**Units**

**Thermal Properties of Matter**

**Q.1: Define the term temperature.**

Ans: The degree of hotness or coldness of a body is called its temperature.”

It is a scalar quantity. .-Unit: The SI unit of temperature is Kelvin (K). t

**Q.2: Define the term heat**

Ans: “Heat is the energy that is transferred from one body to the other in thermal contract with each other as a result of the difference of temperature between them.”

**Q.3: What is meant by thermometery? What is the principle of thermometry? **

Ans: “The art of measuring temperature is termed as therinjometery,”

According to the principle – of thermometry, if two objects at different

temperatures are joined together, after a certain time they attain the same

temperature. ,

**Q.4: What is meant by thermal equilibrium?**

Arts: “The property of a system all parts of which have attained the uniform temperature atongwithits surroundb^ is kiK*m as therrrori equilibrium.”

**Q.5: Temperature determines the directiottof flow of heat Explain**

Ans: When you place a cup of hot tea or w^ter in a room,, it coote down gradually. It stops cooling as it reaches the room temperature. Thus, temperature determines

199

of flow of heat. Neat flows from a hot body to a coW body i.e. iperature to lower temperature until thermaf equilibrium is reached.

**Q.6: Wha|,r|f^^:^wb^-t^**

Ans: The temperature of hot body decreases. It looses energy. This energy enters! the cold body at tower temperature. Gold body gains energy and its temperaturs?! increases. The transfer of energy continues untMl both the bodies are at the samll temperature. This state of the system is known as thermal equilibrium.

**Q.7: Why is the heat energy called as the energy in transit?**

Ans: Once heat enters a body, it becomes^ts internal eiiergy and f»6 linger exist| as heat energy. Therefore^ heat energy is called a$energyinj*wsit; ..

**Q..8: What is internal energy of a body? r**

Ans: The sum of Mnetic energy and potential energy aaso^ molecules and particles of a body is calted its internal energy.”

Kinetic enengy of an-atorrt or rnolecufe is due to its motion which depends • upon the temperature. Potential energy of atoms or molecules is the stored energy due to intermoiecular attractive forces.

**Q.9: Why does heat flow from**

Ans: Heat ftow$ due to (^erel^ higher temjjjeraitufe to lower ter^erattire.^s the temperature of hot body is rnorestrian the eo$ bc^ Therefore, heat flows from hot body to cold body.

**Q.10: How does heating affect the motton of motecu)es-of a gas?**

Ans; The rnote^ttes of a gas have random motion awd inove vwth Wgh

a gas is heated, fe temperature rises ^and ^ itSr^fnoteKaaJes will mo

very high velpcities, The /rralecules . iexert rnore pressure on the walls of the , ]

container

**Q.11: What is a therrnometer? Why mercury is preferred as a thermometric substance?**

Ans: “A device that is Qsed to measure the temperature of a body is called thermometef.”

; Thermal Properties of Matter

20.Mercury is preferred as a thermometric substance because it has the following properties which;.a thermometfic isubstance should have;

(i) It te visible, ; (H)

(iii) It f®s a low freezing point. (iv)

(v) It foes not vi^t the glass. (vi)

It has uniform thermal expansion It has a high boiling point. It is a good conductor of heat.

(vii) It has a srriafl specifictieat capacity.

**Q.12: What Is liquid-in-glass thermometer? What are its uses? .**

Ans: A liquid-in-glass thermometer has a bulb with a long capillary tube of uniform and fine bore. A suitably liquid is filled in the bulb. When the;bulb contacts a hot object, the liquid in it expsirtds and ri^es in ttM§ tube, The glass stem erf anemometer is thick and acts as a cylindrical lens, this makes it easy to see the liquid level in the

vBult>

Mercury thread

Temperature scale

–. I

Mercury

Fig. 8.1: A mercury – in – glass thermometer’

Uses: Mercury-in-glass thermometers are widely used in laboratories, clinics and houses to measure temperatures in the range fr6riri-ld0C to 150*6.

**Q.13: What dp you mean by lower and upper fixed points in a thermometer?**

Ans: A thermometer has ‘a scale on its stem. This scale has two fixed points. The (ower fixed point is marked to show the position of liquid in the thermometer when it is placed in ice. Similarly, upper fixed point is marked to show the position of liquid in the thermometer^^Heri it is placed in steam at standard, pressure above boiliriig

water.

**Q.14: What are different scales Of temperature? Explain .**

Ans: Three, scales of temperature are in common use. These are:

1. Celsius scale or centigrade scale

2. Fahrenheit scale

3. Kelvin Scale

202

0ri Celsius scale, the interval between lower and upper fixed points is divided into 100 equal parts. The lower fixed point is marked as 0°C and the upper fixed point is marked as

100°C.

On Fahrenheit scale, the interval between lower and upper fixed points is divided into 180 equal parts. The lower fixed point is marked as 32° F and the upper fixed poirrt is markedas212°F. .

On Kelvin scale, the interval between lower and upper fixed points is divided into 100 equal parts. The lower fixed point is. marked as 273 K and the upper fixed point is marked as

373K.”

**Q.15: How can you convert tiwjjperaturo Oil**

Ans:The tempefatM^ on^Ke^ spjtte canrbe

obtained by abiding 213*1 the te^iwature on Rg. 8.2: various scales of temperature; Celsius scale, thus

^ (K)==273 + C ……(1)

From Kelvin to Celsius Scale

The ternperatureMon Celsius scale can be obtained by subtracting 273 from;| the temperature on Kelvin scale. Thus

.’ • . /’-.’•”‘. ‘ •- ‘..”•’•’

From Celsius to Fahrenheit Scale

The temperature on Fahrenheit scale from Celsius scale can be obtained from the following formula:

F=1.8C + 32 (3)

Untt-8: Thermal Properties of Matter

From Fahrenheit to Celsius Scale . ,

Ttoe temperature on Celsius scale from Fahrenheit scale can tMI obtained from the following formula; \

•M8C = F-32 ……(4)

**Q.16: What Is a clinical thermometer?**

Arts: A clinical therrjiometer is used to measure the temperature of human body. Its range is from 35°C to 42°C. It has a constriction that prevents the mercury to return. Thus, its reading does not change until reset; .

**Q.17: What temptrature scales are used measure ttMTiperature in ordinary life.**

**to**

Ans: Fahrenheit scale arid Celsius scale are used to measure temperature in ordinary life.

**Q.18: What is meant by absolute zero?**

Ans: The temperature -273°C or OK is known as \ absolute zero. At this temperature, the volume of any gas reduois to zero.

**Q.19: Define specific heat capacity. How would you find the specific heat of a**

Fig. 8.3

“The amount of heat required to raise the temperature of 1 kg mass of a substance through 1 K is known as specific heat capacity of that substance.”

As the quantity of heat required to rise the temperature of a body is proportional to the mass m 6f the body. Thus

AQ qc mAT1 or AQ oq cmAT …… (1)

Here AQ is the amount of heat absorbed by the body and c is the constant of proportionality called the specific heat capacity or simply heat capacity. ,

From Eq. (1), we get –

c =

AQ mAT

(2)

‘tinit of specific heat capacity is joule -amme

TabteB.1

. .

**Q.20: What is meant by heat capacity? Explain**

Ans: Heat Capacity ,,,The f j»«antity of merni^^eriei$^ -abs^itied-by a body for 4 K rise in its temperature is called heat ‘ ‘

If temperature of a body increases through AT on adcjg AQ amount ctf heat, then its heat

Heat capacity =

SubMance

•’ ”

Alcohol

Atonfciutn

• Carfeori

Ctey

£jlass

tern

Lead

Mercury

Sand

Soitpy)

Steam

Tungsten

Tmpentine.

Water

Zirib –

Spccff (cheat

2S50.0

903.0

920.0

‘-3074),

2010.0

840.0

128.0

790.0

• 128.0

138:6

83S.O

810.0

1760,3

4200,0

Putting the value of AQ in Eq. (1), we get

‘ Heat capacity = mcAT … ,v . Heat capacity = rnt; ‘ .i…(2)

Equation (2> shows that heat capacity of a body is equal to the product of its mass a$d specific heat capacity.

Unit: The SI unit of heat capacity is joute per kelyin(JK”1). ‘ , Q.21: How heat capacity relates with the mass of a body?

Ans: A body with greater mass wM absort* m nfease jp^e heat^ and vice-wersa. Thus a body with greater mass has .more heat capac8ty;:and a body with srnaHer, mass has low heat capacity.’ ,

**Q.22: How hea^capacity relates with the change in temperature? **

Untt-a* Thermal Properties of Matter

Ants: The greater .’the change vn temperature of a body, greater wll be the heat capacity.

**Q.23: Describe the importance of 4arge specific heat capacity of water.**

Ans: Importance of Large Specific Heat Capacity of Water

1 . Due to large specific heat of ,^tef ^t^ t^^^rature variations f rorfi , sumrner

to winter are much smaller at places hear the sea than land far away from the

. ‘ sea/

2. Due to large specific heat ^oapacii^ jwater is very useful in storing and carrying thermal energy . “Fhe coding, system of automobiles uses water to carry away unwanted thermal energy. . ,

3. In central heating system, hot water is used to carry ..thermal .energa! through pipes from boiler to radiator. These radiators are fixed inside the house at suitable places. . ,

**Q.24: Show the change of state of ice into water and steam with the help of temperature-time graph.**

Ans: For a change of state, thermal energy is added to or removed frort»iifc substance. ‘

Take a beaker and pla^d§4t over® stafM. Pirt ^^ some pieces of ice in the beaker thermometer in the beaker to measure the temperature of ice.

and suspend a

Now place a burner und-er the .beaker. The ice witt start matting. The temperature of the mixture containing ice ad water will not increase abcwepW’Gwvtil all the ice melts and we get water at 0°C. If water at 0°C is fufthfr heated, its temperature begins to rise above 0°C as shown in the graph.

Part AB: On this portion of the curve, the temperature of ice increases from -30°C toO’C.

Part BG: Wheri the tetnperaUire ,of ice reaches 0°G, the ice water mixture remains at this temperature until all the ice melts.

Water D S steam

t,

fime-

Fig. 8.4: A graph of temperature and •finrieshowmgchaf^e of state of ice into water and steam. .

temperature of the substance gradually Jncreases from O’C to nt of energy so added is used up in increasing the temperature of n

Part DE: At 100^0

remain

>We

Q&& Define and explain latent heat of fusion. Ans: Latent Heat of Fusion

“Heat energy required to change unit mass of a substence from solid to I state at its meiting point without change in ite temperatore is called its latent fusion,*

Hf

m

or

^•=’:mWf

Ic^ changes at 0°C into water. L i.e., ,3-x 36. 105J heat is ^^

: Define and explain latent heat of vaporization.

n of ice is 3.36 x105

Ans: “Heat enirgy re^utfed fo^-.change unit^ mass^ of a fe^ complwely hito gas i its boiling point witrksut .any change in its temperature te caHed its latent heat <

vaporization.” v’.\;.-”’—

It is denote^ by Hv. ”,. •

or

Hv =

AQv *

AQy m

(D”

When water is heated, it boils at 100c under standard pressure. Its remains 100°C until it is changed epi7ip|el^.^iRto—-slel^.”L^r*’heM-of of ice is 2.26 x 106 Jkg”1 i.e. 1 kg of wate^ re^ujress 2J^ )< 108 J heat to diangtf compieteiy into steam at ite boiling point. *iu ^;?

Table 8^: Melting point boiHng point, latent heat of fusion and latent heat of vaporization of some comrnon’substanees. – ,

Sub«tance

Mating

H«at of fusion

Aluminum

660

2450

39-7

••*«*&

Copper

1083

2595

205.0

4810

GoW

toes

1580

WWiiih

Lead

327

Mercury

,39

357

Nitrogen

-210

-T96

25.S.

Oxygen

-219

Water

Ana: “The changing of a liquid into vapours from the surface of the liquid without

The esc^pir^ out erf fast mo\4r^ mote<^tes frc^ trw surface of a liquid without heating is known as evaporation.” 7i

**Q.28: Explain how GCMrfteig tofirodiiced by evaporation?**

Ans: During evaporation^ fast moving molecules escape otrt from*® swrlace of tfre liquid. Molecules that possess lower kinetic energy are left behind. This lowers the average kinetic energy of the liquid molecules and the temperature of the liquid. Since temperature of a substance depends on the average kinetic energy of its nrx)lecutes, so O)Q^ is orcduced by evaporation. . ‘ :

**Q.29: On what factors the rate of evaporation bf a liquid depends?**

Ans: The rate of evaporation is affected by tfie foikswing factors: -1. Temperature

dry up r?^^^^ temperature, more molecules of a Hquki are rrKWir^ with molecules escape from its surface. Thus, evaporation is faster at higher temperature than at lower temperature.

2.

evaporates faster when spread,over large area. Larger is the.sc area of a liquid, greater number of molecules has the chance to escape but from!

surface Thus rate of evaporation increases. 4 –

Wind blowing in over the surface of^aJiquid sweeps away the liquid mqlecute£f that have just escaped out. This increases the chance for more liquid i escape^yt and the rate of evaporation incareases.

4. Nature of the Liquid

Liquids differ tit the rate at which they evaporate. For example, spirit evaporates quickly than water.

**Q.30: How evaporation differs from vaporization?**

Ans: Thechanging of a; liquid inti>vapours from the surface of the liquid without heating is known as evaporation. Whereas the changing of a liquid, or solid into’ gaseous state by the. application of heat |§ known as vaporization.

**Q.3I: What is meant by thermal expansion?**

Ans: Usually, bodies expand on heating, this ‘expansion of bodies on heating is called thermal expansion. It is different for different substances.

**Q.32: What is meant by Imear thermal expansion? Explain.**

Ans: “The expansion along length or in one dimension is called linear thermal expansion.”

Consider a metal rod of length lo at temperature T0. Let its-length on heating to a temperature Tbe9omesL. Thus

Increase in length of the rod AL-L-^ La. Increase in temperature AL = T-T0

It is found that change in length AL ts directly pmportidnal to the original length lo and ^e change in temjaerature AT. Thus

AL oc L<jAL

Untt-8: Thermal Properties of Matter

or

dr

AL * ctL0AL l-lo =

(1)

or L . – lo x

V L = Lo(1 taAL) FromEq. (1)

AL . •

(2)

a =

L0AT

lv. ^<3>

where a is called the coefficient of linear thermal expansion of the substance. It is definedas: .

“The fractional increase in the length of a substance per Kelvin rise in temperature is called coefficient of linear thermal expansion.”

Unit: The SI Unit of coefficient of flnear tHfrmaj expansion is per keMn(K~\)^

**Q.33: What is meant by voJumetric th^nnal expansion? Explain**

Ans: “The expansion of a solid in three dimensions is called volumetric thermal expansion or cubical thermal expansion.”

Consider a solid of initial volume V0 at certain temperature T0. On heating the solid to a temperature T.i^i vdurne becomes V. Thus 7

Change in volume AL = y-V0 and Change in temperature v AT = T-T0

The change in volume AV is proportibnal to its original volume V0 arid change in temperature AT. Thu§*

or or or or

AV

V-A/0=pV0AT ‘ V – vo-k 0%^T

V = V0 (1 .+ p AT)

(.1)

(2)

FromEq. (1)

AV

V0AT

.(3)

called the coefficient of volume therrml expansion. It is defined•>

fractional change in volume of a substance per Kelvin change perature is called coefficient of volume thermal expansion.*

The coefficient of linear thernial expansion and volumetric thermal are related by the equation:

• P = 3oc ..,.(4)

**Q.34: What are the consequences of thermal expansion?**

Ans: Some consequences of thermal expansion are:

Railway tracks buckled on the hot summer day due to expansion if gaps not left between sections, So provision is made during construction fof expansion and contraction with temperatur,e. « ,,

night. They will bend if their ends are fixed. To allow thermal expansion, one end is fixed white the other end of th« girders rest on rollers in the gap left fcf

3. Overhead transmission lines are alsc- given a certain amount of sag so that

**Q.35: Describe some everyday applications of thermal *xpansion**.

Ans: Sc>rne applications of thermal expansion are:

1. Thermal j^panslon is Msejd

2 The cap of a bottle that is tight enough, is immersed in a hot water for a minute or so. Metal cap expands and becomes loose. Nksw it can easily be ‘• ‘opened.’

3. To fix iron rims on wooden wheels of carts, they are heated. Thermal expansion allows them to slip over the wooden wheel. Water is poured on it

to cool. The rim contracts and becomes tight over the wheel.

4 To join steel plates tightly together, red hot rivets are forced through holes in the plates. The ends of/hot rivets is^^n hammered. Ort cooling, the rivets contract and bring the plates tightly gripped.

: Thermal Prop«i1t»« of Matter.

strip? Describe Its working and some

jAns: A bimetal strip [consists of two thin strips of metals • sueh as ttorass and iron joined (Fig 8.5*>>. This strip bends on iting due to difference in thermai , expansion

1.

I

Bimetal stripes areused for variouspurposes.

Bimetal thermometers are used to measure temperature especially in furnaces and ovens.

Bimetal strips are used in thermostats. Bimetal thermostat switch is used to *»ntro1 ttm terT»p€*ature of heater cott in an electric iron.

Ans: A thermostat is an automatic switch which contr^s the^ tejri^aratyre fl<rany

.device. It usually consists of a strip made of two different metals Such as iron and

brass known as

**Q.38: What Is thermal expansion of Hquids? Explain** ^

Ans: Thermal Expanskwi of Liquids ‘

The thermal expansion in liquids is greater than solkls due to the. weak irrtermolecular forc»s. Therefore, the coefficient of volume expansion of Hquids is -greater than solids. ” -,

When a liflttW *s> hea^ ta©titf «a^g»d^e «fiMwW«*|fi|e ^^hgnge ‘m volume. Thus^ iheff ^*ff»Sfc oftheraiiBi jii^inj^lp^

1. Apparent volume expansion 2. Real

Activity: Take a long necked flask and fill coloured water in it up to mark Aas;shc ift-Rg»”&6. Now start heating the flask from the bottom. The liquid level first falls tbl and then rises to C.

The heat first reaches the flask which expands and its volume increases. As a result liquid descends in the flask and its level falls to B. After sometime, the liquidf begins to rise above B on getting hot. At certain temperature it reaches at G.

The rise in level from A to C is dye to the apparent expansion in the yolujffflif of the liquid. Actual expansion of the liquid is greater than that due to the expansion; because of the expansion of the glass flask. Thus, reaj expansion of the I«juid4s equal to the volume difference between A and C in addition to the volume expansion!

of the flask. Hence .Real expansion of the liquid = Apparent expansion of the liquid + Expansion of theflasfc”‘

or BC = AC + AB ……(1) ”

**Q.39: What do you know about anomalous expansion of water?**

Ans: Water on cooling below 4°C expands until it reaches 0°C. On further cooling, its volume increases suddenly as it changes into ice at 09C. When ice is below 0°C, it contracts. This unusual expansion of water is called anomalous^ expansion of water.

# mcqs

Encircle the correct answer from the following choices:

i. Water freezes at

(a) 0°F (b> 32?P (c) 273 K (d) OK ii. Normal human body temperature is

(a) 15°C ” ” (b)37°C. (c) 37 °F (d)98.6°C

Hi. Mercury is used as thermometric material became it has

(a) uniform thermal e>^pansion'” (b) low^eegfBgpoint

.^,j.fs ;

(c) small heat capacity a (^1^ the above properties

L>nit-8: Thermal Properties of Matter _______ 213 iv. Which of the following material has large specific heat?

(a) copper (b) ice –

(c) water (d)> mercury , m

v. Which of the following material has large value of temperature

e^ttWe^

(a) aluminum (b)gold ‘ • (c) brass (d)steel

vi. What will be the value of p for a solid for which a has a value of 2 x10~ 5 . K-1? .

(a)2×10″5K”1 .••.;v:<b).6x.lOs»C1 (c) 8×10″15K”1 (dJSxIO^K’1

vii. A large water reservoir keeps the temperature of nearby land moderate due to

(a) low temperature of water (b) low specific heat of water

(c) tess, absorption .’-of hell (d>’farge specific heat of water

viii. Which of the following affects evaporation?

(a) temperature (b) surface area of the liquid

(c)wind ‘ (d) all of the above

ANSWERS ‘ V

(i)’b’ ..(») b (Hi) d (iv) c (v) a |vi) o

(viii)d 4 v

ADDITIONAL MCQ’S

Encircle the correct answer from the following, choices:

1 . A quantitative study of thermal phenomena requires a careful definition

of ‘

(a) friction – i(s)heat. o« cq ^T > (c) temperature and inteimat energy (d) both (b) and (c) , fc 2. To understand the concept 0f temperature, it is useful to understand

(a) specific heat , , ^(bj^e^l^corita^^, (c) thermal equilibrium (d) both (b) and (c)

of K. E. and P. E.

‘. (b) teat capacity –

(a) heat , ,-

(c) internal 6n©ipy 4

WhteH of the following st^timoe^ liew orMier

of tts

(a)steef (c) water

Mercury freezes at (a)-9eC (c)-29*C -Mercury boils at

(c)387°C

(a) *5°C to 135*C ;s ; ^ (|j

(c)-15°G to liS’C (d)-20°C to

On Fahrenheit scale, the interval between lower and upper fixed points

isdfvWedlnta

(a)IOOparts’ ?^?(b>1SO parts

(c)270parts M^Tfparte ; ‘?

On KeMn scale, freezing point of ice is :

(a)OK (b)32K

(c)273K {d)373K

The absolute zero is equal to •

(c)-273°C (d)-373’C The temperature of ice to fraezer Is .

(a)-rc ; .’•:•’-:•• : ^’-v:’….-, ”a;v

!l *>;’

The temperattire of Hquld iowygen is

(a)-180°C “;;;-“‘;=;;'”. “”s!’;’•”•” ;:”.. (b) (c)-273K (dj

ut Properties of Matter

215

13. To change temperature from Cetelus W^rif*e* *a*le, we use -the

‘ ‘ K ••• *• •_• •> *'<§)H’j’.fK “-i •’• **’*Vi**TJHB^

“‘formula: .

T v . /\• /iy\ _ /•* .r *>TTO /W\ /^ « TT/lrfr\ O^O ^4c*

(a) T (r\) a C + 273 (b) C = T(K)-273 *• (c) F = 1.8 C * 32 (d) (4) botti (a) and (b)

‘•’ l\* Vfl;U -S {‘H’^lfTtiiJ’i K -‘^-^-‘J-fi’-

14. Todange temperature from Celsius to Kervln scales, we use the ‘ •’. formula. \ ” (a) T (K)* C•+ 273 (b) C = T(K)-273 (c) F =1.8G-«-32(d)^’-V” ^ ._^d)’-b^.^i^^}^

15. The range of clinical thermometer is 4 (a)35°Cto42’C. (c) 37’C to 98.e°C

16. 20C Is equal to (a)253K (c>273K fe,, (d)293K

17. 300 K Is equal to v -(a)27°C (bV37°C (c) 42’C,

18. 50°Clsequalto (a)112°F (c)132°F

19. 100°F is equal to (a)27.8°C (c)45.6°C

20. The scale which is (a) Celsius scate (c) Kelvin scate

21. The temperature of pure fccatted

(d)45’C

(bj 122°F (d>142’F

(d)98^°C

(b)f ahrenheit scate (d)8ttof these tee at standard atmospheric pressure

(c) absolute zero (d^tti»fTrK»netfy ‘

The temperature of pure boWngwatsr at standard atnw«»*ieric pressure

Is called –

(a) tower fixed point (b) upper fixed point

(c) absolute zero

of heat required to raise tfre temperature of 1 kg of I substance through 1 K is called

(a) heat capacity (b) specific heat capacity

(c) thermal energy (d) thermal conductivity

For the same change In temperature, Hie substance with greater

wiflemit •

(a) more heat (b) tess heat

(c) same heat (d) none of these

The greater the change in temperature of a body, greater will be the I

(a) absorbed (b) released :

(c) both (a) and (b) (d) none of these

The materials which take longer time to become cool or hot

specific heat capacity

(a) low (b.) very low

(c)high (d) none of these –

The SI unit of specific neat capacity is

(a).Jfcg v, (b)Jkg-1

(c).J.-kg-

(d)JK

The specific heat capacity of water is ;

(a)420 J kg’1 K’1 (b)210 Jkg’1 K’1 ‘

(c)2100J’%T1-K”1 (d) 4200 J kg’11C1

The specific heat capacity of ice is

(a) 420 J Kg-1 K’1 (b) 210 J Kg”1 K^1

(c)2100JKg-1K-1 V. (d) 4200 J Kg’1 K~1

The specific heat capacity of soil is

(a)810Jkg-1K-1 * ‘ . (b) 903 j kg’1 K”1

(c) 840 J kg’1 K-1 (d) 920 J kg’1 K’1

The specific heat capacity of aluminum is

(a)810JKg-1K-1 (b) 903 J Kg”1 K-1

(c) 840 J Kg’1 K’1 (d) 920 J Kg’1 K’1

The .quantity o| thermal energy absorbed by a body for 1 K rise in its

temperature is called

(a) specific heat capacity (b) heat capacity

(c) heat of fusion – (d) heat of vaporization

Unrtrft: Thermal Properties of Matter

Heat capacity of 5 kg water Is t w

(a)420JK’1 (b)210JK-1 (C)2100JK-1 • < (d)4200JrC1 Latent heat of fusion of ice Is

(a)2.26x1d5Jkg-t (b>3.36x 10s J kg-1 . (c) 2.26×1<PJ kg’1 (d)3.36x 106 Jkg’1 Latent heat of vaporization of water is (a) 2.26×1 Q5J kg’1 • ‘ (b) 3.36x105Jkg~r (c)2.26x106Jkg-1 ‘ (d) 3.36 X 106 J kg’1 The changing of a liquid into gaseous state from the surface of the liquid without heating is caited

(a) vaporization (b) latent heat of vaporization (c) latent heat of fusion, (d) evaporation Therateof evaporation is affected by –

(a) temperature;

(c)wWd ^

Freon gas is now replaced by

(a) nitrogen i;,

(c) ammonia

The coefficient of linear thermal expansion is equal to

AL

U’ ; ..[:’

al :”

a=

(b) surface area (d) all of these

(b) chlorine

(d) carbon monoxide

f (b) a =

40. The coefficient of iinear thermal expansion of aluminum is

41.

42.

(c) 2.4×1 Q-5rC1 ‘ . (dlBJxIQ^ K:1 , The coefficient of linear thermal expansion of platinum is

(c) 2.4×1 O^IC1 (d) 8.6×1 0^ K”1 » The coefficient of volumetric thermal expansion is equal to

AV VoAT

ent of volume thermal ex^nsion

fr -.

(c)21×10’6K-1

The coefficient of volume thermal e)^||^o^p|r:fci:^,,^r^. ^

(a) 7,2x1O*JCr {b) 18X1<J^ JC1 ^

(c) 21x-10*/IC* (d) 27xir5Kt .,v,,

The coefficients of linear thermal expansion and voiume

expansion are reined by the equation

46. Wires on electric poles are given some sag to prevent breaking

(a)« samffBr ; ^b) in winter (c) in daytime (d)atnig^t

47. A bimetal strip consists of two trHn strips of different metais »ueh as

(a) copper and iron > (b)alun^nuniand^%»ir (c) brass and iron i^cl) plajtowfn and ten \^ –

48. Birnetal thermostat breaks the electricaJcircuR at preset

(a)length • (b)vofurne (d) densjty • (d) temperature

49. ‘ When we heat a bimetal strip, it ;’ fj-—;;;

(a)conttacts ^- (b)expand (c) bends

50. Bimetal thermostat switch is used in

(c)ovens ^

At what temperature, water on cooling begins to expand?

‘

(c) above 4°C ;*

At what temperature, when ice is cooled, tt contracts?

^:]:-.^,-^^^

(c)abdve4°C .

The unusual expansion of water is catted

(a) linear thermal expansion (b) volumetric thermaJ expansion (c) anomalous thermal e)^insk>n

Untfc«:Th»fmalProp«rtt«« of Matter

54. 55.

The density of water is maximum at . • :

(a) 100’C ..(bM’C •”• (c)4°C ” ‘ : . r- ; .(<qr

When we increase the temperature of water from 4°C to its boiling point,

(a) contracts (b) expands ‘ (c) vanishes (d)vWurtTe does not change

ANSWERS .’•

1. 2.

d d

8. 9.

b c

14. 15.

a a

20. c 21. a

26. c 27. e

32. b 33. c

38. c 39:” c

44. c

Af\ /*

*TW. \f

50. d 51. b

3.

c

10

c

16.

d

22. b

28. d

34^ b

4% C

)4& b

52, a

5.

d

11.

b

17.

a

23. b

29: c

35. C

41. d

47. c

53. c

6.

a

12.

a

18.

d

24, a

30. a

36. d

42; d

jt^O-“” • ft

^rO”. Q

54. c

7.

b

13.

c

19.

b

25, c

3-1: b

37. d

43. a

49. c

55. b

P.8.1 Temperature of water in a beaker is 50°C. What is its value in Fahrenheit

Given Data –

Temperafyre in Cefsius scate « 50’C To Find: Temperature in ^ Fahrenheft scate -,=

PuWng the values

F – 1:8×50+32

* 90* 32 v. ‘ ‘•

; f *’

P.8.2 Nomial human body temperature is 98.6°F. Convert tt Into Celsius scale and Kelvin scaie- ••. “j v–;r ‘ – .-,; .-‘ ,,,,-. ,:q; ,,,’: _.’ ..•• ..:;.,- •/.•••

Given Data :• Temperature in Fahrenheit scate , =98.6°F

Ptil

To Find

(i) Temperature in Celsius scale , (ii) Temperature in Kelvin scale Calculations: (i) We knowthatiu c

F = 1.8 C + 32 or 1.8C .= F – 32 s; : Putting the values

1.8C -‘ 9&6-S2 –

1,8 C > 66.6

66.6 v5

-• ?

c –

1.8

‘^^Kfm..^. •^ft’^’SAx.ijSi

It

Putting the values 3

‘T(K) . = 37 +\273’ : :-c -—•

T(K) =310K Ans. – *- ; yw

P.8.3 Calculate the increase in the length of an aluminum bar 2m long when heated from 0°C to 20 °C. If the thermal Coefficient of linear expansion of aluminum Is 2.5 x 10″5 K”1. •

Given Data:

Original length of the rod L, Initial temperature T

Final temperature \ pT Change in temperature A

Coefficient of linear expansion of aluminum

= 2.m

= T- to = 20

0 °C * 26 °C = 20 K

a = 2.5 x 10

,-5

Unit-8: Thermal Properties of Matter 221

‘ To Find: Increase in length Of the bar AL = ? a

Calculations: Using the formula L – lo (1+aAT)

L = 2 m (1 + 2.5 x 10-5 JT1 x 20 K) = 2m(1 + 50+.X-KT5)

= 2 m (1 +0.00050)

_ ‘ . ;-: • “- 2 rrv(f .00050) •>• ‘M:

: L = 2.001 m : •

. • AL= 1-U •

AL= 2.001 m-2m

= 0.001 m

= 1 x 10~3m A L= 0.1 cm Ans.

P.8.4 A balloon contains 1.2 m3 air at 15 °C. Find its volume at 40 °C. Thermal coefficient of volume expansion of air is 3.67×10″^ K”V ‘l¥?

Given Data

Original volume of the balJoon Vo = 1.2m3

, Initial temperature T0> t5°C

Final temperature T = 40 °C

Change ih-tejrtjpera^ure AT =*T-T0«408C-1&°C = 25°C==25K

Coefficient of volume expansion of air p = 3.67 x 10″3 K~1

To Find: Votume at 40°C V = ?

Calculations: Using the formula V = V0(1 +PAT)

222. . _. _____ ;______ , , . …„, . , •’ •• Putting the values

V = 1.2m3 (1+3,67 x lO^K’1 x 25 K) = 1.2ma(1+-9t.75x xlOT3) = 1.2m3 (1+0.09175) = 1.2m3 x 1.09175 .’. V = 1.3m3 Ans. ,« t

P.8.5 How much heat is required to increase the temperature of 0.5 kg of water from10°Cto65°C?

Given Data

Mass of water m = 0.5kg

Initial temperature T2 – 10°C • > Final temperature Tr;= 65°C

Change in temperature AT = tt – T2 = 65°C -10°C = 55 °C = 55 K To Find: Heat required AQ = ? Calculations: Using the formula

AQ = meAT Putting the values

AQ = 0.5 kg x 4200 J kg’11C1 x 55 K

AQ = 115500 J Ans.

P.8.6 An electric heater supplies heat at the rate of iOQO joule per second. How much time is required to raise the temperature of 200 g of water from 20 “C to 9<*°C?

Given Data:

Rate of heat supply or Power P= tOOOJs’1* – ‘• Massbf water m = 200g =0.2 kg

Unrt^: Thermal Propgrties of Matter_ ; .223 ‘ Initial temperature T2 = 20°C –

Final temperature •”•TY= 90°C

Change in temperature AT = T, -T2 =.90*C – 2STC = 70°C = 70 K

Specific heat of water c = 4200 J kg”1 K”1 to Find: Time t = ? Calculations: We know that

or Pxt .* E or Pxt = mcAT

_ mcAT Putting the values

or’ * p

• 0.2 kg x 4200 J kg”1 K”1 x 70 K . V ‘ =10QOJST1

t = 58.8s Ans,

P.8.7 How much ice will melt by 50000 J of heat? Latent heat of fusion of ice is 336000 JkgV

Given Data

Amount of heat required Qf = 50000 J -to melt ice

Latent heat ol fasten of ice Hf * 336000 J kg-1 To Find: Mass of ice m = ? Calculations: Using the formula .

.

or m – rf . • -rif

224

Putting the values

50.000. J.

336000 J kg”

m =

= 0.1488 x 1000g : . “‘:”Q^

‘•’ .’: • m = 150g Ans. P.8.8 Find the quantity of heat needed to melt 100g of ice at -10 °C ir

water at 10 °C. (Note: Specific heat of ice is 2100

, specific heat of Water

~ 1

420p;Jkg~’K~\ Latent heat of fusion of ice is 336000 Jkg’1). r Given Data ‘

V’ ‘

‘Massof.ice.., m = 100g = 0.1kg Specific heat capacity of ice Ci = 2100 J kg”1 K”1 T

Latent heat of fusion of ice Hf = 336000 J kg”1

Specific heat capacity of water c2 = 4200 J kg”1 K”1 4 ,

Temperature of ice T3 .-.= -1-0°G ,r :

Temperature of ice and T2, = 0/°C ‘

water mixture :~

Change in temperature AT = .-Tj-Tg = 00C-(-10°C) = 10°C = 10K temperature of water Ti •’.”= 10*5”

Change in temperature AT =’ ti -T2 = 100C-0°C- 10°C = 10 K

To Find: Amount of heat required Q = ?

Calculations

Heat gain by ice from -10°C to 0°C ;<

Heat required for ice to melt

, ( , :

= 0-1 kg x 2100 J kg”1 K”1 x 10 K Q, = 2100 J Q2 – mHf

= 0.1 kg x 336G^0Jkg”1

– 33600 U

Unit-8: Thermal Properties of Matter

— ———

Heat required to raise the temperature, Q3 = mc2AT .o-‘S^ r

of water from 0°C to 10°C ^ ,,

Qa = 0.1 kg x 42QO J kg-r K-1 x 10 K v Q3 = 4200 j

Total heat required ‘ “Q = Q + Q2 + Q3

= 2100 J + 33600 JJ +: 4200 J Q * 39900 J Ans.

P.8.9 How much heat is required to change 100 g of water at 100°C into steam? Latent heat of vaporisation of water is2.26x1Q6 Jkg”1.

Given Data

Mass of water . m = 100 g = 0.1kg Catent heat of vaporization Hv = 2.26 x t06Jkg”1 of water . To Find: Heat required AQV = ? Catculation: Using the formula

… ; AQV = rnHv–” Putting the values ;

AQV = 0.1 kg x 2.26 x 106J kg’1 , –

– 0.226 x 106J : … AQV = -2.26. x,10^J Ans,.,, ;•’.

P.8.10 Find the temperature of water after passing 5 g of steam at 100 °C through 500 g of water at 10 °C.

: •’ ‘ ‘. • ‘ , ‘ • ‘ ‘ ‘ ;• , ;t’ •.- • ‘ ;,:-••’•••’

(Note: Specific heat of water is 4200 Jkg~1K~1, Latent heat of vaporization of water is 2.26 xl06JkgV

Mass of steam m, = 5 g = 0.005 kg Temperature of steam T, = lOO-‘C” •-•’•• • Mass of water m2 – 500 g = 0.5kg Temperature of water T2 = 10°C

v

226 _______________:••. _______. ••;_____Phy»fc»tf Latent heat.of vaporization Hv = 2.26 x* of water

To Find: Final temperature of water T3 * ?

Calculations

Latent heat lost by steam Qi = mHv

Qi = 0.005kg x 2.26 x 106Jkg~1

Qi -11.3 x to3.

Heat lost by steam to attain “final qz – mi CAT

temperature Qz = 0.005 kg x 4200 J kg’11C1 x (100-T3) K

Q2 = 21J|100-T3) Heat gained by water Q3 = m2cAT

Q, = 0.5 Hg X 4200 J kg”11C1 x (T3 – 10) K

Q3 = 2lbOJ(T3-1Q) Since Heat lost by steam = heat gained by water

or Qt + Q2 = Q3 Putting the values

11.3 x 103J -»-21J(100-T3) = 2100J (T3-16) or 11300\J + 2100 J-21JT3 = 2100 JT3-21000 J or 1130bU+ 2100 J + ^1000 Je= ^fdOJts •»• Zi’JTs. or 34400 J = 2121JT3 . ,

34400 or ‘3 = 212t

T3 = 16.2°C Ans. :