success magnet

UNIT

Heat

2

Section A : Straight Objective Type 1.

Answer (4) Work done by a paddle wheel on a quantity of liquid is non-zero. While dV = 0

2.

Answer (3) Average momentum is independent of temperature. It is equal to zero)

3.

Answer (3) It is assumed that collision is elastic. Therefore Δp = 2 pcosθ Δp = 2pcosθ ≤ 2p

 

∴ Δp ≠ 3p 4.

Answer (3)

K trans 3nRT 3   2nRT 2 . K rot 5.

Answer (1) 1

PV 3  constant 1

nRT .V 3 = C V

TV

2

3

=C 2

T ∝ V3

T 3 ∝ V 2 . As volume is increasing, therefore temperature is increasing. 6.

Answer (3) P = m.V or PV–1 = m. Now, C = Cv 

7.

By comparison with PV α = constant, α = –1 R R 3R R  CV     2R .  1  1 1 2 2

Answer (3) 2 Vrms

169v 2 V12  V2 2  V3 2 (3 2  4 2  12 2 )V 2  =  3 3 3

 Vrms 

13V 3

= 7.51 V .

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8.

Heat

Answer (2) Pressure increases in a direction opposite to that of acceleration.

9.

Answer (1) P 2V

1 3

 PV

=C 1 6

1 6

=C ⇒α=

C = Cv –

= Cv –

R  1 R R = Cv – 1 1 5 6 6

C > Cv 10. Answer (1) ΔQ = ΔU + W for isobaric process nCpΔT = nCvΔT + nRΔT Cp = Cv + R Cp – Cv = R. 10a.

Answer (2, 4)

(IIT-JEE 2009)

(CP  CV )Diatomic 

7R 5R   6R 2 2

(CP  CV ) Monoatomic 

(CP .CV ) Diatomic 

35R 2 4

(CP .CV ) Monoatomic  Cp

5R 3R   4R 2 2

15R 2 4

2 is smaller for diatomic gas f Cp – Cv = R = constant Cv

= γ = 1

11. Answer (1) Pressure ∝ momentum transferred to the walls. If T is increased, v is increased, hence Δp will increase. 12. Answer (3) U = nCVT = n

fR T 2

=

nRT .( f ) 2

=

PV (f ) . 2

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Heat

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13. Answer (1)

VH2 3RT

=

VO2 3R  320

–3

=

3RT 2

=

3R × 10

T

=

20 K

2  10

32  10 – 3

14. Answer (3) At a given temperature, energy of A is more than that of B. So degrees of freedom for A should be more than B. 15. Answer (2)

P x

P

100 100–x P = Patm + pressure due to Hg = (76 + x) V = (100 – x) A

Area of cylinder = A PV = 76 × 100 × A ⇒ PV = constant for isothermal process 76 × 100 × A = (76 + x) (100 – x) A 7600 = 7600 + 100 x – 76x – x2 x = 24, x = 0 x = 0, 24. 16. Answer (3)

N (no. of molecules)

Nmax

Vm.p

V

At most probable speed, number of molecules is maximum 17. Answer (4) Vmp 

1 M

18. Answer (3)

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

P.M RT

Heat

ρtop =

70  M R  280

ρbottom = 

76  M R  300

 top bottom

 0.99

19. Answer (2) The gas in the tube should have an acceleration towards centre. Therefore it should experience a net force inwards. So P2 > P1. 20. Answer (2) As P 

E=

3E 2V

3PV 3nRT  . ∴ E must be translational kinetic energy 2 2

21. Answer (4) Collision is elastic Hence total kinetic energy and momentum remains conserved. 22. Answer (1) Let the water rises upto a height of H. PA = PB

...(i)

P0L0 = PA (L0 – H)

...(ii)

PB = P0 + ρg (L0 – H)

...(iii)

From equation (ii) PA 

L0

A B

(L0–H) H

P0L0 (L0 – H )

P0 L0  P0  g (L0 – H ) (L0 – H )  L0  P0  – 1  g (L0 – H ) ( L – H )  0 

P0H = ρg (L0 – H)2 ∴ 105 × H = 103 × 10 (20 × 10–2 – H)2 ⇒ H = 0.38 cm 23. Answer (3) B 

P . When P is constant, ΔP = 0.  – V     V 

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24. Answer (3) Q1 = n CP ΔT = 1×

5 5 RT R (2T – T) = 2 2

V1 V2   T2  2T T1 T2

 P1  Q2 = W = nR (2T) n  P   2



P1 2 P2

= 2 RT n 2 QTotal = (2.5 + 2 n 2) RT 25. Answer (1) P1 P  2 ( volume is constant) ⇒ T1 T2

Apply

T2 

P2T1 P1

26. Answer (1) dQ = dU + dW CΔT = Cv ΔT + W. T  T2 – T1  W = Area  C 

1 (P0  2P0 )  2V0 2

P2V2 PV 6P0V0 PV – 0 0 – 1 1  R R R R



5P0V0 R

 3P0V0

5P0V0 5P V 3R  CV  0 0  3P0V0  C  CV  5 R R 21R  C  3 R  V 2  10 

⇒ C 

27. Answer (2) dQ = 3dU  C  3 CV 

9 R 2

⇒ C = 4.5 R 28. Answer (1) ΔQ = ΔU + W. V2

W 

Q 

V 2 V 2  RT0 PdV  C  2 – 1   ( P1V1 = RT0 to P2V2 = 2RT0 ) 2 2 2   V1



RT0 3 RT0  2 2

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Heat

29. Answer (3) PV = RT P

R (T0  V 2 ) V

RT dP  0   20  R  0 dV V

⇒

V 

T0 

2

 T  T  T0  V  T0    0   2T0      2

Minimum value of pressure is Pmin 

RT R(2T0 )   2R T0  V T0 

30. Answer (4) As PV = nRT ⇒ U = 2 + 2nRT ⇒ ΔU = 2nRΔT ⇒ CV = 2R Since CV = 2R for system, 3 5 R  CV  R 2 2

∴ So it is mixture of monoatomic and diatomic gas 31. Answer (2) As P   P1 

2V 2 1 V 2 8 2  1 and P2  5 2

P2V2 – P1V1 = nRΔT  T 

11 K 5R

32. Answer (2) PV γ = constant

 

m V

Pρ–γγ = constant

V 

m 

33. Answer (1) ΔW = – Area of cycle =

J

1  100  103  200  10 6  10 2

W 10   4.17 Q 2.4

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34. Answer (3) ΔQ = ΔW

(cyclic process)

ΔQ = 3.14 × 100 × 10 × 100 × 10–6 3

ΔQ = 31.4 J 35. Answer (2) W = Area of rectangle = Area of triangle 1 (2V0 × 2P0) 2

W = (2V0 × P0) + = 4 P0 V0 36. Answer (4)

As initial and final temperatures are same, ΔU = 0 37. Answer (3) P 1 – γ T γ = constant 300 P

3

2

T



1 2

3

2

(27P )

1 2

⇒ T = 900 K 38. Answer (3) Using v 

RT , m

we get  

mv 2 32  10 –3  315  315   1 .4 RT 8.31  273

39. Answer (4) λ = 6.6 cm As v = f λ ⇒ v = 5 × 103 × 6.6 × 10–2 m/s

R RT and Cv   –1 M

Aslo, v  40. Answer (1) CP = CV + R 41. Answer (3)

For a process PV n = k,

C  CV  As Cv  

R 1– n

3R 2

2R 

R 1– n

 n 

1 2

1

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Heat

42. Answer (1) C  CV 

PdV dT

PV = nRT 2e 2VV = nRT 

dV nR  dT 2e 2V (1  2V )



PdV nR  dT 1 2V

 C  CV 

nR (1  2V ) n

⇒ C  CV 

R 1 2V

43. Answer (1) T = 300 + 2V

PV = 300 + 2V R P 

300R  2R V V2  4

Work done W 



V1 

 300R   2R  dV  V   2

⇒ W = 300 Rn2 + 4R 44. Answer (3)

P  200 

1 V

Calculate P1V1 and P2V2 where V1 = 2m3 and then nR Δ T = P2V2 – P1V1 = 400  nRT So ΔU = 3 × 400  using  U   –1 

  

= 1200 J 45. Answer (3)

C

R R    1 1 n

Pv   k (a constant)

C

R R    1 1 

O

R R    1 1 

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46. Answer (2)

U  nCV T 

U = V2

nR  T  –1

⇒ 3RT = V 2 3RdT = 2VdV Now, C  CV  P

C  3R  C  3R   3R 

dV dT

RT 3R V 2V 3R 2T

 C  3R 

2V 2

3R 2T 2  3RT

R 2

C = 3.5 R 47. Answer (4) ΔQ = ΔU + W

W = 2J nRΔT = 2J

ΔU = nCV ΔT =

5 (nRT ) 2

=5J ∴ ΔQ = 7 J 48. Answer (1)

P 2P0

B

A

P0

V0

W AB  U 

C 2V0

V

1 3P0V0 (3P0 )V0  2 2

(4P0V0  P0V0 )  3P0V0  1

9 3  Q AB    3 P0V0  P0V0 2 2 

Wnet 



1 P0V0 2

Wnet 1  Q AB 9

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Heat

49. Answer (2)

V 

1 T

V 

Constant PV ⇒n=2

PV2 = Constant

 C  CV 

R 1– n

⇒ C = CV – R

C = 2R

ΔQ = CΔT = 200 R 50. Answer (1) Q1 : Q2 : Q3

T

900

900 : T : 400

Q1

Q2 – Q1 = Q3 – Q2

400 K Q3

Q2 W

W

T – 900 = 400 – T 2T = 1300 T = 650 K 51. Answer (3) ΔW = 25 = PΔV = RΔT   1

2 2 1 4  1  1  f 6 3 3

CV 

R  3R  –1

ΔU = CV ΔT = 3R ΔT

...(i)

...(ii)

or ΔU = 3(RΔT) = 3 × 25 = 75 ΔQ = ΔU + ΔW = 75 + 25 = 100 J 52. Answer (4) Since dW = 0

dV = 0

and dQ = dU > 0 ∴ ΔT > 0 53. Answer (4) PV n = constant PV = constant is an Isothermal process PV ∞ = constant V 

C 1

P

⇒ V = Constant is an Isochoric process

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54. Answer (2) T2 T1

 1 

T2 since T is same so η will be same. 1

55. Answer (4)

   1    100    T1  – 1   T2   56. Answer (2) PT = Constant P(PV) = K P 2V = K

PV

1 2

 K R

C  CV 

1–

1 2

37.35 = CV + 2R

fR = 37.35 – 2R 2 f=5 57. Answer (1)

W = nRT0 ln

Vf Vi

 1 W = 3R × 300 ln    2

W = – 5188 J 58. Answer (3) Molar heat capacity depend on molecular structure 59. Answer (1) dQ = ΔU + W W = – ΔU = – (U2 – U1) = U1 – U2 60. Answer (1) Isochoric process

∴P∝T

P T  P T

⇒

So,

1 1  100 T

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Heat

61. Answer (3) θH – θR = k (θR – θ0) θH – 20 = k (20 + 20) θH – 10 = k (10 + 40) or

H  20 4  H  10 5

or 5θH – 100 = 4θH – 40 θH = 60° 62. Answer (1) dT dx

i

= kA

So

dT 1  dx k

or K1 > K2 63. Answer (1)  ( Tsun ) 4  4R sun 2  2   Re Energy received by earth per second is  4d 2  

Energy radiated by earth per second is [Tearth 4  4Re2 ] . Equating the two values, we get Tearth = 290 K 64. Answer (2) P

= σ εT 4A

(A = πdl)

8 4 100 = 5.67  10  0.7 T 

or T 4 =

22  8  10 5  0.6 7

100  7  1013 5.67  0.7  22  8  0.6

⇒ T = 2018 K

65. Answer (3) dQ – dW = dU which is an exact differential, since U does not depend on path 66. Answer (4) PVn = constant ⇒ P1/n V = constant For

n= ∞ V = constant

67. Answer (3) W = 0 and Q = 0 So

ΔV = 0

or

V = constant

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68. Answer (1) E1 = ∈1σT14(4πr12) E2 = ∈2σT24(4πr22) E1 1  T1     So E2 2  T2 

4

 r1     r2 

2

69. Answer (4) P = σT 4 (πr 2) 450 = σT 4 πr 2 4 r  P =  (2T )    2

2

450 1  ⇒ P = 1800 W P 4

69(a). Answer (9)

IIT-JEE 2010

As per Wein’s displacement law TA B  3 TB  A

As per Stefan’s law Also,

PA AATA4 (6)2    (3)4  9 4 2 PB ABTB (18)

70. Answer (3) A 

L 0.05   2  105 º C1 LT 25  100

L 0.04   105 º C1 LT 40  100 1 + 2 = 50 cm ... (i) B 

1    2   50.03 cm

αA1ΔT + αB2ΔT = 0.03 2 × 10–5 × 1 × 50 + 10–5 × 2 × 50 = 0.03 cm 21 + 2 = 60 cm

... (ii)

From (i) and (ii) 1 = 10 cm 71. Answer (2) k1x1 = k2x2 and x1 + x2 = L α ΔT x1 = 3x2 x1 =

( x1  x 2 ) k 2 ( k1  k 2 )

=

L  T 3K (K  3K )

=

3L  T 4

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Heat

72. Answer (4) ρ1 = ρ2 [1 + γ (T2 – T1)] v2 = v1[1 + β(T2 – T1)] The loss in weight at T1 = v1ρ1g ⇒ w0 – w1 = v1ρ1g The loss in weight at T2 = v2ρ2g ⇒ w0 – w2 = v2ρ2g w 0  w1 1   (T2  T1 )  w 0  w 2 1   (T2  T1 )

73. Answer (1) 50g + 2g = Vρ1g 50g + xg = Vρ2g where ρ1 is density of water at 10°C and ρ2 is density of water at 30°C ⇒

1 52 1 = = 50  x 2 (1  T )

⇒

52 – 52 × 10–4 × 20 = 50 + x

⇒

(2 – x) = 52 × 10–4 × 20 x = 0.104 kg

74. Answer (1) 30 = (50 – 20) α mc

d = (40 – 20) α dt

mc

d 20  30 = dt 30

⇒

[where α is constant)

mc 

0 .1  20 10

Heat capacity = mc = 2000 J/ºC 75. Answer (4) For gas A temperature is maximum and A is polyatomic gas. [Slope is proportional to the γ]

P A B V1

V2

76. Answer (2) 4Q = nCPΔT 3Q = nCVΔT ⇒ ⇒

4 3 f=6 

C

77. Answer (1)

kA(T 3  TC ) kA(TC  T 3)    TC 1  3  ⇒ 2 T

A

(T3)

B

(T)

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78. Answer (2) Work done for isobaric process is nRΔT R (T0 – T1) + R (T0 – T2) = W T0 =

W  R (T1  T2 ) 2R

78(a). Answer (1, 2)

IIT-JEE 2010

V

AB is an isothermal process ⇒ Internal energy at A = Internal energy at B. Applying

PV = constant between A and B T

V0

P0V0 PB 4V0 P   PB  0 T0 T0 4

B

4V0

C

A T0

T

As it is not clearly indicated that BC is passing through origin, prediction about point C is not possible. 79. Answer (2) P = σT 4A so 4T14 r12  4T24 r22 ⇒

r12 T24  r22 T14

⇒

r1  T2    r2  T1 

2

79(a). Answer (9)

IIT-JEE 2010

As per Wein’s displacement law TA B  3 TB  A

As per Stefan’s law Also,

PA AATA4 (6)2    (3)4  9 4 2 PB ABTB (18)

80. Answer (2)  2  1    1   k 2  0  t 2  

θ1 = 50

θ2 = 40

θ0 = 20

t=5

10  k ( 45  20 ) 5 40    40     k  20  5 2   Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623456 Fax : 47623472

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Heat

10 25  2  40    10θ = 2000 – 50θ 60θ = 2000



100 C 3

81. Answer (2) Under steady state constant thermal current flows. 82. Answer (3) Rate of cooling is proportional to

1 A or V r

V1 = 2V2

4 3 4 r1  2 r23 3 3 1 r1  2 3 r2

Rate of cooling of P r2  1     Rate of cooling of Q r1  2 

1 3

1



2

1 3

83. Answer (4) Solar constant ∝

1

r2 Where r is distance from sun

r

=

D 

D  diameter of sun

D

 Planet

Sun

So Solar constant ∝ θ2 84. Answer (1)  T = 2 xiˆ  2yjˆ  2zkˆ  (T )(1,1, 2 ) = 2iˆ  2 ˆj 85. Answer (3) Req = R1 + 2R2 = TCu 

10 2(0.01) 30   A A(0.001K ) AK 10 100 100    33.4 3 AK Req

86. Answer (1) P 2T = constant P 3V = C x

1 3

C

5 R 2

R 1

1 3

C = 4R Q = 4P0V0

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Success Magnet (Solutions)

87. Answer (2)

T Heat current per unit area = k x 88. Answer (1)

k  constant (Weidmann-Franz law) T 89. Answer (1) Black spot is good absorber of heat, so according to Kirchhoff’s law this will be good emitter also. 90. Answer (4) Fraunhoffer lines are absorption lines which are characteristic of the elements in the sun atmosphere. They are also observed in the emission spectra of the elements. This is based on Kirchhoff’s law. 91. Answer (3) En = nhv [energy of oscillator] 92. Answer (2) In laboratory, rate of cooling of liquids is observed to measure their heat capacity. 93. Answer (2) O

R

TR  TP  TQ

N A

and TO  TN  TS

B

Q P

S

So A

B (P,Q,R)

⇒

R/3

⇒ Req 

R/6

(O,N, S)

R/3

R R R 5R    3 6 3 6

94. Answer (3) Air is bad conductor of heat and non conducting medium between two surfaces reduces transfer of heat. 95. Answer (1) When temperature of two bodies are same then there will be no transfer of heat. 96. Answer (3) x1 + x2 + x3 = LαΔT +

L 3 T  LT 2 2

Kx1

2 Kx2

2 Kx2

3 Kx3

x1 = 2x2 = 3x3 ∴ x1 + ⇒

x1 x3 3   L T 2 3 2

9 LT 1 1 81 2 2 2 81 E1  KL  T  KL2  2 T 2 2 121 242 x1 

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