, Final temperature = θ₂ = 42 ^oC, Time taken t = 7.5 min

Dividing equation (2) by (1)

∴    72 – 1.5 θ_o = 56 – θ_o

∴    16 = 0.5 θ_o

∴    θ_o = 16/0.5 = 32 ^oC

Surrounding temperature is 32 ^oC

substituting in equation (1)

1.6 = C( 56 -32)

C = 1.6/24 = 1/15 min^-1.

Consider further cooling from 44 ^oC to θ₂ ^oC :

Initial temperature = θ₁ = 44 ^oC, Final temperature = θ₂, Time taken t = 10 min

Ans: The temperature after 10 minutes is 38 ^oC

Examples – 13:

A body cools from 60 ^oC to 52 ^oC in 10 minutes and to 46 ^oC in the next 10 minutes. Find the temperature of the surroundings.

Solution:

Let θ_o be the temperature of surroundings.

Consider a cooling from 60 ^oC to 52 ^oC :

Initial temperature = θ₁ = 60 ^oC, Final temperature = θ₂ = 52 ^oC, Time taken t = 10 min

By Newton’s law of cooling

Consider a cooling from 52 ^oC to 46 ^oC :

Initial temperature = θ₁ = 52 ^oC, Final temperature = θ₂ = 46 ^oC, Time taken t = 10 min

Dividing equation (2) by (1)

∴    196 – 4θ_o = 168 – 3θ_o

∴     θ_o = 28

Ans: Surrounding temperature is 28 ^oC

Example – 14:

The rate of cooling of a body is 2 ^oC/min at temperature 60 ^oC and 1 ^oC/min at 45 ^oC. What will be the temperature of the surroundings?

Solution:

Let θ_o be the temperature of the surroundings.

Consider the cooling when temperature θ₁ = 60 ^oC:

Rate of cooling (dθ/dt)₁= 2 ^oC per minute,

By Newton’s law of cooling

Consider the cooling when temperature θ₂ = 30 ^oC:

Rate of cooling (dθ/dt)₁= 1 ^oC per minute,

Dividing equation (1) by (2)

∴  90 – 2θ_o= 60 – θ_o

∴  θ_o = 30 ^oC

Ans: Surrounding temperature is 30^oC

Example – 15: 

A metal sphere cools from 60 ^oC to 50 ^oC in 5 minutes. How much more time will it take to cool from 50 ^oC to 40 ^oC if the temperature of the surroundings is 30 ^oC.

Solution:

Let θ_o be the temperature of surroundings.

Consider a cooling from 60 ^oC to 50 ^oC :

Initial temperature = θ₁ = 60 ^oC, Final temperature = θ₂ = 50 ^oC, Time taken t = 5 min

By Newton’s law of cooling

Consider a cooling from 50 ^oC to 40 ^oC :

Initial temperature = θ₁ = 50 ^oC, Final temperature = θ₂ = 40 ^oC, Time taken t = ?

Ans: Time taken to cool from 50 ^oC to 40 ^oC is 8.33 min

Example – 16:

A copper sphere is heated and then allowed to cool while suspended in an enclosure whose walls are maintained at a constant temperature. When the temperature of the sphere is 86 ^oC, it is cooling at the rate of 3 ^oC/min; at 75 ^oC, it is cooling at the rate of 2.5 ^oC/min. What is the temperature of the sphere when it is cooling at the rate of 1 ^oC/min? Assume Newton’s law of cooling.

Solution:

Let θ_o be the temperature of surroundings.

Consider the cooling when temperature θ₁ = 86 ^oC:

Rate of cooling (dθ/dt)₁= 3 ^oC per minute,

By Newton’s law of cooling

Consider the cooling when temperature θ₁ = 75 ^oC:

Rate of cooling (dθ/dt)₁= 2.5 ^oC per minute,

Dividing equation (2) by (1)

∴  450 – 6θ_o = 430 – 5 θ_o

∴  θ_o = 20 ^oC

Substituting in equation (1)

3 = C(86 – 20)

∴   C = 3/66 = 1/22 min^-1

Consider the cooling when temperature = θ₃:

Rate of cooling (dq/dt)₁=  1 ^oC per minute,

Ans: At 42 ^oC it is cooling at the rate of 1 ^oC/min

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In this article, we shall study Stefan’s law of radiation, Newton’s law of cooling, derivation of Newton’s law of cooling from Stefan’s law and to calculate the temperature of the solar surface.

Stefan’s Law:

Statement:

The heat energy radiated per unit time per unit area of a perfectly black body is directly proportional to the fourth power of its absolute temperature.

Explanation: 

Let E_b, the heat radiated per unit time per unit area of a perfectly black body whose absolute temperature is T.

So by Stefan’s Law,

E_b ∝ T⁴

E_b   = σ T⁴

Where σ is a constant known as Stefan’s constant.

The value of σ in S.I. system is 5.67 × 10^-8 Jm^-2 K^-4s^-1 or 5.67 x 10^-8 Wm^-2 K^-4. The value of σ in c.g.s system is 5.67 × 10^-5 erg cm^-2 C°^-4s^-1. Dimensions of σ are [M¹L⁰T^-3K^-4]

Expression for the Rate of Loss of Heat to the Surrounding:

Let T be the absolute temperature of a perfectly black body. Let T_o be the absolute temperature of the surrounding.

So by Stefan’s Law,

Heat radiated per unit time per unit area of a perfectly black body   = σ T⁴

Let A be the surface area of the perfectly black body. Then,

Heat lost by the body per   unit   time = A σ T⁴

Where σ is a constant known as Stefan’s constant.

Heat   received  from the surrounding per unit time = A σ T_o⁴

The net rate of loss of heat = A σ T⁴ – A σ T_o⁴

= A σ(  T⁴ – T_o⁴)

This is an expression for the rate of loss of heat to the surrounding.

Newton’s Law of Cooling:

Statement:

The rate of loss of heat by a body is directly proportional to its excess temperature over that of the surroundings provided that this excess is small.

Explanation: 

Let θ and θ_o, be the temperature of a body and its surroundings respectively. Let dQ / dt be the rate of loss of heat, So from Newton’s Law of Cooling,

Where k is a constant.

Alternate Statement:

By Newton’s law of cooling, mathematically

Where, θ and θ_o, are the temperature of the body and its surroundings respectively and

dQ / dt is the rate of loss of heat. K is constant.

Let ‘m’ be the mass of the body, c be its specific heat.

Thus, the rate of fall of a temperature of a body is directly proportional to its excess temperature over that of the surroundings.

Derivation of Newton’s Law of Cooling from Stefan’s Law:

Let us consider a body whose surface area is A having absolute temperature T and kept in the surrounding having absolute temperature T_o.  Let e be the emissivity (or coefficient of emission) of the surface of the body.

Let (T  -T_o) =  x,  where  x  is Small.

∴ T   =  T_o  +    x.

Let dQ/ dt be the rate of loss of heat by the body. We know that

E / E_b = e

∴ E = e E_b

Where E & E_b, are the emissive powers of the body and perfectly black body respectively.

Using Stefan’s Law we know that for a perfectly black body rate of loss of heat = Aσ(  T⁴   –  T_o⁴ )

Therefore, for a given body,

As x /T_o is small so higher powers of x /T_o will be very small and hence those terms can be neglected.

This is Newton’s Law of cooling i.e. the rate of loss of heat of a body is directly proportional to its excess temperature over the surroundings provided the excess is small. Thus Newton’s Law of Cooling is derived (or deduced) from Stefan’s Law.

Limitations of Newton’s Law of Cooling:

• This law is applicable when the excess temperature of a body over the surroundings is very small (about 40^oC)
• When the body is cooling the temperature of the surrounding is assumed to be constant. Which is not true.
• The law is applicable for higher temperature using forced convection.

Solar Constant:

The solar constant is the rate at which solar radiant energy is intercepted by the earth per unit area at the outer limits of the earth’s atmosphere at the earth-sun mean distance.

The solar constant, S = 1353 W/m².

Calculation of Surface Temperature of the Sun:

The central portion of the sun is very hot. It has a temperature of 10⁷ K. It can be estimated using concepts of nuclear reactions.

The outer surface of the sun is comparatively cooler this region is called the photosphere. Its temperature can be estimated using solar constant.

Let T be the absolute temperature of the surface of the sun. Let Rs be its radius. By Stefan’s law, the total power radiated per second is given by

Where σ = Stefan’s constant

Let r be the earth-sun mean distance. r = 1.496 × 10¹¹ m,

Now the energy radiated by the sun is distributed over a sphere of surface area 4πr². By definition of the solar constant

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