Grade 11 Physics Note

Transfer of heat

Convection current:

Heat transfer in a gas or liquid by the circulation of currents from one region to another. A current in a fluid that results from convection, The Pacific Plate is floating, propelled by convection currents deep in the mantle.

 

Distinguish between conduction, convection and radiation:

conduction

convection

radiation

Material medium is essential.

Material medium is also essential.

Material medium is not essential.

Transfer of heat can be in any direction.

Transfer of heat can be in vertically upward.

Transfer of heat can be in any directions in straight line.

Molecules do not leave their mean positions.

Molecules move from one place to another place.

Electromagnetic waves move from one place to another place.

 

Coefficient of thermal conductivity of good conductor by Searle’s Method;

Searle’s apparatus for thermal conductivity

 

Searle’s apparatus consist of a metal bar AB with cavities C1 and C2. One end and A of the bar is enclosed inside the steam chamber and other and B is wounded by a copper tube.

Thermometer T1 and T2 are fitted in cavities C1 and C2 respectively;

A and V is used to collect out coming water.

When steam is passed in to the stem chamber, in A gets heated and this heat is conducted towards and B. After sometime, steady state is reached and all the thermometers T1 T2 T3 T4 show constant reading. At steady state, the amount of heat flowing in to the bar at end A is equal to the amount of heat flowing out from the end B. If θ1 θ2 θ3 and θ4 respectively. Then amount of heat flowing in to the bar at end A where

${\rm{Q}} = \frac{{{\rm{KA}}\left( {{\theta _1} - {\theta _2}} \right){\rm{*t}}}}{{\rm{x}}} \ldots  \ldots .1$

Where,

x is the distance between the cavities C1 and C2

 A is the area of cross section of metal bar

K is the thermal conductivity and

T is the time where heat is flowing in to the bar.

 Let,

M be the mass of water collected in t vessel V in time T. then amount of heat absorbed by water at end B in steady state is

$\theta  = {\rm{ms}}$$\left( {{\theta _3} - {\theta _4}} \right) \ldots  \ldots 2$

Where s is specific heat capacity of water

Hence, from equation 1 and 2

$\frac{{{\rm{KA}}\left( {{\theta _1} - {\theta _2}} \right)}}{{\rm{x}}}{\rm{t}} = {\rm{ms\: }}\left( {{\theta _3} - {\theta _4}} \right)$

${\rm{k}} = \frac{{{\rm{ms\: }}\left( {{\theta _3} - {\theta _4}} \right){\rm{x}}}}{{{\rm{A}}\left( {{\theta _1} - {\theta _2}} \right){\rm{t}}}}$

From this relation we can calculate thermal conductivity of a good conductor.

 

Coefficient of thermal conductivity:

The coefficient of thermal conductivity of a material is defined as the amount of heat flowing in one second across the faces of a cube of sides of unit length maintained at temperature difference of unit degree.

Its unit is ${\rm{Wm}}{.^{ - 1}}{{\rm{K}}^{ - 1}}$

Its dimension K is $\left[ {{\rm{M\: L\: }}{{\rm{T}}^{ - 3}}{{\rm{K}}^{ - 1}}} \right]$

 

An expression for the thermal conductivity of a good conductor in steady state;

Experimentally it is found that under steady state condition, the quantity of heat Q flowing through the section of the bar is

                                              ${\rm{Q}} \propto {\rm{A}} \ldots  \ldots  \ldots  \ldots 1$

Directly proportional to the temperature difference

                        $\left( {{\theta _1} - {\theta _2}} \right){\rm{\: between\: the\: two\: faces}}.$

                      ${\rm{Q}} \propto \left( {{\theta _1} - {\theta _2}} \right) \ldots  \ldots  \ldots ..2$

Directly proportional to the time t for the heat follows

                                               ${\rm{Q}} \propto {\rm{t}} \ldots  \ldots  \ldots  \ldots 3$

Inversely proportional to the perpendicular distance between hot and cold faces.

                                    ${\rm{Q}} \propto \frac{1}{{\rm{x}}} \ldots  \ldots  \ldots  \ldots .4$

Combining both 1,2,3,4 we get,

          ${\rm{Q}} \propto \frac{{{\rm{A}}\left( {{\theta _1} - {\theta _2}} \right){\rm{t}}}}{{\rm{x}}}$

${\rm{or}},{\rm{\: Q}} \propto \frac{{{\rm{kA}}\left( {{\theta _1} - {\theta _2}} \right){\rm{t}}}}{{\rm{x}}}$…………5

This is required expression

 

 

Perfect black body:

A perfect black body is the one which absorbs heat radiation of all wave lengths that fall upon it. Such a body neither reflects nor transmits the incident heat radiation and hence the body appears black. But when heated, it emits radiation which is called black radiation. The wave length range of emitted radiation is independent of the material of the body and depends on temperature of the blackbody.

http://qpdb.eu/wp-content/uploads/2014/10/blackbody-radiation.jpg

black body

In practice no material has been found to absorb all incoming radiation, but carbon in its graphite form absorbs all but about 3%. It is also a perfect emitter of radiation. At a particular temperature the black body would emit the maximum amount of energy possible for that temperature. This value is known as the black body radiation. It would emit at every wavelength of light as it must be able to absorb every wavelength to be sure of absorbing all incoming radiation. The maximum wavelength emitted by a black body radiator is infinite. It also emits a definite amount of energy at each wavelength for a particulartemperature, so standard black body radiation curves can be drawn for each temperature, showing theenergy radiated at each wavelength. All objects emit radiation above absolute zero. 

 

Stefan’s law of black body radiation;

It states that the total amount of heat energy radiated per second per unit area of a perfectly black body is directly proportional to fourth power of its absolute temperature.

If E be the heat energy radiated per second per unit area by a black body of absolute temperature T1 Then,

T$ \propto {\rm{\: }}\sigma {\rm{\: }}{{\rm{T}}^4}$ …..1

If the body is not perfectly black, then

${\rm{E}} = {\rm{e\: }}\sigma {\rm{\: }}{{\rm{T}}^4}$…..2

Where,

E is the emissivity of the body.

The heat energy radiated per second or power radiated by a body is given by,

                                                                                ${\rm{P}} = {\rm{E*A}}$

${\rm{P}} = {\rm{\: e\: }}\sigma {\rm{\: }}{{\rm{T}}^4}{\rm{*A}}$……….3

When a black body of absolute temperature T1 is placed inside an in closer of absolute temperature T0, Then black body radiates as well as absorbs heat energy.

The heat energy radiated per second per unit area by the black body is,

${{\rm{E}}_{{\rm{emit}}}} = \sigma {\rm{\: }}{{\rm{T}}_1}{{\rm{\: }}^4}$………4

Heat energy absorbed per second per unit area by the black body is given by,

${{\rm{E}}_{{\rm{abs}}}} = \sigma {\rm{\: }}{{\rm{T}}_{\rm{o}}}{{\rm{\: }}^4}$………5

Hence,

Net heat energy radiated per second per unit area by the body is

                         ${{\rm{E}}_{{\rm{net}}}} = {{\rm{E}}_{{\rm{emit}}}} - {{\rm{E}}_{{\rm{abs}}}}$

     $ = \sigma {\rm{\: }}{{\rm{T}}_1}{{\rm{\: }}^4} - \sigma {\rm{\: }}{{\rm{T}}_{\rm{o}}}{{\rm{\: }}^4}$

$ = {\rm{\: \: }}\sigma ({{\rm{T}}_1}{{\rm{\: }}^4} - {\rm{\: }}{{\rm{T}}_{\rm{o}}}{{\rm{\: }}^4}) \ldots  \ldots  \ldots  \ldots 6$

If body is not perfectly black, then

${{\rm{E}}_{{\rm{net}}}} = \sigma {\rm{e}}({{\rm{T}}_1}{{\rm{\: }}^4} - {\rm{\: }}{{\rm{T}}_{\rm{o}}}{{\rm{\: }}^4}) \ldots  \ldots  \ldots  \ldots 7$

This is required expression

 

Heat transferred in solids by conduction;

Heat transferred in solids by conduction due to increase in kinetic energy of vibration. When molecules at one end get heat energy and began to vibrate. When molecules vibrate about their means position, they collide with neighboring molecules and the kinetic energy of vibration is transmitted to the neighboring molecules.

 

Gases and liquids mainly heated by convection;

Thermal energy is transferred from hot places to cold places by convection. Convection occurs when warmer areas of a liquid or gas rise to cooler areas in the liquid or gas. Cooler liquid or gas then takes the place of the warmer areas which have risen higher. This results in a continuous circulation pattern. 

Woolen clothes are warmer than cotton clothes:

Woolen clothes have large layer of dry air in its texture. As air is bad conductor of heat. The air does not allow the heat produced in the heater, so woolen warmer than cotton clothes.

 

Mud houses colder in summer and warmer in winter;

Mud is a bad conductor of heat and thatched roof traps air in between the straw which itself is a bad conductor of heat. Hence, mud houses with thatched roofs are warmer during winter and cooler during summer.

Good absorbers of heat also good emitters:

The body must emit radiant energy at the same rate at which it absorbs, when a body reaches in thermal equilibrium with its surroundings. Hence, a body that is a good emitter must also be good absorber of radiation and vice versa.

 

Water in well cooler in summer and warmer in winter:

Temperature of underground water remains almost constant throughout the year because water level is far below from the surface of Earth. Sun is not able to heat up such deep water. The temperature of underground water is almost 10-20 degree Celsius. It can be different for different geographical regions. So, if temperature outside is suppose 40 degree Celsius but underground water has only 15 °C, feel it as cold. In winter, if temperature on surface of Earth is 0 or below 0 °C, well water with approx. 15 degree Celsius temperature will definitely be felt as warm.

 “Ventilations are provided near the ceilings in a room”:

Ventilations are provided near the ceilings in a room because hot vitiated air which always ascends towards the ceiling can escape more easily.

 

Three practical applications of thermal conductivity in everyday life:

Three practical applications of thermal conductivity in everyday life are as

1.  Woolen clothes are warmer because they have fine pores filled with air.

2. Ice is packed in saw dust

3. In winter, iron chairs appear to be colder then the wooden chairs.


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