Heat always gets transferred from the body and higher temperature to a body at lower temperature heat transfer can take place in three ways a) Conduction b) Convection and c) Radiation. In this article, we shall study the heat transfer by the conduction.

Conduction:

If one end of a metal rod is heated, the other end also gets heated up.  This is due to conduction. When one end of a metal rod is heated, the kinetic energy of the molecules at that end increases.  The molecules start vibrating with a higher amplitude.  These molecules start vibrating with a higher amplitude.  These molecules during vibration collide with the neighbouring molecules and transfer part of their energy to the neighbouring molecules.  Thus the kinetic energy of the neighbouring molecules increases hence their amplitude of vibration increases and during the collision the energy transfers to the next molecule.  Thus heat transfer takes place by conduction.

The mode of heat transfer between two parts of a body or between two bodies in contact which are at different temperatures without actual migration of particles of the body is called conduction.

Depending upon easiness of heat transfer by conduction the substance are classifieds into types a) Good Conductors and b) Bad conductors

Good Conductors:

The substances which allow the heat to pass through them very easily are called good conductors. Examples. Aluminum, copper, Silver, Steel, Bronze, Brass, all metals

Bad Conductors:

The substances which do not allow the heat to pass through them are called bad conductors. Bad conductors of heat are also called as insulators. Examples: wood, rubber, Plastic, paper, glass, air, ebonite , bakelite.

Use of conduction:

• Metals are used for making utensils because the metals are good conductors of heat they allow heat to pass through them easily.
• Cooking vessels have plastic handles because plastic a bad conductor of heat it does not allow the heat to pass through from hot vessel to hands and thus danger of burning can be avoided.
• Tea-cups, Teapots, coffee jugs are made of porcelain.
• Mountaineers use sleeping bags in polar regions.
• People wear woolen cloth in winter.
• Nowadays cooking vessels are made with copper bottoms.
• In winter, the metal lock feels colder than the wooden door on touch.

Characteristics of conduction:

• In this type of heat transfer, there is no actual migration of the medium particles from one point to another.
• For conduction, there must be a material contact between the two bodies.

Concept of Steady-State and Temperature Gradient:

Heat conduction may be described quantitatively as the time rate of heat flow in a material for a given temperature difference.

Consider a metallic bar AB of length L and uniform cross-sectional area A with its two ends maintained at different temperatures. The temperature difference between the ends can be obtained by keeping the ends in thermal contact with large reservoirs having temperature differences. Some holes are drilled on this rod to insert thermometers (say T₁, T₂, T₃, and T₄) in the rod. For better thermal contact between the rod and thermometers mercury is poured into the holes. The sides of the bar are fully insulated so that no heat is exchanged between the sides and the surroundings.

Let θ₁, θ₂, θ₃, and θ₄  be the temperatures recorded by the thermometers T₁, T₂, T₃, and T₄ respectively. Initially, the temperature rises and after some time every thermometer shows its own constant reading such that (θ1 > θ2 > θ3> θ4). This state is called the steady-state.

Due to the insulation of the rod, no heat is lost due to surroundings. At a steady-state, at every cross-section of the rod, the quantity of heat entering the section in one second is equal to the quantity of heat leaving the section due to conduction.

Let us consider two sections separated by distance  Δx and let Δθ be the temperature difference between these two sections. then the quantity Δθ / Δx is called the temperature gradient.

The temperature gradient is defined as the rate of change of temperature with the distance when the material is in steady-state.

Thermal Conductivity:

It is found experimentally that in this steady state, the rate of flow of heat (or heat current)H is proportional  to the temperature difference (θ2 – θ1) and the area of cross-section A and is inversely proportional to the length L 

Where K = Constant called the thermal conductivity or the coefficient of thermal conduction the material. The greater the value of K for a material, the more rapidly will it conduct heat.

The SI unit of K is J S^–1 m^–1 K^–1 (joule per second per metre per kelvin) or W m ^–1 K^–1 (watt per metre per kelvin).

The value of thermal conductivity varies slightly with temperature but can be considered to be constant over a normal temperature range. Good thermal conductors have very high values of thermal conductivity while thermal insulators have negligible values of thermal conductivity.

Houses made of concrete roofs get very hot during summer days because the thermal conductivity of concrete (though much smaller than that of metal) is still not small enough. Therefore, a layer of earth or foam insulation is put on the ceiling so that heat transfer is prohibited and the room remains cooler.

Searle’s Experiment:

Apparatus:

Apparatus consists of the thermally insulated box housing a metallic bar of a uniform cross-sectional area with its one end kept in contact with steam in a steam chamber. Two holes are drilled to insert thermometers T₁ and T₂, in the rod separated by distance x. For better thermal contact between the rod and thermometers mercury is poured into the holes. Cooling water is circulated around the rod whose initial and final temperatures are measured by the thermometers T₃ and T₄.

Working and Calculations:

At steady state, the heat lost by rod = heat gained by the water

Where, m_W = Mass of water, S_W = Specific heat of water,  t = time for which heat is flowing

Measuring all values on R.H.S. of the formula value of K can be found.

Values of thermal conductivity in J S^–1 m^–1 K^–1  for different materials are given below

Science > Physics > Heat Transfer > Conduction

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In this article, we shall study different modes of heat transfer.

Convection:

Convection is a mode of heat transfer through a material medium in which heat energy is carried from one place to another by actual motion “migration” of heated matter.

Consider a beaker containing water small quantity of sawdust in added to this water.  The beaker is then heated it was found that the sawdust particles start moving from bottom to top and then from top to bottom in a circular way.

When we heat the liquid in the vessel the particles at the lower level get heated first hence there is an expansion of water at that layer.  Hence the density of water at that layer decreases this lowe density water starts rising upward to float and the cold water with the higher density moves downward this process continue until the boiling point of water.

Phenomena Associated with Convection:

Sea Breeze:

During the daytime, the landmass having higher sp. heat gets were heated up than the seawater having lower sp. heat the air near landmass will get more heated up (since its specific heat is higher than water) than that near seawater the heated air near landmass having low density starts.  Rising up thus creating low pressure gone on the landmass and thus the wind starts, blowing from sea i.e. high-pressure region to land i.e. low-pressure region these winds are called sea breezes.

Land Breeze:

During night time the landmass having higher sp. heat loses, heat faster than the seawater having less sp. heat.  The air near the surface of seawater will get more heated up than that near landmass.   The more heated air near seawater rises upward due to low density.  Thus a low-pressure zone is created on seawater.  Thus wind starts blowing from land having high-pressure region to sea having low-pressure region.  These winds are called the land breeze.

Trade Wind:

Trade wind is the steady surface wind on the earth blowing in from north-east towards the equator.

The equatorial and polar regions of the earth receive unequal solar heat. Air at the earth’s surface near the equator is hot while the air in the upper atmosphere of the poles is cool. A convection current would be set up, with the air at the equatorial surface rising and moving out towards the poles, descending and streaming in towards the equator.

Due to the rotation of the earth, modifies the direction of a convection current. Because of the rotation of the earth air close to the equator has an eastward speed of 1600 km/h, while it is zero close to the poles. As a result, the air descends not at the poles but at 30° N (North) latitude and returns to the equator. This flow of wind is called the trade wind.

Other Examples:

• The exhaust fans, ventilators are always kept at the top portion of the wall.
• The freezer region is the topmost portion in the freeze

Radiation:

Radiation is a process of transfer of heat in the form of electromagnetic waves for which material medium is not necessary. The thermal energy which is transferred by radiation is called radiant heat or radiant heat or simply radiations

Characteristics of Radiation:

• In the process of radiation thermal energy or heat energy is transferred from one point to other in the form of electromagnetic waves.
• As radiation is due to electromagnetic waves and electromagnetic waves are capable of passing through a vacuum, there is no necessity of material medium for radiation.
• Due to the electromagnetic nature of radiation has the same properties as that of light, such as rectilinear propagation, reflection, refraction, interference etc.
• The velocity of radiant energy in air or vacuum is the same as that of light in vacuum i.e  3 × 10⁸ m/s. Due to this high-speed radiation is the most rapid process of heat transfer.
• When radiant heat is incident on a matter, it is partly absorbed and converted into heat.
• Radiations have a wavelength greater than that of red colour and thus radiation form infrared region of the electromagnetic spectrum.

Diathermanous Substances:

The substances which can transmit the radiant heat incident upon their surfaces are called diathermanous substances. e.g. glass, quartz, gases

Adiathermanous  (Athermanous) Substances:         

The substances which cannot transmit the radiant heat incident upon their surfaces are called adiathermanous (athermanous) substances. e.g. wood, iron copper etc.

Perfectly Black Body:

A body which absorbs all the radiant heat incident upon it is called a perfectly black body.

No body exists in nature, which can be called a perfectly black body. For practical purposes, lamp black which absorbs nearly 98 % of the heat incident upon it is considered as a perfect black body.

Characteristics of Perfectly Black Body:

• A perfectly black body which absorbs all the radiant heat incident upon.
• For a perfectly black body the coefficient of absorption is equal to 1.
• The blackness of such a body is due to the fact that it does not reflect or transmit any part of heat incident upon it. Thus the coefficient of reflection and coefficient of transmission are zero.

Applications of Radiations:

• Black bodies absorb and emit radiant energy better than bodies of lighter colours. We wear white or light coloured clothes in summer so that they absorb the least heat from the sun. However, during winter, we use dark coloured clothes which absorb heat from the sun and keep our body warm.
• The bottoms of the utensils for cooking food are blackened so that they absorb maximum heat from the fire and give it to the vegetables to be cooked.
• A Dewar flask or thermos bottle is a device to minimise heat transfer between the contents of the bottle and outside. It consists of a double-walled glass vessel with the inner and outer walls coated with silver. Radiation from the inner wall is reflected back into the contents of the bottle. The outer wall similarly reflects back any incoming radiation. The space between the walls is evacuated to reduce conduction and convection losses and the flask is supported on an insulator like a cork. The device is, therefore, useful for preventing hot contents (like milk) from getting cold, or alternatively to store cold contents (like ice).

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.

Explanation:

Let θ and θo, 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.

Thus Newton’s law of cooling states that the rate of loss of heat by cooling body is directly proportional to its excess of temperature over the surrounding, provided this excess is very small.

The alternate statement of the law is that the rate of fall of temperature of a cooling body is directly proportional to its excess of temperature over the surrounding, provided this excess is very small.

dθ/dt  ∝ (θ – θo)

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 °C)
• 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.

Verification of Newton’s Law of cooling:

Newton’s law of cooling can be verified with the help of the experimental set-up shown in the figure. The set-up consists of a double-walled vessel (V) containing water in between the two walls. A copper calorimeter (C) containing hot water is placed inside the double-walled vessel. Two thermometers through the corks are used to note the temperaturesT₂ of water in calorimeter and T₁ of hot water in between the double walls respectively.

The temperature of hot water in the calorimeter is noted after equal intervals of time. A graph is plotted between loge (T2–T1) and time (t). The nature of the graph is observed to be a straight line having a negative slope as shown in the figure.

Science > Physics > Heat Transfer > Convection and Radiation

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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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Science > Physics > Radiation > Numerical Problems on Newton’s Law of Cooling

The post Numerical Problems on Newton’s Law of Cooling appeared first on The Fact Factor.

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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Science > Physics > Radiation > Stefan’s Law of Radiation

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In this article, we shall study the concept of a black body and its realization in practice.

Perfectly Black Body:

A body which absorbs all the radiant heat incident upon it is called a perfectly black body. Thus the coefficient of absorption of a perfectly black body is equal to 1. In fact, the blackness of such a body is due to the fact that it does not reflect or transmit any part of the heat incident upon it.

No body exists in nature, which can be called a perfectly black body. For practical purposes, lamp black which absorbs nearly 98 % of the heat incident upon it is considered as a perfectly black body.

Characteristics of Perfectly Black Body:

• A perfectly black body which absorbs all the radiant heat incident upon it.
• The coefficient of absorption for it is equal to 1.
• The blackness of such a body is due to the fact that it does not reflect or transmit any part of the heat incident upon it. Thus the coefficient of reflection and coefficient of transmission are zero.

Ferry’s Black Body:

A body which absorbs all the radiant heat incident upon it is called a perfectly black body.

• Construction: It can be artificially constructed by taking a double-walled, hollow metal sphere having a small hole. The inner surface of the sphere is coated with lamp black and it has a conical projection on the opposite side of the hole.
• Working: The radiation entering the sphere through this hole suffers multiple reflections. During every reflection, about 98% of the incident radiant heat is absorbed by the sphere. Therefore the radiation is completely absorbed by the sphere within a few reflections. In this way, the sphere acts as a perfectly black body whose effective area is equal to the area of the hole.

Spectrum of a Black Body:

A black body emits radiations of all possible wavelengths from zero to infinity. These radiations are of electromagnetic nature. These radiations do not depend on the nature of the surface of the black body but depend only on its absolute temperature. Black body radiations extend over the whole range of wavelengths of electromagnetic waves. The distribution of energy over this entire range of wavelength or frequency is known as the black body radiation spectrum.

A sensitive instrument called bolometer is used to find energy density between the wavelengths λ and λ  + dλ, By rotating the prism of the instrument this energy density is found for all the range of wavelengths at a constant high temperature of the perfectly black body.

Graphical representation:

Characteristics of the Spectrum of a Black Body:

• The emissive power of a perfectly black body increases with an increase in its temperature for every wavelength.
• Each curve has characteristic form and each of them has a maxima i.e. maximum emissive power corresponding to a certain wavelength.
• The position of maxima shifts towards the ultraviolet region (shorter wavelength) with an increase in the temperature.
• λ_m T = Constant (Wien’s displacement law)
• The area under each curve gives the total radiant power per unit area of the black body at that temperature and it is directly proportional to T⁴ (Verification of Stefan’s law)

Wien’s Displacement Law:

For a black body, the product of its absolute temperature and the wavelength corresponding to maximum radiation of energy is constant.

Thus, λ_m T = Constant

The value of the constant of Wien’s displacement law is 2.898 x 10^-3 mK.

Significance of Wien’s Displacement Law:

• This law can be used to surface the temperature of stars. This is the only method to determine the temperature of celestial bodies.
• It explains the colour change in solid on heating from dull red (longer wavelength) to yellow (shorter wavelength)to white (all wavelengths of visible spectra).

Simple Radiation Correction:

The specific heat of a solid or liquid is determined by the method of mixtures. The solid is heated to a high temperature. It is dropped in a calorimeter containing water (or liquid) at room temperature. Finally, the maximum temperature of the mixture is noted. Now as the temperature of the mixture begins to increase, the mixture begins to lose heat by conduction and radiation. Loss of heat by conduction can be minimized by surrounding the mixture by a bad conductor of heat such as cotton, wool, etc. However, the loss of heat by radiation cannot be stopped.

Therefore the maximum temperature of the mixture is always less than the temperature it would reach if radiation were absent. This correction to be made in the final temperature of the mixture is called radiation correction.

Method of Applying Radiation Correction:

A stopwatch is started at the moment the solid is dropped into the liquid and time t taken by the mixture to reach the maximum temperature is θ noted.

The mixture is then allowed to cool for time t / 2. Let ‘θ’ be the temperature of the mixture after time t / 2.

Then, radiation correction = Δθ =  ½ (θ – θ ’ )

Thus corrected maximum temperature of the mixture = θ + Δθ

Greenhouse Effect:

Earth’s surface absorbs thermal energy from the sun and becomes a source of thermal radiation. The wavelength of the radiation lies in the infrared region. A large part of the radiation is absorbed by greenhouse gases like carbon dioxide, methane, nitrous oxide, Chlorofluorocarbons, troposphere ozone. Due to which the atmosphere of the earth heats up and the atmosphere gives more energy to the earth resulting in a warmer surface.

The above process repeats until no radiations are available for absorption. This heating of the surface and the atmosphere of the earth is called the greenhouse effect. The significance of the greenhouse effect is that it keeps the earth warmer which leads to biodiversity. In the absence of this effect the temperature of the earth would be -18° C.

But due to human activities, the quantities of greenhouse gases are increasing rapidly making the earth warmer. This increase may disturb the life of plants and animals. It may result in the melting of ice in Polar Regions, which may lead to a rise in sea level submerging the coastal regions.

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Science > Physics > Radiation > Concept of Black Body

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In this article, we shall study Kirchhoff’s law of radiation and its theoretical and experimental proof.

Statement:

The ratio of the emissive power to the coefficient of absorption is constant for all substances at a given temperature and is equal to the emissive power of a perfectly black body at that temperature.  OR At any given temperature, the emissivity (or coefficient of emission) of a body is equal to the coefficient of absorption

Explanation: 

If ‘E’ is the emissive power of a substance and ‘a’ is its coefficient of absorption then by Kirchhoff’s law of radiation

i.e a = e

Theoretical Proof of Kirchhoff’s Law of Radiation:

Let us consider two bodies A and B suspended in a constant temperature enclosure. B is a perfectly black body. After some time both A and B will attain the same temperature as that of the enclosure. By Prevost heat exchange theory in this state also every body will emit and absorb thermal radiations.

Let E be the emissive power of A and ‘a’ be its coefficient of absorption. Let E_b be the emissive power of B. Let Q be the radiant heat incident per unit time per unit area of each body.

Heat absorbed per unit time per unit area of   A =   a Q

Heat emitted per unit time per unit area of A =  E.

As the temperature remains constant so the heat emitted will be equal to the heat absorbed

∴   E = a Q      …………(1)

Perfectly black body B will absorb all the radiant heat incident on it.

Heat absorbed per unit time per unit area of B  =   Q.

The heat emitted per unit time per unit area of B = E_b

As the temperature of B remains constant so in case of B also heat emitted is equal to the heat absorbed.

E_b = Q   …………..(2)

Dividing equation (1) by (2) we get,

Thus, the coefficient of emission is equal to the coefficient of absorption.  This proves Kirchhoff’s law of radiation theoretically.

Experimental Proof of Kirchhoff’s Law or Ritchie’s Experiment:

Apparatus:

The apparatus consists of a U – tube manometer containing some coloured liquid. The two arms of the manometer are connected to two identical cylinders P and Q, having the same axis (co-axially arranged). The same face (either left or right) of each cylinder is coated with lamp black while the other face is kept polished. A third cylinder R can be placed between P  and  Q co-axially.  One face of the cylinder R is coated with lamp black while the other face is kept polished. The cylinder R can be rotated about a vertical axis.

Working:

The cylinder R is kept as shown in the figure such that all black surface point in the same direction. Hot water is poured into the cylinder R due to which its temperature will increase. No change will be observed in the liquid levels in the manometer. This shows that the quantity of heat absorbed by both P and Q from R is the same. Therefore pressure exerted by the air in P and Q on the liquid is the same on both sides.

Let E and E_b be the emissive powers of the polished and black surfaces and ‘a’ be the coefficient of absorption of the polished face. Let A be the area of the cross-section of each cylinder.

The amount of heat radiated per unit time by the black face of R = A E_b.

A part of this heat is the incident on the polished face of P.

Heat incident per unit time on the polished face of P = k A E_b,

The constant k depends upon the distance between P and R.

Heat absorbed per unit time by P per second =   a k A E_b.

Heat radiated per unit time by the polished face of R = A E

Heat incident per unit time on the black face of Q = k A E

Heat absorbed per unit time by the black face of Q = k A E.

The level of coloured liquid in both the arms of the apparatus is the same.

Hence both P and Q absorbed same quantity of heat per unit time.

∴   a k A E_b = k A E

∴   a E_b =  E

∴    a   = E / E_b = e

Thus, the coefficient of absorption = coefficient of emission.

This is Kirchhoff’s law of radiation. Thus the Kirchoff’s Law is experimentally verified.

Numerical Problems:

Example – 01:

512 J of radiant heat are incident on a body which absorbs 224 J. What is its coefficient of emission?

Given: Radiant heat incident = Q = 512 J, radiant heat absorbed = Q_a = 224 J

To Find: Coefficient of emission = e =?

Solution:

Coefficient of absorption = a = Q_a/Q = 224/512 = 0.4375

By Kirchhoff’s law of radiation

Coefficient of emission (e) = Coefficient of absorption (a)

∴ e = 0.4375

Ans: Coefficient of emission = 0.4375

Example – 02:

A body of surface area 15 × 10^-3 m² emits 1260 J in 40 s at a certain temperature. What is the emissive power of the surface at that temperature?

Given: Surface area = A = 15 × 10^-3 m², radiant heat emitted = Q = 1260 J, time taken = t = 40 s.

To Find: Emissive power = E =?

Solution:

E = Q/At = 1260 /(15 × 10^-3× 40) = 2100 J/m²s

Ans: Emissive power of surface = 2100 J/m²s

Example – 03:

The emissive power of a sphere of area 0.02 m² is 2100 J/m²s. What is the amount of heat radiated by the spherical surface in 20 seconds?

Given: Surface area = A = 0.02 m², Emissive power = 2100 J/m²s, time taken = t = 20 s.

To Find: Heat radiated = Q =?

Solution:

E = Q/At

∴ Q = E A t

∴ Q = 2100 × 0.02 × 20

Ans: Heat radiated = 840 J

Example – 04:

The energy of 6000 J is radiated in 5 minutes by a body of surface area 100 cm2. Find the emissive power of the body.

Given: Radiant heat emitted = Q = 6000 J, Time taken = 5 min = 5 × 60 = 300 s, Surface area = 100 cm² = 100 × 10^-4 m²

To Find: Emissive power = E =?

Solution:

E = Q/At = 6000 / (100 × 10^-4× 300)

∴ Q = 2000 J/m²s

Ans: Emissive Power = 2000 J/m²s

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Science > Physics > Radiation > Kirchhoff’s law of Radiation

The post Kirchhoff’s law of Radiation appeared first on The Fact Factor.

The different modes of heat transfer are conduction, convection, and radiation. In this article. we shall study radiation.

Radiation:

Radiation is a process of transfer of heat in the form of electromagnetic waves for which material medium is not necessary.

Radiant Heat:

The thermal energy which is transferred by radiation is called radiant heat or radiant heat or simply radiations

Characteristics of Radiation:

• In the process of radiation thermal energy or heat energy is transferred from one point to other in the form of electromagnetic waves.
• As radiation is due to electromagnetic waves and electromagnetic waves are capable of passing through a vacuum, there is no necessity of material medium for radiation.
• Due to the electromagnetic nature of radiation has the same properties as that of light, such as rectilinear propagation, reflection, refraction, interference, etc.
• The velocity of radiant energy in air or vacuum is the same as that of light in vacuum i.e  3 × 10⁸ m/s. Due to this high-speed radiation is the most rapid process of heat transfer.
• When radiant heat is incident on a matter, it is partly absorbed and converted into heat.
• Radiations have the wavelength greater than that of red colour and thus radiation form infrared region of the electromagnetic spectrum.

Relation Between a, r, and t:

Coefficient of Absorption (a):

The coefficient of absorption of a body is defined as the ratio of the quantity of radiant heat absorbed by it to the total quantity of radiant heat incident upon it.

Let Q  be the radiant heat incident on the body and Q_a be the radiant heat absorbed. Thus, 

a = Q_a /Q

Coefficient of Reflection (r):

The coefficient of reflection of a body is defined as the ratio of the quantity of radiant heat reflected by it to the total quantity of radiant heat incident upon it.

Let Q  be the radiant heat incident on the body and Q_r be the radiant heat reflected. Thus,

r = Q_r /Q

Coefficient of Transmission (t):

The coefficient of transmission of a body is defined as the ratio of the quantity of radiant heat transmitted by it to the total quantity of radiant heat incident upon it.

Let Q  be the radiant heat incident on the body and Q_t be the radiant heat transmitted. Thus,

t = Q_t /Q

Relation:

Let Q be the total radiant heat incident on the surface of a body. Let Q_a, Q_r, Q_t be the radiant heat absorbed, reflected and transmitted respectively by the body.

Then, Q_a +   Q_r +   Q_t  =    Q

Dividing both sides of the equation by Q

∴ a    +   r   +  t   = 1

Thus the sum of the coefficient of absorption, the coefficient of reflection and the coefficient of transmission is unity.

Classification of Substances:

Diathermanous Substance:

The substances which can transmit the radiant heat incident upon their surfaces are called diathermanous substances e.g. glass, quartz, gases

Characteristics of Diathermanous Substance:

• The substances which can transmit the radiant heat incident upon their surfaces are called diathermanous substances.
• For such a substance, the value of the coefficient of transmission (t) is not zero.

Adiathermanous Substance:

The substances which cannot transmit the radiant heat incident upon their surfaces are called adiathermanous (athermanous) substances e.g. wood, iron copper, etc.

Characteristics of Adiathermanous Substance:        

• The substances which cannot transmit the radiant heat incident upon their surfaces are called adiathermanous (athermanous) substances.
• For such a substance, the value of the coefficient of transmission (t) is zero.

Numerical Problems:

Example – 01:

The coefficients of reflection and transmission of the surface of a thin plate are 0.22 and 0.04 respectively. If 250 J of radiant heat is incident on the surface of the plate, how much heat is absorbed by the surface?

Given: Coefficient of reflection = r = 0.22, coefficient of transmission = 0.04, radiant heat incident = Q = 250 J

To Find: Radiant heat absorbed = Q_a =?

Solution:

We have, a + r + t = 1

∴   a + 0.22 + 0.4 = 1

∴   a = 1 – 0.26 = 0.74

∴   Coefficient of absorption = a = 0.74

Now, Q_a = aQ = 0.74 × 250 = 185 J

Ans: Radiant heat absorbed = 185 J

Example – 02:

Coefficient of absorption of a body is 0.54 and coefficient of reflection is 0.16. The amount of radiant heat incident on a body is 4000 J. Calculate the amount of heat transmitted through the body.

Given: Coefficient of absorption = r = 0.54, coefficient of reflection = 0.16, radiant heat incident = Q = 4000 J

To Find: Radiant heat transmitted = Q_t =?

Solution:

We have, a + r + t = 1

∴   0.54 + 0.16 + t = 1

∴   t = 1 – 0.70 = 0.30

∴   Coefficient of transmission = t = 0.30

Now, Q_t = tQ = 0.30 × 4000 = 1200 J

Ans: Radiant heat transmitted = 1200 J

Emissive Power:

Emissive power of a surface is defined as the amount of heat radiated per unit time per unit area of that surface at a given temperature.

It is found that the emissive power of a perfectly black body is greater than the emissive power of any other surface at the same temperature. It is denoted by the letter ‘E’ and its S.I. unit is J/m²s or W/m².

The emissive power of a body depends on

• the temperature of a body
• the nature of the body
• the surface area of the body
• the nature of the surroundings.

Coefficient of Emission (Emissivity):

The coefficient of emission of a surface is defined as the ratio of heat radiated per unit time per unit area of the surface to the heat radiated per unit time per unit area of a perfectly black surface at the same temperature. OR Coefficient of emission of a surface is the ratio of the emissive power of the surface to the emissive power of a perfectly black surface at the same temperature.

The coefficient of emission is also known as emissivity. It is denoted by letter ‘e’ and it is a unitless quantity.  For perfectly black body e = 1, for perfect reflector e = 0 and for all other bodies 0 < e < 1.

Let E be the emissive power of the body and E_b be the emissive power of the perfectly black body at the same temperature, then

E = e E_b  = a E_b

Prevost’s Theory of Heat Exchanges:

The exchange of heat between a body and its surroundings is explained on the basis of Prevost’s theory. The main points of Prevost’s heat exchange theory are

• Every body emits and absorbs thermal radiations at all temperatures except at absolute zero.
• The rate at which thermal radiations are emitted depends upon the absolute temperature of the body.
• The rate of radiation is independent of the temperature of the surrounding.

• Case -1: When a body is at a higher temperature than that of the surroundings then it radiates heat at a faster rate than it absorbs. Therefore it loses heat and its temperature falls.
• Case – 2: When a body is at a lower temperature than that of surroundings then it absorbs heat at a faster rate than the rate at which it radiates heat. Therefore, it gains heat and its temperature rises.
• Case – 3: When the temperature of the body is the same as that of the surroundings then the body radiates heat at the same rate at which it absorbs. Therefore, the body neither loses nor gains heat and hence its temperature remains constant. So, even in this state the processes of radiation and absorption continues.

Next Topic: Kirchhoff’s Law of Radiation

Science > Physics > Radiation > Introduction to Radiation

The post Introduction to Radiation appeared first on The Fact Factor.