Title 14. Transmission of Heat Easy Ans.pdf ✅

1. (a) Cu is better conductor than Al and Ag is better conductor than Cu . Hence conductivity in increasing order is Al  Cu  Ag . 2. (d) l KA t Q  =  l r l A t Q 2    l r 2 is maximum in option (d), hence it will conduct more heat. 3. (d) Q t = KAΔθ l  Q t ∝ A l ∝ d 2 l (d = Diameter of rod)  (Q/t)1 (Q/t)2 = ( d1 d2 ) 2 × l2 l1 = ( 1 2 ) 2 × ( 1 2 ) = 1 8 4. (a) Q t = KAΔθ l = Δθ (l/KA) = Δθ R (R = Thermal resistance)  t ∝ R (∵Q and Δθ are same)  tP tS = RP RS = R/2 2R = 1 4 tP = tS 4 = 4 4 = 1min. (Series resistance RS = R1 + R2 and parallel resistance RP = R1R2 R1+R2 ) 5. (d) For cooking utensils, low specific heat is preferred for it’s material as it should need less heat to raise it’s temperature and it should have high conductivity, because, it should transfer heat quickly. 6. (d) In steady state there is no absorption of heat in any position. Heat passes on or is radiated from it’s surface. Therefore, in steady state the temperature of the body does not change with time but can be different at different points of the body. 7. (d) It is the property of material. 8. (d) Because steady state has been reached. 9. (c) KA KA t Q 50 0.6 (90 60) 1 = − = and KA KA t Q 50 0.8 (150 110 ) 2 = − = 10. (d) Given A1 = A2 and 4 5 2 1 = K K  R1 = R2  K A l K A l 2 2 1 1 =  . 4 5 2 1 2 1 = = K K l l 11. (c) x KA t Q   =     Thermal gradient x  C cm KA Q t 25 / 0.4 ( / ) 10 = =    = 12. (a) It is given that K1 K2 = 1 3  K1 = K then K2 = 3K the temperature of the junction in contact θ = K1θ1+K2θ2 K1+K2 = 1×100+3×0 1+3 = 100 4 = 25°C 13. (a) Q = KA(θ1−θ2)t l ; in both the cases , A, l and (θ1 − θ2 ) are same so Kt = constant  K1 K2 = t1 t2 = 30 20 = 3 2 = 1.5. 14. (b) ( Q t ) 1 = K1A1(θ1−θ2) l and ( Q t ) 2 = K2A2(θ1−θ2) l given ( Q t ) 1 = ( Q t ) 2  K1A1 = K2A2 15. (d) In variable state K t Q  and t c Q  1   c K t Q   (K = thermal conductivity,  = density, c = specific heat) 16. (b) K1:K2 = l1 2 :l2 2 ⇒ l1 l2 = √ K1 K2 = √ 10 9 = √10 3 17. (c) l KA t Q () =  0.2 5 1 0.4 5 20 50  = =  = K K 18. (c) 19. (a) Thermal resistance = l KA = [ L MLT−3K−1×L 2 ] = [M−1L −2T 3K] 20. (a) When a piece of glass is heated, due to low thermal conductivity it does not conduct heat fast. Hence unequal expansion of it’s layers crack the glass. 21. (a) In series both walls have same rate of heat flow. Therefore 2 2 2 1 1 1 ( ) ( ) d K A T d K A T dt dQ − = − =   ( ) ( )  K1d2 T1 − = K2d1  − T2 A B Junction temperature  Q Q 100°C 0°C d1 d2 K1 T1  K2 T2
1 2 2 1 1 2 1 2 1 2 K d K d K d T K d T + +   = 22. (a) Temperature of interface  = 1 2 1 1 2 2 K K K K +  +  =  4 1 ( 2 1 K K  If K1 = K then K2 = 4K)  K K K 5  0 + 4 100  = = 80°C 23. (b) C l o 80 10 0.5 30 80 2 1 2 2 =  = − − =  −     24. (d) dx d KA dt dQ  = − ; when = , = 0 dx d K  i.e.  is independent of x i.e. constant or uniform. 25. (a) Air is poor conductor of heat. 26. (b) 27. (b) 28. (d) Let the heat transferred be Q. When rods are joined end to end. Heat transferred by each rod 12  = = l KA Q  .....(i) When rods are joined lengthwise, t l KA Q 2  = .....(ii) From equation (i) and (ii) we get t = 48 s 29. (d) Q t = KAΔθ l ⇒ KA KB = AB AA = ( rB rB ) 2 = 1 4 ⇒ KA = KB 4 30. (b) Thermal conductivity of composite plate 2 3 2 2 2 3 1 2 1 2 +   = + = K K K K Keq 5 12 = =2.4 31. (b) Q ∝ A l ∝ r 2 l  Q2 Q1 = r2 2 r1 2 × l1 l2  Q2 Q1 = 4 1 × 1 2  Q2 = 2Q1 32. (c) Q At = K Δθ l  K Δθ l = constant  Δθ l ∝ 1 K Hence If Kc > Km > Kg, then ( Δθ l ) c < ( Δθ l ) m < ( Δθ l ) g ⇒ Xc < Xm < Xg because higher K implies lower value of the temperature gradient. 33. (b) In series Req = R1 + R2  2l KeqA = l K1A + l K2A  2 Keq = 1 K1 + 1 K2  Keq = 2K1K2 K1+K2 34. (b) dl d KA dt dQ  =  dl d dt dQ   (Temperature gradient) 35. (a) dQ dt = K(πr 2)dθ dl  ( dQ dt ) s ( dQ dt ) l = Ks×rs 2×ll Kl×rl 2×ls = 1 2 × 1 4 × 2 1  ( dQ dt) s = ( dQ dt ) l 4 = 4 4 = 1 36. (d) Q = KA(Δθ)t l ∵ Q and Δθ are same for both spheres hence K ∝ l At ∝ l r 2t  Klarger Ksmaller = ll ls × ( rs rl ) 2 × ts tl . It is given that rl = 2rs , ll = 1 4 ls and t1 = 25min, ts = 16min.  klarger ksmaller = ( 1 4 ) ( 1 2 ) 2 × 16 25 = 1 25 37. (d) Q t = kA(Δθ) l  Q t ∝ A l ∝ r 2 l  (Q/t)1 (Q/t)2 = ( r1 r2 ) 2 × l2 l1 = ( 2 1 ) 2 × ( 4 1 ) = 16 1 38. (b) Temperature of interface 1 2 1 1 2 2 K K K K + + =    where K1 = 2K and K2 = 3K         = 3 2 2 1 K K    = K K K K 2 3 2 100 3 0 +  +  C K K = = 40 5 200 39. (a) K1 K2 = l1 2 l2 2  K2 = K1l2 2 l1 2 = 0.92×(4.2) 2 (8.4) 2 = 0.23 40. (c) Mud is bad conductor of heat. So it prevents the flow of heat between surroundings and inside. 41. (b) Temperature gradient 4 C / cm 20 100 20 =  − = temperature at centre =100 − 4 10 = 60°C 42. (c) Temperature of interface θ = K1θ1l2+K2θ2l1 K1l2+K2l1 = K×0×2+3K×100×1 K×2+3K×1 = 300K 5K = 60°C 43. (c) Δθ = Q×l KAt = 4000×0.1 400×10−2 = 100 oC l l l
44. (b) Heat passes quickly from the body into the metal which leads to a cold feeling. 45. (c) Heat energy always flow from higher temperature to lower temperature. Hence, temperature difference w.r.t. length (temperature gradient) is required to flow heat from one part of a solid to other part. 46. (a) When the temperature of an object is equal to that of human body, no heat is transferred from the object to body and vice versa, Therefore block of wood and block of metal feel equally cold and hot if they have same temperature as human body. 47. (c) 48. (b) Temperature of water just below the lower surface of ice layer is 0°C. 49. (b) Q t = KA(θ1−θ2) l = 100×100×10 −4(100−0) 1 ⇒ Q t = 100 Joule/sec = 6 × 10 3 Joule/min 50. (a) Temperature of interface θ = K1θ1l2+K2θ2l1 K1l2+K2l1 It is given that KCu = 9KS. So if KS = K1 = K then KCu = K2 = 9K   = 9K×100×6+K×0×18 9K×6+K×18 = 5400K 72K = 75°C 51. (c) Q t = KA(θ1−θ2) l  Q t ∝ A l ∝ r 2 l [As (θ1 − θ2) and K are constants] ⇒ ( Q t ) 1 ( Q t ) 2 = r1 2 r2 2 × l2 l1 = 4 9 × 2 1 = 8 9 52. (b) In parallel combination equivalent conductivity K = K1A1+K2A2 A1+A2 = K1+K2 2 (As A1 = A2) 53. (b) Q = KA(θ1−θ2) l t = ⇒ K1t1 = K2t2 ⇒ K1 K2 = t2 t1 = 35 20 = 7 4 (As Q, l , A and (θ1 − θ2) are same) 54. (c) A lake cools from the surface down. Above 4C, the cooled water at the surface flows to the bottom because of it’s greater density. But when the surface temperature drops below 4C (here it is 2C ), the water near the surface is less dense than the warmer water below. Hence the downward flow ceases, the water at the bottom remains at 4C until nearly the entire lake, is frozen. 55. (a) Temperature gradient C cm cm C dx d o 2 / 50 (125 25) =  − =  56. (a) K ∝ l 2 ⇒ K1 K2 = l1 2 l2 2 = ( 10 25) 2 = 1 6.25 57. (a) Thermal resistance of Cu is lesser than the thermal resistance of steel. Hence only in option (a) thermal resistance is minimum so heat current is maximum. 58. (c) At steady state, rate of heat flow for both blocks will be same i.e., 2 2 2 1 1 1 ( ) ( ) l K A l K A    − = − (given 1 2 l = l ) ( ) ( )  K1A  1 − = K2 A  − 2 1 2 1 1 2 2 K K K K + +  =    59. (c) K = 2K1K2 K1+K2 = 2.K.2K K+2K = 4 3 K 60. (a) Temperature of interface 1 2 1 1 2 2 K K K K + + =    It is given that 3 5 2 1 = K K  K1 = 5K and K2 = 3K K K K K 5 3 5 100 3 20 +  +   = K K 8 560 = =70°C 61. (c) In winter, the temperature of surrounding is low compared to the body temperature (37.4 C) o . Since woolen clothes are bad conductors of heat, so they keep the body warm. 62. (b) Rate of heat flow ( Q t ) = kπr 2(θ1−θ2) L ∝ r 2 L  Q1 Q2 = ( r1 r2 ) 2 ( l2 l1 ) = ( 1 2 ) 2 × ( 2 1 ) = 1 2  Q2 = 2Q1 63. (b) l KA t Q  =  1 200 0.75 6000    =  = C    = 40 200 0.75 6000 1  64. (b) In series rate of flow of heat is same  KAA(θ1−θ) l = KBA(θ−θ2) l  3KB(θ1 − θ) = KB(θ − θ2 ) 1  2 K1 K2 l1 l2 1 A  B KA KB
 3(θ1 − θ) = (θ − θ2 )  3θ1 − 3θ = θ − θ2 4θ1 − 4θ = θ1 − θ2  4(θ1 − θ) = (θ1 − θ2 )  4(θ1 − θ) = 20 (θ1 − θ) = 5°C 65. (c) Let  be temperature middle point C and in series rate of heat flow is same  K(2A)(100 −) = KA( − 70)  200 − 2 =  − 70  3 = 270   = 90C 66. (b) Thermal resistances are same  l1 K1A1 = l2 K2A2  l1 K1 = l2 K2 (∵ A1 = A2)  l1 l2 = K1 K2 = 5 3 67. (b) l r t Q 2  ; from the given options, option (b) has higher value of l r 2 . 68. (c) Convection significantly transferring heat upwards (Gravity effect). 69. (a) Heat flows from hot air to cold body so person feels comfort. 70. (c) No flow of heat by convection in vacuum. 71. (a) 72. (b) Density of hot air is lesser than the density of cold air so hot air rises up. 73. (a) 74. (c) In convection hot particles moves up ward (due to low density) and light particle moves downward (due to high density). 75. (b) 76. (a) Natural convection arises due to difference of density at two places and is a consequence of gravity. 77. (d) 78. (a) Convection is not possible in weightlessness. So the liquid will be heated through conduction. 79. (c) In forced convection rate of loss of heat ( ) A T T0 t Q  − 80. (c) 81. (b) Because of uneven surfaces of mountains, most of it's parts remain under shadow. So, most of the mountains. Land is not heated up by sun rays. Besides this, sun rays fall slanting on the mountains and are spread over a larger area. So, the heat received by the mountains top per unit area is less and they are less heated compared to planes (Foot). 82. (a) The velocity of heat radiation in vacuum is equal to that of light. 83. (c) Radiation is the fastest mode of heat transfer. 84. (d) A thermopile is a sensitive instrument, used for detection of heat radiation and measurement of their intensity. 85. (d) The polished surface reflects all the radiation. 86. (c) Heat radiations are electromagnetic waves of high wavelength. 87. (d) When element and surrounding have same temperature. There will be no temperature difference, hence heat will not flow from the filament and it’s temperature remains constant. 88. (d) Every body at all time, at all temperatures emits radiation (except at T = 0 ). The radiation emitted by the human body is in the infra-red region. 89. (c) 90. (b) Infrared radiations are detected by pyrometer. 91. (b) 92. (a) In vacuum heat flows by the radiation mode only. 93. (c) Good absorbers are always good emitters of heat. 94. (a) A perfectly black body is a good absorber of radiations falls on it. So it’s absorptive power is 1. 95. (d) According to Kirchoff’s law in spectroscopy. If a substance emit certain wavelengths at high temperature, it absorbs the same wavelength at comparatively lower temperature. 96. (b) A person with dark skin absorbs more heat radiation and feels more heat. It also radiates more heat and feels more cold.