Thermometry, Thermal Expansion and Calorimetry

C F − 32 25 F − 32
17. (b) =  =  F = 77 F .
5 9 5 9
18. (c) Thermoelectric thermometer is used for finding rapidly
varying temperature.
19. (c) Due to evaporation cooling is caused which lowers the
Thermometry temperature of bulb wrapped in wet hanky.
F − 32 K − 273 x − 32 x − 273
1. (d) T = 273 .15 + tC  0 = 273 .15 + tC 20. (c) =  =  x = 574 .25
9 5 9 5
 t = −273 .15 C
C F − 32 C (140 − 32 )
C F − 32 −183 F − 32 21. (c) =  =  C = 60 
2. (b) =  =  F = −297 F 5 9 5 9
5 9 5 9
C F − 32 t t − 32
22. (a) =  =  t = − 40 
F − 32 K − 273 F − 32 95 − 273 5 9 5 9
3. (a) =  =  F = −288 F
9 5 9 5 23. (d) Standardisation of thermometers is done with gas
4. (c) Temperature change in Celsius scale = Temperature thermometer.
change in Kelvin scale = 27 K 24. (a) For gases  is more.
5. (b) Change in resistance 3.70 − 2.71 = 0.99  corresponds 25. (c) The boiling point of mercury is 400°C. Therefore, the
to interval of temperature 90°C. mercury thermometer can be used to measure the
temperature upto 360°C.
So change in resistance 3.26 − 2.71 = 0.55 
(Pt − P0 ) (60 − 50 )
Corresponds to change in temperature 26. (a) t =  100 C =  100 = 25 C
(P100 − P0 ) (90 − 50 )
90
=  0 .55 = 50 C 27. (b) By filling nitrogen gas at high pressure, the boiling
0 .99
point of mercury is increased which extend the range
6. (d) – 200°C to 600°C can be measured by platinum upto 500°C.
resistance thermometer.
28. (a) Pyrometer is used to measure very high temperature.
7. (c) Pyrometer can measure temperature from 800°C to
F − 32 K − 273 F − 32 0 − 273
6000°C. Hence temperature of sun is measured with 29. (c) =  =
9 5 9 5
pyrometer.
 F = −459 .4 F  −460 F
8. (a) v 2  T 30. (c) Initial volume V1 = 47 .5 units
9. (b) Thermoelectric thermometer is based on Seeback Effect. Temperature of ice cold water T1 = 0C = 273 K
10. (b) Maximum density of water is at 4°C
Final volume of V2 = 67 units
C F − 32 4 F − 32
Also =  =  F = 39 .2F V1 V2
5 9 5 9 Applying Charle’s law, we have =
T1 T2
11. (c) Production and measurement of temperature close to
0 K is done in cryogenics (where temperature T2 is the boiling point)

12. (c) V2 67  273

or T2 =  T1 = = 385 K = 112 C
V1 47 .5
13. (c) At absolute zero (i.e. 0 K) vrms becomes zero.
14. (c) 31. (a) Temperature on any scale can be converted into other
x − LFP
15. (c) We know that P = P0 (1 + t) and V = V0 (1 + t) scale by = Constant for all scales
UFP − LFP
and  = (1 / 273 ) / C for t = −273 C , we have P = 0 x − 20 60
=  x = 98 C
and V = 0 150 − 20 100

Hence, at absolute zero, the volume and pressure of the C F − 32 C 140 − 32

32. (d) =  =  C = 60 C
gas become zero. 5 9 5 9
33. (a) Rapidly changing temperature is measured by
16. (d) Zero kelvin = −273 C (absolute temperature). As no
thermocouple thermometers.
matter can attain this temperature, hence temperature
34. (d) Difference of 100°C = difference of 180°F
can never be negative on Kelvin scale.
180
 Difference of 30° =  30 = 54 
100

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 Thermometry, Thermal Expansionand and Calorimetry

35. (a) 12. (b) In summer alcohol expands, density decreases, so 1

litre of alcohol will weigh less in summer than in
winter.
Thermal Expansion
13. (b) Similar to previous question, benzene contracts in
1. (c) When a copper ball is heated, it’s size increases. As winter. So 5 litre of benzene will weigh more in winter
Volume (radius)3 and Area  ((radius)2, so than in summer.
percentage increase will be largest in it’s volume. 14. (d) Water has maximum density at 4°C.
Density will decrease with rise in temperature.
15. (a) Since coefficient of expansion of steel is greater than
h  (1 +  1 )  0  that of bronze. Hence with small increase in it's
2. (a) 1 = 1 =   = 
h2  2 (1 +   2 )  (1 +  )  temperature the hole expand sufficiently.
50 1 +   50 A L A
 =   = 0 .005 / C 16. (d) A  L2  = 2  = 2  2 = 4% .
60 1 +   100 A L A

3. (b)  r =  a +  v ; where  r = coefficient of real expansion, V1 1 +  t1 100 1 +   20

17. (d) =  =   = 0.0033/°C
 a = coefficient of apparent expansion and V2 1 +  t 2 125 1 +   100

 v = coefficient of expansion of vessel.  2  10 −5

18. (d)  = = = 10 − 5 / C
For copper  r = C + 3 Cu = C + 3 A 2 2
For silver  r = S + 3  Ag 19. (d) Coefficient of volume expansion
 ( −  2 ) (10 − 9 .7)
C − S + 3A  = = 1 = = 3  10 − 4
 C + 3 A = S + 3 Ag   Ag =  .T  .( ) 10  (100 − 0)
3
4. (d) Fractional change in period Hence, coefficient of linear expansion
T 1 1 
=  =  2  10 − 6  10 = 10 − 5 = = 10 − 4 / C
T 2 2 3
T
% change =  100 = 10 − 5  100 = 10 − 3 % 20. (a)  = 0 (1 −  . ) = 13 .6[1 − 0.18  10 −3 (473 − 273 )]
T
L 1 +  ( )1 = 13 .6[1 − 0.036 ] = 13 .11 gm / cc .
5. (c) L = L0 (1 +  )  1 =
L2 1 +  ( )2 21. (b) As we know  real =  app. +  vessel
10 1 + 11  10 −6  20   app. =  glycerine −  glass
 =  L2 = 9.99989
L2 1 + 11  10 − 6  19
= 0.000597 − 0.000027 = 0.00057 / C
 Length is shorten by 22. (c) Water has maximum density at 4°C, so if the water is
10 − 9.99989 = 0.00011 = 11  10 −5 cm heated above 4°C or Density
cooled below 4°C density
6. (c) Stress = Y  ; hence it is independent of length. decreases i.e. volume
7. (c) Solids, liquids and gases all expand on being heated as increases. In other words,
result density (= mass/volume) decreases. it expands so it overflows
in both the cases. 0°C 4°C Temp.
8. (c) As coefficient of cubical expansion of liquid equals
coefficient of cubical expansion of vessel, the level of V 0 .24
23. (a)  = = = 6  10 − 5 / C
liquid will not change on heating. V .T 100  40
T 1 1 
9. (b) Loss in time per second =  =  (t − 0) = = 2  10 − 5 / C
T 2 2 3
 loss in time per day  
24. (a) As  =   :  : =1: 2 : 3
=
23
1  1 1
t =  t  t = t  (24  60  60 ) = t  86400 Mass expelled
2  2 2 25. (a)  app. =
Mass remained  T
10. (c) A bimetallic strip on
x / 100 1
being heated bends in = = = 1 .25  10 − 4 / C
the form of an arc with x  80 8000
A >B
more expandable metal B 26. (b) In anomalous expansion, water contracts on heating
B and expands on cooling in the range 0°C to 4°C.
(A) outside (as shown) A A
correct. B Therefore water pipes sometimes burst, in cold
A B countries.
A
27. (c) On heating the system; x, r, d all increases, since the
11. (a) When the ball is heated, expansion of ball and cavity
expansion of isotropic solids is similar to true
both occurs, hence volume of cavity increases.
photographic enlargement.

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 Thermometry, Thermal Expansion and Calorimetry

L 0 .01 = 1680  10 3 J = 1680 kJ

28. (d)  = = = 2  10 − 5 / C
L0   5  100
11. (b) Melting point of ice decreases with increase in pressure
L 0 .19 (as ice expands on solidification).
29. (a)  = = = 1 .9  10 − 5 / C
L0 ( ) 100 (100 − 0) 12. (c) Conversion of ice (0°C) into steam (100°C) is as follows
Now  = 3 = 3  1.9  10–5/°C = 5.7  10–5/°C
ice
30. (d) Since, the coefficient of linear expansion of brass is 0°C (Q1 = mLi)
greater than that of steel. On cooling, the brass
contracts more, so, it get loosened. Water at 0°C
31. (b) Increase in length L = L0   (Q2 = mcW)
= 10  10  10–6  (100 – 0) = 10–2 m = 1 cm
L (1 − 0 . 9997 )
32. (a)  = = = 25 C
L0  0 . 9997  12  10 − 6 (Q3 = mLV)

33. (c) The densest layer of water will be at bottom. The Steam at 100°C Water at100°C
density of water is maximum at 4°C. So the Heat required in the given process = Q1 + Q2 + Q3
temperature of bottom of lake will be 4°C.
34. (c) Given l1 = l2 or l1 a t = l2 s t = 1  80 + 1  1  (100 − 0) + 1  536 = 716 cal

l1  s l1 s 13. (a) If m gm ice melts then

 = or = . Heat lost = Heat gain
l2  a l1 + l2  a +  s
80  1  (30 − 0) = m  80  m = 30 gm
Calorimetry 14. (c) At boiling point saturation vapour pressure becomes
equal to atmospheric pressure. Therefore, at 100°C for
1. (b) In vapor to liquid phase transition, heat liberates.
water. S.V.P. = 760 mm of Hg (atm pressure).
2. (b) Pressure inside the mines is greater than that of
15. (b) Thermal capacity = Mass  Specific heat
normal. Pressure. Also we know that boiling point
increases with increase in pressure. Due to same material both spheres will have same
specific heat. Also mass = Volume (V)  Density ()
Q
3. (c) Q = m .c.  c = ; when  = 0  c =   Ratio of thermal capacity
m .
4 3
4. (c) Mass and volume of the gas will remain same, so m1 V1 
r1  r 3 3
1
density will also remain same. = = = 3 =  1  =   = 1 : 8
m 2 V2  4 3  r2 
r2 2
5. (d)
3
6. (a) The latent heat of vaporization is always greater than
16. (a) Ice (–10°C) converts into steam as follows
latent heat of fusion because in liquid to vapour phase
change there is a large increase in volume. Hence more (ci = Specific heat of ice, cW = Specific heat of water)
heat is required as compared to solid to liquid phase
change. ice ice
–10°C (Q1 = mci) 0°C (Q2 = mLf)
7. (c) When state is not changing Q = mc.
8. (a) Heat taken by ice to melt at 0°C is Water at 0°C
Q1 = mL = 540  80 = 43200 cal
(Q3 = mcW)
Heat given by water to cool upto 0°C is
Q2 = ms  = 540  1  (80 − 0) = 43200 cal
Hence heat given by water is just sufficient to melt the (Q4 = mLV)
whole ice and final temperature of mixture is 0°C. Steam at 100°C Water at 100°C
Short trick : For these type of frequently asked Total heat required Q = Q1 + Q2 + Q3 + Q4
questions you can remember the following formula
mL  Q = 1  0 .5(10 ) + 1  80 + 1  1  (100 − 0) + 1  540
mWW − i i
cW = 725 cal
 mix = (See theory for more details)
m i + mW
Hence work done W = JQ = 4.2  725 = 3045 J
Li 80
W − 80 −
cW 1 = 0 C 17. (b) When water is cooled at 0°C to form ice then 80
If mW = m i then  mix = =
2 2 calorie/gm (latent heat) energy is released. Because
9. (d) Due to large specific heat of water, it releases large heat potential energy of the molecules decreases. Mass will
with very small temperature change. remain constant in the process of freezing of water.
10. (a) Q = m .c. = 5  (1000  4 .2)  (100 − 20 )

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 Thermometry, Thermal Expansionand and Calorimetry

18. (a) Steam at 100°C contains extra 540 calorie/gm energy 1

 mv 2 = J .[m.c. + mL ] = J [m S (475 − 25 ) + mL ]
as compare to water at 100°C. So it’s more dangerous 2
to burn with steam then water.
mv 2
19. (a) Same amount of heat is supplied to copper and water  mS (475 − 25 ) + mL =
2J
so m c c c  c = m W cW  W
1 gh
34. (a) As W = JQ  (mgh) = J  mc    =
m c ( )c 50  10 −3  420  10 2 2 Jc
  W = c c = = 5 C
m W cW 10  10 − 3  4200 9 .8  84
 = = 0 .098 C
2  4 .2  1000
 Ac A +  BcB
20. (c) Temperature of mixture  mix = cal
c A + cB (cwater = 1000 )
kg  C
32  c A + 24  c B
 28 = g
c A + cB Short trick : Remember the value of = 0 .0023 ,
JcW
cA 1 1 1
 28 c A + 28 cB = 32 c A + 24 cB  = here  =  (0 . 0023 )h =  0 .0023  84 = 0 .098 C
cB 1 2 2
21. (b) Heat lost by hot water = Heat gained by cold water in 35. (a) W = JQ  mgh = J  Q
beaker + Heat absorbed by beaker
mgh 5  9 .8  30
 Q= = = 350 cal
 440 (92 −  ) = 200  ( − 20 ) + 20  ( − 20 ) J 4.2
  = 68 C 36. (b) W = JQ = 4.18  400 = 1672 joule
22. (b) Q = m .c. ; if  = 1 K then Q = mc = Thermal 37. (c) Energy supplied = 0.93  3600 joules = 3348 joules
capacity. Heat required to melt 10 gms of ice
23. (a) Latent heat is independent of configuration. Ordered = 10  80  4.18 = 3344 joules
energy spent in stretching the spring will not
contribute to heat which is disordered kinetic energy of Hence block of ice just melts.
molecules of substance. 38. (b) Suppose person climbs upto height h, then by using
24. (d) Temperature of mixture W = JQ  mgh = JQ
m
m  c  2T +  28 
m c  + m 2 c2 2
(2c)T
3  60  9.8  h = 4 .2  10 5    h = 200 m
 mix = 1 1 1 = 2 = T  100 
m1c1 + m 2 c2 m
m.c + (2c) 2
2 39. (a) When water falls from a height, loss of potential energy
causes rise in temperature.
25. (a)
Li 40. (a) W = JQ  mg h = J (m.c. )
W − 80
100 −
cW
26. (a)  mix = = 1 = 10 C
  =
gh
= 0 .0023 h = 0 .0023  84 = 0 .196 C
2 2 Jc
27. (b) When pressure decreases, boiling point also decreases.
41. (a) Suppose m kg ice melts out of m kg then by using
28. (a) Boiling occurs when the vapour pressure of liquid
becomes equal to the atmospheric pressure. At the W = JQ  mgh = J (m L) . Hence fraction of ice melts
surface of moon, atmospheric pressure is zero, hence
boiling point decreases and water begins to boil at m  gh 9 .8  1000 1
= = = =
30°C. m JL 4 .18  80 33
29. (d) Thermal capacity = mc = 40  0.2 = 8 cal / C . W Joule
42. (b) J = =
Q Q cal
30. (b) Q = m .c.  c =
m . 43. (d) W = JQ  (2m)gh = J  m c
In temperature measurement scale  F   C so  2  5  10  10 = 4 .2(2  1000   )
(c) F  (c)C .
  = 0.1190 C = 0.12 C
31. (a) Increasing pressure lowers melting point of ice.
32. (b) Work done changes into heat energy, when the 1 1  V2
44. (b) W = JQ   mV 2  = J  mS    =
temperature of palm becomes above the atmosphere so 22  4 JS
it starts losing heat to the surroundings.
45. (c) ‘J’ is a conversion
33. (b) Firstly the temperature of bullet rises up to melting
point, then it melts. Hence according to W = JQ . 46. (c)  = 0.0023 h = 0.0023  210 = 0.483 C  0.49 C .

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 Thermometry, Thermal Expansion and Calorimetry

47. (a) According to energy conservation, change in kinetic Li

energy appears in the form of heat (thermal energy). W − 80 −
336
cW 4 .2 = 0 C
mi = mW   mix = =
1   2 2
 i.e. Thermal energy = m(v 12 − v 22 )  W = Q 
2  (Joule) (Joule)  62. (a) When the relative humidity is low (approx. 25%), the
evaporation from our body is faster. Thus we feel
1 colder.
= (100  10 − 3 )(10 2 − 5 2 ) = 3 .75 J
2 Q J
63. (a) c = →
48. (b) Work done to raise the temperature of 100 gm water m. kg  C
through 10°C is
Li 80
W − 80 −
W = JQ = 4.2  (100  10 −3  1000  10 ) = 4200 J cW
64. (a)  mix = = 1 =0
49. (b) Among all the option, latent heat of steam is highest. 2 2

50. (a)  = 0.0023 h = 0.0023  100 = 0.23 C 65. (a) Freezing point of water decreases when pressure
increases, because water expands on solidification
51. (c) Since specific heat of lead is given in Joules, hence use while “except water” for other liquid freezing point
W = Q instead of W = JQ . increases with increase in pressure.
Since the liquid in question is water. Hence, it expands
1 1  v2 (300 )2
   mv 2  = m.c.   = = = 150 C . on freezing.
2 2  4 c 4  150
66. (c) Partial pressure of water vapour PW = 0 .012  10 5 Pa ,
52. (d) At boiling point, vapour pressure becomes equal to the
external pressure. Vapour pressure of water PV = 0.016  10 5 Pa .
53. (c) When pressure increases boiling point also increases. The relative humidity at a given temperature is given
54. (b) Calorimeters are made by conducting materials. Partial pr essure of water v apour
by =
55. (b) Triple point of water is 273.16 K. Vapour pre ssure of water
56. (a) 0 .012  10 5
= = 0 .75 = 75 %
57. (d) W = JQ  W = 4.2  200 = 840 J . 0 .016  10 5
m1c11 + m 2 c 2 2 1 1 
58. (d) Temperature of mixture  = 67. (a) W = JQ   Mv 2  = J (m.c. )
m1c1 + m 2 2 22 

m 1  0.2  40 + 100  0 .5  20 
1
 1  (50 ) 2 = 4 .2[200  0 .105   ]   = 7.1C
 32 =  m 1 = 375 gm 4
m 1  0.2 + 100  0.5
cal 536  4.2 J
59. (d) Suppose m gm ice melted, then heat required for its 68. (a) 536 = = 2.25  10 6 J / kg
melting = mL = m  80 cal gm 10 −3 kg

Heat available with steam for being condensed and 69. (a) Water has maximum specific heat.
then brought to 0°C Li
W − 100 −
80
= 1  540 + 1  1  (100 − 0) = 640 cal 70. (a)  mix =
CW
= 1 = 10 C
2 2
 Heat lost = Heat taken
71. (b) Suppose m kg of ice melts then by using W = H
 640 = m  80  m = 8 gm ( Joules ) (Joules)

Short trick: You can remember that amount of steam  Mgh = mL  3 .5  10  2000 = m  3 .5  10 5
(m') at 100°C required to melt m gm ice at 0°C is  m = 0.2 kg = 200 gm
m
m' = . m iL i 100  80
8 mWW − 300  25 −
SW 1
Here, m = 8  m ' = 8  1 = 8 gm 72. (d)  mix = = = −1 .25 o C
m i + mW 100 + 300
m iL i
mWW − Which is not possible. Hence  mix = 0 o C
cW
60. (b) For water and ice mixing  mix = 73. (c) Ice (0°C) converts into steam (100°C) in following three
m i + mW
steps.
5  80
20  40 −
1 ice
= = 16 C
5 + 20 0°C (Q1 = mLi)
m iL i Water at 0°C
mWW −
cW (Q2 = mcW)
61. (c)  mix =
m i + mW

(Q3 = mLV)
Steam at 100°C

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 Thermometry, Thermal Expansionand and Calorimetry

 153 × 10–6 + (vessel)glass = (144 × 10–6 + vessel)steel

Water at 100°C Further, (vessel)steel = 3 = 3 × (12 × 10–6) = 36 × 10–6/ºC

 153 × 10–6 + (vessel)glass = 144 × 10–6 + 36 × 10–6
Total heat required Q = Q1 + Q2 + Q3
 (vessel)glass = 3 = 27 × 10–6/ºC   = 9 × 10–6/ºC
= 5  80 + 5  1  (100 − 0) + 5  540 = 3600 cal
3. (d) The expansion of solids can be well understood by
74. (a) Let L be the latent heat and using principle of potential energy curve for two adjacent atoms in a
calorimetry. crystalline solid as a function of their internuclear
2L + 2 (100 – 54.3) = 40  (54.3 – 25.3) separation (r).
 L = 540.3 cal/gm. U
m iL i
m W W − 100  50 − 10 
80
cW 1  38 . 2 C
75. (d)  mix = = r
m i + mW 10 + 100
P3
76. (b) Let the final temperature be T °C. E P2 F T3
Total heat supplied by the three liquids in coming down C D T2
P1
to 0°C = m 1 c 1 T1 + m 2 c 2 T 2 + m 3 c 3 T3 ..... (i) A BT
r0 1

r1 T3> T2 > T1
Total heat used by three liquids in raising temperature r2
from 0oC to ToC
At ordinary temperature : Each molecule of the solid
= m 1 c1 T + m 2 c 2 T + m 3 c 3 T .....(ii) vibrate about it' s equilibrium position P1 between A
By equating (i) and (ii) we get and B (r0 is the equilibrium distance of it from some
other molecule)
(m 1 c 1 + m 2 c 2 + m 3 c 3 ) T
At high temperature : Amplitude of vibration increase
= m 1 c 1 T1 + m 2 c 2 T 2 + m 3 c 3 T3 (C  D and E  F). Due to asymmetry of the curve, the
equilibrium positions (P2 and P3) of molecule displaced.
m 1 c 1 T1 + m 2 c 2 T2 + m 3 c 3 T3
 T = . Hence it's distance from other molecule increases (r2 >
m 1 c1 + m 2 c 2 + m 3 c 3 r1 > r0).
77. (c) At triple point all the phases co-exist Thus, on raising the temperature, the average
Q equilibrium distance between the molecules increases
78. (b) c = ; as  = 0, hence c becomes . and the solid as a whole expands.
m .
4. (c) Initial diameter of tyre = (1000 – 6) mm = 994 mm, so
79. (d) Let final temperature of water be  994
initial radius of tyre R = = 497 mm
Heat taken = Heat given 2
110 1 ( – 10) + 10 ( – 10) = 220  1 (70 – ) and change in diameter D = 6 mm so
  = 48.8°C  50°C. 6
R = = 3 mm
80. (c) We know that thermal capacity of a body expressed in 2
calories is equal to water equivalent of the body After increasing temperature by  tyre will fit onto
expressed in grams. wheel
Increment in the length (circumference) of the iron tyre
t
m s (2 t) + 1 . 5 (m s)   
m c  + m 2 c2 2 3 =t L = L     = L    [As  =
81. (b)  mix = 1 1 1 = 3 3
]
m1c1 + m 2 c 2 m s + 1 . 5 (m s)
  3 R 33
82. (d) We know that when solid carbondioxide is heated, it 2 R = 2 R      = =
becomes vapour directly without passing through its 3  R 3 . 6  10 − 5  497
liquid phase. Therefore it is called dry ice.   = 500 o C
5. (b) Due to volume expansion of both liquid and vessel, the
Critical Thinking Questions change in volume of liquid relative to container is given
by V = V0 [ L −  g ]
1. (c) Due to volume expansion of both mercury and flask, the
change in volume of mercury relative to flask is given Given V0 = 1000 cc, g = 0.1×10–4/°C
by V = V0 [ L −  g ] = V [ m − 3 g ]
  g = 3 g = 3  0.1  10 −4 / C = 0.3  10 −4 / C
−6 −6
= 50 [180  10 − 3  9  10 ] (38 − 18 ) = 0.153 cc
 V = 1000 [1.82 × 10–4 – 0.3 × 10–4] × 100 = 15.2 cc
2. (a) real = app. + vessel
6. (a) With temperature rise (same 25°C for both), steel scale
So (app. + vessel)glass = (app. + vessel)steel and copper wire both expand. Hence length of copper

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 Thermometry, Thermal Expansion and Calorimetry

wire w.r.t. steel scale or apparent length of copper wire 12. (b) Substances are classified into two categories
after rise in temperature (i) water like substances which expand on
Lapp = L'cu − L' steel = [L0 (1 +  Cu  ) − L0 (1 +  s  ) solidification.
(ii) CO2 like (Wax, Ghee etc.) which contract on
 Lapp = L0 ( Cu −  s )
solidification.
= 80 (17  10 −6 − 11  10 −6 )  20 = 80.0096 cm Their behaviour regarding solidification is opposite.
7. (b, d) Let L0 be the initial length of each strip before heating. Melting point of ice decreases with rise of temp but that
of wax etc increases with increase in temperature.
Length after heating will be
Similarly ice starts forming from top downwards
L B = L0 (1 +  B T ) = (R + d ) whereas wax starts its formation from bottom.
L C = L 0 (1 +  C T ) = R d mL
13. (b) Heat lost in t sec = mL or heat lost per sec = . This
t
R + d 1 +  B T must be the heat supplied for keeping the substance in
 = R 
R 1 +  C T molten state per sec.
d mL Pt
 1+ = 1 + ( B −  C )T  = P or L =
R t m

d 1 1 14. (a) Heat is lost by steam in two stages (i) for change of
 R=  R and R  state from steam at 100ºC to water at 100ºC is m × 540
( B −  C )T T ( B −  C )
(ii) to change water at 100ºC to water at 80ºC is
8. (d) Thermostat is used in electric apparatus like m × 1 × (100 – 80), where m is the mass of steam
refrigerator, Iron etc for automatic cut off. Therefore condensed.
for metallic strips to bend on heating their coefficient of
linear expansion should be different. Total heat lost by steam is m × 540 + m × 20 = 560 m
(cals) Heat gained by calorimeter and its contents is
9. (c) As the coefficient of cubical expansion of metal is less
as compared to the coefficient of cubical expansion of = (1.1 + 0.02) × (80 – 15) = 1.12 × 65 cals.
liquid, we may neglect the expansion of metal ball. So using Principle of calorimetery, Heat gained = heat lost
when the ball is immersed in alcohol at 0ºC, it displaces
some volume V of alcohol at 0ºC and has weight W1.  560 m = 1.12 × 65, m = 0.130 gm

 W1 = W0 – V0g 15. (b) Initially ice will absorb heat to raise it's temperature to
0oC then it's melting takes place
where W0 = weight of ball in air
If mi = Initial mass of ice, mi' = Mass of ice that melts
Similarly, W2 = W0 – V50g
and mW = Initial mass of water
where 0 = density of alcohol at 0ºC
By Law of mixture Heat gained by ice = Heat lost by
and 50 = density of alcohol at 50ºC water  mi  c  (20 ) + mi '  L = mW cW [20 ]
As 50 < 0,  W2 > W1 or W1 < W2
 2  0.5(20 ) + mi '80 = 5  1  20  m i ' = 1kg
10. (d) V = V0(1 +  )  Change in volume
V − V0 = V = A.l = V0 So final mass of water = Initial mass of water + Mass of
ice that melts = 5 + 1= 6 kg.
V0 . 10 −6  18  10 −5  (100 − 0) 16. (a) Heat gained by the water = (Heat supplied by the
 l = =
A 0.004  10 −4 coil) – (Heat dissipated to environment)
= 45 × 10–3 m = 4.5 cm  mc  = PCoil t − PLoss t
11. (b) Loss of weight at 27ºC is
 2  4.2  10 3  (77 − 27 ) = 1000 t − 160 t
= 46 – 30 = 16 = V1 × 1.24 l × g …(i)
Loss of weight at 42ºC is 4 .2  10 5
 t= = 500 sec = 8 min 20 sec
840
= 46 – 30.5 = 15.5 = V2 × 1.2  l × g …(ii)
17. (a) If mass of the bullet is m gm,
16 V 1 . 24
Now dividing (i) by (ii), we get = 1  then total heat required for bullet to just melt down
15 . 5 V2 1 .2
Q1 = m c  + m L = m  0.03 (327 – 27) + m  6
V2 15 . 5  1 . 24
But = 1 + 3 (t2 – t1) = = 1.001042 = 15 m cal = (15 m  4 .2)J
V1 16  1 . 2

 3 (42º – 27º) = 0.001042   = 2.316 × 10–5/ºC.

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 Thermometry, Thermal Expansionand and Calorimetry

Now when bullet is stopped by the obstacle, the loss in With temperature rise height of liquid in each arm
1 increases i.e. l1 > l and l2 > l
its mechanical energy = (m  10 − 3 )v 2 J
2 l1 l2
Also l = =
−3 1 +  t1 1 +  t 2
(As m gm = m  10 kg )
l1 − l2
As 25% of this energy is absorbed by the obstacle,  l1 +  l1t2 = l2 +  l2 t1   =
l2 t1 − l1 t 2
The energy absorbed by the bullet
22. (c) V = V0 (1 +  )
75 1 3
Q2 =  mv 2  10 −3 = mv 2  10 −3 J
100 2 8 L3 = L0 (1 + 1  )L20 (1 +  2  )2 = L30 (1 + 1  )(1 +  2  )2
Now the bullet will melt if Q 2  Q1
Since L30 = V0 and L3 = V
3
i.e. mv 2  10 −3  15 m  4 .2  v min = 410 m / s
8 Hence 1 +  = (1 + 1 ) (1 +  2  )2
1 ~= (1 +   ) (1 + 2  ) ~= (1 +   + 2  )
18. (c) Energy = mv 2 = mc ;    v 2 1 2 1 2
2
  = 1 + 22
Temperature does not depend upon the mass of the
2
balls. l
23. (d) (OR )2 = (PR)2 − (PO)2 = l 2 −  
19. (c) Heat gain = heat lost 2
C 3
CA(16 –12) = CB (19 – 16)  A =
2
l 
CB 4 = [l(1 +  2 t)]2 −  (1 + 1 t)
2 
CB 5
and CB(23 – 19) = CC (28 – 23)  = l2 l2
CC 4 l2 − = l 2 (1 +  22 t 2 + 2 2 t) − (1 + 12 t 2 + 21 t)
4 4
C A 15
 = ...(i) Neglecting  22 t 2 and 12 t 2
C C 16
l2 21
If  is the temperature when A and C are mixed then, 0 = l 2 (2 2 t) − (21 t)  2 2 = ; 1 = 4 2
4 4
CA 28 − 
C A ( − 12) = C C (28 −  )  = ...(ii) 24. (c) It is given that the volume of air in the flask remains the
CC  − 12
same. This means that the expansion in volume of the
On solving equation (i) and (ii)  = 20.2ºC. vessel is exactly equal to the volume expansion of
mercury.
20. (b) Suppose m kg steam required per hour
Heat released by steam in following three steps i.e., Vg = VL or Vg g  = VL L 

(i) When 150°C steam ⎯⎯→

⎯ 100 C steam Vg g 1000  (3  9  10 −6 )
 VL = = = 150 cc
Q 1
L 1 . 8  10 − 4
Q1 = mcSteam  = m  1 (150 – 100) = 50 m cal
25. (c) Heat given by water Q1 = 10  10 = 100 cal.
(ii) When 150°C steam ⎯⎯
⎯→ 100 C water
Q 2
Heat taken by ice to melt
Q2 = mLV = m  540 = 540 m cal
Q2 = 10  0.5  [0 – (– 20)] + 10  80 = 900 cal
(iii) When 100°C water ⎯⎯
⎯→ 90 C water
Q As Q1  Q2 , so ice will not completely melt and final
2

Q3 = mcW  = m  1  (100 – 90) = 10 m cal temperature = 0°C.

As heat given by water in cooling up to 0°C is only just
Hence total heat given by the steam Q = Q1 +Q2 + Q3 =
sufficient to increase the temperature of ice from –
600 mcal ... (i) 20°C to 0°C, hence mixture in equilibrium will consist
Heat taken by 10 kg water of 10 gm ice and 10 gm water at 0°C.

Q' = mc W  = 10  10 3  1  (80 − 20 ) = 600  10 3 cal 26. (b) L = L0 

Hence Q = Q  600 m = 600  103 75

Rod A : 0.075 = 20  A  100   A =  10 − 6 / C
2
 m = 103 gm = 1kg.
45
21. (a) Suppose, height of liquid in each arm before rising the rod B : 0.045 = 20  B  100   B =  10 − 6 / C
2
temperature is l.
For composite rod : x cm of A and (20 – x) cm of B we
t1 have x (20 – x)
t2
l1
l l A A B B
l2
20cm

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 Thermometry, Thermal Expansion and Calorimetry

0.060 = x A  100 + (20 – x) B  100 L

 2m C(90 − 80 ) = m L  = 20
 75 45  C
= x   10 −6  100 + (20 − x )   10 −6  100 
 2 2  6. (b) In the given graph CD represents liquid state.
On solving we get x = 10 cm. 7. (a) Density of water is maximum at 4°C and is less on
27. (a) Let m gm of steam get condensed into water (By heat either side of this temperature.
loss). This happens in following two steps. C F − 32 9
8. (a) We know that, = or F = C + 32
100 180 5
100°C 100°C
Equation of straight Y
Steam (H1 = m  540) Water
line is, y = mx + c
F
Hence, m = (9 / 5) ,
[H2 = m  1  (100 – 90)]
positive and c = 32
90°C
positive. The graph is
X
shown in figure. O C
Water
C F − 32 5 20
Heat gained by water (20°C) to raise it’s temperature 9. (a) = C = F− . Hence graph between
5 9  9.  3
upto 90° = 22  1  (90 − 20 )
°C and °F will be a straight line with positive slope and
Hence, in equilibrium heat lost = Heat gain negative intercept.
10. (bc) The horizontal parts of the curve, where the system
 m  540 + m  1  (100 − 90 ) = 22  1  (90 − 20 )
absorbs heat at constant temperature must depict
 m = 2 .8 gm changes of state. Here the latent heats are proportional
to lengths of the horizontal parts. In the sloping parts,
The net mass of the water present in the mixture
specific heat capacity is inversely proportional to the
= 22 + 2.8 = 24 .8 gm. slopes.
11. (c) Since specific heat = 0.6 kcal/gm  °C = 0.6 cal/gm °C
Graphical Questions From graph it is clear that in a minute, the temperature
is raised from 0°C to 50°C.
1. (b) Relation between Celsius and Fahrenheit scale of
C F − 32 5 160  Heat required for a minute = 50  0.6  50 = 1500 cal.
temperature is = C= F−
5 9 9 9 Also from graph, Boiling point of wax is 200°C.
Equating above equation with standard equation of line
12. (c)
5
y = mx + c we get slope of the line AB is m = 13. (c) Substances having more specific heat take longer time
9
to get heated to a higher temperature and longer time
2. (c) Since in the region AB temperature is constant to get cooled.
therefore at this temperature phase of the material
changes from solid to liquid and (H2 – H1) heat will be T A
absorb by the material. This heat is known as the heat B
of melting of the solid.
C
Similarly in the region CD temperature is constant
therefore at this temperature phase of the material
changes from liquid to gas and (H4 – H3) heat will be t
tA tB tC
absorb by the material. This heat as known as the heat
If we draw a line parallel to the time axis then it cuts
of vaporisation of the liquid.
the given graphs at three different points.
3. (a) Initially, on heating temperature rises from –10°C to Corresponding points on the times axis shows that
0°C. Then ice melts and temperature does not rise.
tC  tB  t A  CC  CB  C A
After the whole ice has melted, temperature begins to
rise until it reaches 100°C. Then it becomes constant, as 14. (c) From given curve,
at the boiling point will not rise. Melting point for A = 60 C
4. (a) The volume of matter in portion AB of the curve is and melting point for B = 20 C
almost constant and pressure is decreasing. These are Time taken by A for fusion = (6 − 2) = 4 minute
the characteristics of liquid state. Time taken by B for fusion = (6 .5 − 4 ) = 2 .5 minute
5. (d) Let the quantity of heat supplied per minute be Q. Then HA 6  4  60 8
quantity of heat supplied in 2 min = mC (90 − 80 ) Then = = .
H B 6  2 .5  60 5
In 4 min, heat supplied = 2m C(90 − 80 )

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 Thermometry, Thermal Expansionand and Calorimetry

Assertion and Reason temperature (T1 + T2 ) / 2 . They will do so, in case the

1. (a) With rise in pressure melting point of ice decreases. two bodies were of same mass and material i.e., same
Also ice contracts on melting thermal capacities. In other words, the two bodies may
be having different thermal capacities, that’s why they
2. (c) Celsius scale was the first temperature scale and
Fahrenheit is the smallest unit measuring temperature. do not settle to the mean temperature, when brought
together.
3. (e) Melting is associated with increasing of internal energy
without change in temperature. In view of the reason 12. (d) Specific heat of a body is the amount of heat required to
being correct the amount of heat absorbed or given out raise the temperature of unit mass of the body through
during change of state is expressed Q = mL , where m unit degree. When mass of a body is less than unity,
is the mass of the substance and L is the latent heat of then its thermal capacity is less than its specific heat
the substance.
and vice-versa.
4. (a) The temperature of land rises rapidly as compared to
sea because of specific heat of land is five times less 13. (a) Water would evaporate quickly because there is no
than that of sea water. Thus, the air above the land atmosphere on moon, due to which surface
become hot and light so rises up so because of pressure temperature of moon is much higher than earth
drops over land. To compensate the drop of pressure, (Maximum surface temperature of moon is 123ºC).
the cooler air starts from sea starts blowing towards 14. (d) The potential energy of water molecules is more. The
lands, so, setting up sea breeze. During night land as
heat given to melt the ice at 0ºC is used up in increasing
well sea radiate heat energy. The temperature of land
the potential energy of water molecules formed at 0ºC.
falls more rapidly as compared to sea water, as sea
water consists of higher specific heat capacity. The air
above sea water being warm and light rises up and to
take its place the cold air from land starts blowing
towards sea and so et up breeze.

5. (a) Linear expansion for brass (19  10 −4 )  linear expansion

for steel (11  10 −4 ) . On cooling the disk shrinks to a
greater extent than the hole and hence it will get loose.
V
6. (a) As,  = i.e., units of coefficient of volume
VT
expansion is K–1.
C F − 32
7. (c) The relation between F and C scale is, = . If F
5 9
= C  C = – 40°C i.e., at – 40° the Centigrade and
Fahrenheit thermometers reads the same.
8. (a) As  = 2 and  = 3 , i.e., coefficient of volume
expansion of solid is three time coefficient of linear
expansion and 1.5 times the coefficient of superficial
expansion, on heating a solid iron ball, percentage
increase in its volume is largest.
9. (a) Water has maximum density at 4°C. On heating above
4°C or cooling below 4°C, density of water decreases
and its volume increases. Therefore, water overflows in
both the cases.
10. (b) The Latent heat of fusion of ice is amount of heat
required to convert unit mass of ice at 0°C into water at
0°C. For fusion of ice
L = 80 cal / gm = 80000 cal / gm = 8000  4.2 J / kg

= 336000 J / kg .

11. (a) When two bodies at temperature T1 and T2 are

brought in thermal contact, they do settle to the mean

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