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HEAT-2

or

UNIT OF R

From, R = =

= K

From R =

– Hence the SI unit used is Kelvin per watt (K)

DETERMINATION OF THERMAL CONDUCTIVITY OF A GOOD CONDUCTOR OF HEAT

SEARLE’S METHOD

The heater is switched on and water is passed through the copper coil at a constant rate.

If the bar is assumed to be perfectly lagged. Then at steady state (i.e when all four thermometers give stead readings) the rate of flow of heat in distance X is given by:

……………… (1)

Where K = Thermal conductivity of the material of the bar

Since the bar is assumed to be perfectly lagged, all of the heat which flow along the bar is being used to increase the temperature of the water.

……………………….. (2)

Where m = mass of water flowing per unit time

C = specific heat capacity of water

Equation (1) = Equation (2)

―K‖ can be found from this equation

The holes at X contain oil or mercury to ensure good thermal contact between the thermometers and the bar.

DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR OF HEAT

LEES’ DISC METHOD

The thermal conductivity of a bad conductor of heat such as glass, asbestos wool, paper etc can be determined by using Lees’ disc method.

With this method a bad conductor of heat such as asbestos wool is sandwiched between a steam

– chest C and thick brass slab B.

The arrangement is suspended on three strings which are attached to B, Are thermometers placed as shown, and in order to ensure good thermal contact some mercury or Vaseline is applied in the holes in which thermometers and are inserted.

The lower face of C and the upper face of B are made flat for uniform heat flow.

Steam is passed through the chest and the apparatus is left to reach steady state.

At steady state we have: –

Where K = Thermal conductivity of the sample A = Area of one face of the sample l = Thickness of the sample

Here the rate at which heat is flowing through the sample is equal to the rate at which B is losing heat to the surroundings.

Now the sample is removed so that B comes into direct contact with C and is heated by it.

When the temperature of B has risen by about 10, ―C‖ is removed and the sample is replaced on top of B.

Since C is no longer present, B cools and its temperature is recorded at one- minute intervals until it has dropped to about 10 below its steady state temperature

A graph of temperature against time is plotted.

Slope of tangent at = =

Where = mass of brass slab B

Specific heat capacity of brass slab B

The condition under which B is losing heat are the same as those at steady state and hence

Equation (1) = Equation (2)

-Thus K can be determined from this equation.

Problem 16

Calculate the quantity of heat conducted through 2m2 of brick-wall 12cm thick in 1 hour if the temperature on one side is 8 and the other side is 28. Given that thermal conductivity of brick = 0.13Wm-1K-1

Problem 17

Estimate the rate at which ice melts in a wooden box 2cm thick and inside measurements 60cm x

60cm x 60cm, assume that external temperature is 27 and coefficient of thermal conductivity of wood = 0.1674 Wm -1K-1. Specific latent heat of fusion of ice = 336 x 103 Jkg-1

Problem 19

Two cylinders of equal physical dimensions are placed one on top of the other as illustrated below.

The lower surface of the silver cylinder is kept at 0 and the upper surface of the iron cylinder is kept at 100. Given that the thermal conductivity of silver is eleven times that of iron, calculate the temperature of the surface AB

Problem 20

A composite bar is made of a bar of copper 10cm long, a bar of iron 8cm long and a bar of aluminium

12cm long, all having the same cross – sectional area. If the extreme ends of the bars are maintained at 100 and 10 respectively, find the temperature at the two junctions. Given that thermal conductivity of copper, iron and aluminium are 400, 40 and 20 Wm-1K-1 respectively.

Problem 21

A window of height 1.0m and 1.5m contains a double glazed unit consisting of two single glass panes, each of thickness 4.0mm separated by an air gap of 70mm. Calculate the heat energy per second conducted through the window when the temperature difference across the unit is 10K.

Problem 22

An electric heater is used in a room of total wall area 137m2 to maintain a temperature of 20 inside it when the outside temperature is 10. The walls have three layers of different materials. The innermost layer is of wood of thickness 2.5cm, the middle layer is of cement of thickness 1.0cm and the outermost layer is of brick of thickness 25cm. Find the power of electric heater. Assume that there is no heat loss through the floor and ceiling. Thermal conductivity of wood, cement and brick are 1.25 Wn-1K-1,1.5 Wm1 -1 -1 -1

K and 1.0 Wn K respectively.

Problem 23

- What does it mean by thermal conductivity of substance.
- Find the heat lost per square meter through a cavity wall when the temperature difference between the inside and outside is 15, given that each of the two brick layers is 100mm thick and the cavity is also 100mm across.

Brick = 1.0 Wm-1K-1

Air = 0.025 Wm-1K-1

Problem 24

- Assuming you are managing a metal box company what requirements for thermal conductivity, specific heat capacity and coefficient of expansion would you want a material to be used as a cooking utensil to satisfy.
- A hot water boiler consists of iron wall of thickness 2.0cm and effective inner area of 2.5m2. The boiler is heated by a furnace and generates high pressure steam of temperature 170 at the rate of 1.2kgmin-1. The latent heat of steam at 170 is 2.09 x 106 Jkg-1. Assuming the outer face of the boiler to be at a temperature of 178 , what is the coefficient of thermal conductivity of iron?

##### Problem 25

A copper kettle has a circular base of radius 10cm and thickness 3.0mm. The upper surface of the base is covered by a uniform layer of scale 1.0mm thick. The kettle contains water which is brought to the boil over an electric heater. In the steady state condition, 5.0g of steam is produced each minute. Determine the temperature of the lower surface of the base assuming the condition of heat along the surface of the cattle can be neglected. Thermal conductivities: –

Copper 3.8 x 102 Wm -1k-1

Scale 1.34 Wm -1k-1

Specific latent heat of steam = 226 x 103 JKg -1

##### Problem 26

- Define thermal conductivity of a material
- Heat is supplied at the rate of 80W to one end of a well – lagged copper bar of uniform cross – sectional area 10cm2 having a total length of 20cm. The heat is removed by water cooling at the other end of the bar. Temperature recorded by two thermometers and at distance 5cm and 15cm from the hot end are 48 and 28 respectively.

- Calculate the thermal conductivity of copper
- Estimate the rate of flow ( in g/min) of cooling water sufficient for the water temperature to rise by 5K
- What is the temperature of the cold end of the bar.

Problem 27

- (i) The thermal conductivity of a substance may be defined by the end equation

= – A

- Identify briefly each term in this equation and explain the minus sign.
- Describe briefly one method of measuring thermal conductivity of a bad conductor in the form of disc.

- One end of a well lagged copper rod is placed in a steam chest and a 0.6kg mass of copper is attached to the other end of the rod with an area of 2cm2. When steam at 100 is passed into the chest and a steady-state is reached the temperature of the mass of copper rises by 4 per minutes; if the temperature of the surrounding is 15 . Calculate the length of the rod. Given that:

Specific heat capacity of copper

400Jkg-1K-1

Thermal conductivity of copper

= 360Wm -1K-1

U – VALUE/ THERMAL TRANSMITTANCE

The U-value of a structure ( cavity wall or a window) is the heat transferred per unit time through unit area of the structure when there is unit temperature difference across it.

U-value =

This can be obtained from;

or

UNIT OF U – VALUE

From the definition

U = =

Hence, the SI unit used in Wm-1K-1

U-values provide architects and building engineering heat losses from buildings.

They take account not only of heat lost by conduction, but of any lost by convection or radiation.

Architects base their calculations on U- values rather than on coefficients of thermal conductivity because they are concerned with the air temperature inside and outside a room, not with the temperature on the surfaces of a piece of glass, a wall etc.

THERMAL CONVECTION

This is the process by which heat is transferred from one part of fluid to another by the movement of the fluid itself.

Here the molecules of a fluid (liquid /gas) are responsible to carry heat energy from one part of a fluid to another.

TYPES OF CONVECTION

There are two types of convection:

- Natural / free convection

- Forced convection

NATURAL FREE CONVECTION

This is a type of convection in which a heated fluid flows from the hot region to the cold region due to differences in density.

##### Example

When a fluid is heated from below, the lower part of the fluid become hot and therefore expands.

Its density decreases due to the increase in volume of fluid molecules. Its position is displaced by cold fluid from the top . This in turn gets heated and rises to the top and this process continues as shown the figure below

FORCED CONVECTION

This is a type of thermal convection in which the heated fluid is forced to move from the hot region to the cold region by means of a blower or pump. i.e by external agent

GOVERNING LAWS

Thermal convection is governed by two laws: –

(i) Newton‘s law of cooling

(ii)Dulong and Petit law

NEWTON‘S LAW OF COOLING

##### Condition of the law

Newton‘s law of cooling is approximately true in still air only for a temperature excess of about 20k or 30k; but it is true for all excess temperature in conditions of forced convection of the air, i.e in draught.

##### Statement of the law

The rate of loss of heat to the surrounding air is proportional to the excess temperature over the surroundings.

Excess temperature The excess temperature of a body over the surrounding is the difference between the temperature of the body and that of the surrounding.

Let

Excess temperature =

Let = be rate at which a body is losing heat.

From Newton‘s law of cooling: –

– )

Where K = convection coefficient

Negative sign shows that the body is losing heat

If m is the mass of a body and C is its specific heat capacity, then

=

For a given body mc = constant

……………………………. (2)

Using (2) in (1) we have: –

DULONG AND PETIT LAW

Dulong and Petit modified Newton‘s law of cooling and they stated a law which works under natural / free convection.

##### Condition of the law

The law works under natural /free convection where θ – θ ° > 50k.

Where the law works

The law works in still air in the laboratory

Alternative name of the law

Five – fourth power law

Statement of the law Under natural/ free convection the rate at which a body loss heat is proportional to the five Fourth power of its excess temperature over the surroundings.

Where K = Convectional coefficient

Negative sign shows that the body is losing heat.

Problem 28

A body cools from 40 in 5 minutes.The temperature of the room being 15, what will be the temperature of the body after another 5 minutes?

Problem 29

In a room at 15 a body cools from 35 in 4 minutes. Find the further time elapse before the temperature of the body is 20

Problem 30

Wind blows over a hot liquid placed in a beaker in the laboratory whose average room temperature is 27. The liquid rate of cooling is 15 / min when it is at a temperature of 87. Calculate the liquid rate of cooling when it is at a temperature of 57 .

Problem 31

A body initially at 80 cools to 64 in 5 minutes and 52 in 10 minutes. What will be the temperature after 15 minutes and what is the temperature of the surrounding?

Problem 32

- State Newton’s law of cooling and give one limitation of the law.
- A body initially at 70 cool to a temperature of 55 in 5minutes. What will be its temperature after

10minutes given that the surrounding temperature is 31 . (Assume Newton’s law of cooling holds true)

THERMAL RADIATION

This is the transfer of heat energy from one point to another without the requirement of any material medium.

It is like a throw of radiant energy

Thermal radiation consist of electromagnetic waves with a range of wavelengths covering the infra-red and visible regions of the electromagnetic spectrum

All bodies continuously emit and absorb thermal radiation in the form of electromagnetic waves A body at higher temperature than the surrounding units it emits more radiation than it absorbs.

Thus, there is a continuous exchange of radiation between the body and the surrounding with the result that there will be a rise or fall in temperature of the body.

THE BLACK BODY

A perfectly black body is the one which absorbs completely all the radiation falling on it and reflects none

Since a perfectly black body is a perfect absorber, it will also be a perfect radiator.

When a perfectly black body is heated to a high temperature, it emits thermal radiation of all possible wavelengths.

Practical examples of perfectly black body,

- The sun

- A surface coated with lamp- black

This surface can absorbs 96% to 98% of the incident radiation and may be considered as a perfect black body for all practical purposes

HOW TO REALIZE A BLACK BODY

A good black body can be realized simply by punching a small hole in the lid of a closed empty tin

The hole looks almost black, although the shining tin is a good reflector.

Reason

The hole looks almost black ,although the shining tin is a good reflector because the radiation that enters through it is reflected from the inside walls several times and is partially absorbed at each reflection and loses energy until no radiation is reflected back. Hence the hole absorbs all radiation falling on it.

BLACK BODY RADIATION (BBR)

Is that thermal radiation emitted by a black body at a given temperature.

Any object at a temperature greater than absolute zero emits thermal radiation of all wavelengths within a certain range.

The amount of thermal energy radiated for different wavelength intervals is different and depends on temperature and nature of the surface.

INTENSITY OF RADIATION

Symbol I

This is the rate at which radiant energy is transferred per unit area.

I =

I =

The SI unit used is Watt/metre2 (Wm-2)

LAWS OF BLACK BODY RADIATION

Law 1

WIEN’S DISPLACEMENT LAW

The wavelength at which the maximum amount of energy is radiated by a black body is inversely proportional to its absolute temperature

Where K = constant of proportionality called Wien‘s consist (k) whose value is 2.9 x 10-3mk

λmaxT =K

OR

λmax T= 2.9 X10 -3

Law 2

STEFAN’S LAW / STEFAN’S BOLTZMAN’S LAW

The total energy emitted per unit area of a black body in unit time is directly proportional to the fourth power of its absolute temperature.

E

E

Where K = consist of proportionality called Stefan‘s consist (sigma) whose value is 5.67 x 10-8

Wm-2K4

Where E =

E =

E =

E =

##### Substitute eqn (2) in eqn (1)

=

Problem 33

Assuming the total surface area of human body is 1.25m2 and the surface temperature is 30.

Find the total rate of radiation of energy from the human body. Given that Stefan’s constant, = 5.67 x

10-8 Wm-2K-4

Problem 34

A black ball of radius 1m is maintained at a temperature of 30. How much heat is radiated by the ball in 4. Given that Stefan’s constant = 5.67 x 10-8 Wm-2K-4

Problem 35

Two spheres made of the same material have radii 2.0cm and 3.0cm and their temperature are 627 and 527 respectively. If they are black bodies, compare:

i. (a) The rate at which they are losing heat

i. (b) The rate at which their temperature are falling.

EMISSIVITY(ε)

The emissivity () of a surface is the ratio of the power radiated by a surface of a given body to that radiated by a black body at the same temperature

From Stefan’s law

Power / area radiated by a black body at temperature

P = ε x power of black body

P = εςAT4

Where P radiated by surface of a given body.

Problems 36

A tungsten filament of total surface area of 0.45km2 is maintained at a steady temperature of 2227 °c.

Calculate the electrical energy dissipated per second if all this energy is radiated to the surrounding.

Given that emissivity of tungsten at 22270C = 0.3 and Stefan’s constant = 5.67 x 10-8 Wm-2K-4[29.9W] Problem 37

The tungsten filament of an electric lamp has a length of 0.5m and a diameter of 6 x 10-5m.

The power of rating of the lamp is 60W. Assuming the radiation from the filament is equivalent to 80% that of a perfect black body radiator at the same temperature, estimate the steady temperature of the filament given that

Stefan’s constant = 5.67 10-8 Wm -2 K-4

T =935.5K

PROVOSTS THEORY OF HEAT EXCHANGE

A body radiates heat at a rate which depends only on its surface and temperature, and that it absorbs heat at a rate depending on its surface and the temperature of its surroundings

THEORY

When the temperature of a body is constant the body is losing heat by radiation and gaining it by absorption at equal rates

Consider a black body at a temperature T to be placed in an enclosure having temperature T0

Energy radiated /sec =

A T4

Energy absorbed /sec = A T04

If the body is not at the same temperature as its surrounding there is a net flow of energy between the surrounding and the body because of

unequal emission and absorption

unequal emission and absorption

If the temperature of the body is greater than that of the surrounding, then the net energy will flow from the body to the surrounding

Net energy emitted /sec = AT0 4

Therefore

Net energy emitted /sec =A (T4 – T04)

If the temperature of the body is less than that of its surrounding then the energy will flow from the surrounding to the body

Net energy absorbed /sec = A (T04 – T4)

Problem 38

The total external surface area of the dog’s body is 0.8 m2 and the body temperature is 37 °C at what rate is it loosing heat by radiation when it is in a room whose temperature is 17 °C? Assume that dog’s body behaves as a black body and given that Stefan’s constant is

5.67 x 10-8 Wm-2 K-4 [98.086W]

Problem 39

The cathode of a certain diode valve consists of a cylinder 2 x 10-2m long and 0.1 x 10-2 m in diameter. It is surrounded by a co-axial anode of diameter larger than that of the cathode. The anode remains at a rate temperature of 127 °C when the power of 4 watts is dissipated in heating the cathode.

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