Tiêu đề 12. Thermometry Easy Ans.pdf ✅

1. (d) T = 273 .15 + tC  0 = 273 .15 + tC  t = −273 .15C 2. (b) 9 32 5 − = C F  9 32 5 183 − = − F  F = −297 F 3. (a) 5 273 9 32 − = F − K  5 95 273 9 32 − = F −  F = −288 F 4. (c) Temperature change in Celsius scale = Temperature change in Kelvin scale = 27 K 5. (b) Change in resistance 3.70 − 2.71 = 0.99  corresponds to interval of temperature 90°C. So change in resistance 3.26 − 2.71 = 0.55  Corresponds to change in temperature = 0.55 = 50C 0.99 90 6. (d) – 200°C to 600°C can be measured by platinum resistance thermometer. 7. (c) Pyrometer can measure temperature from 800°C to 6000°C. Hence temperature of sun is measured with pyrometer. 8. (a) v  T 2 9. (b) Thermoelectric thermometer is based on See back Effect. 10. (b) Maximum density of water is at 4°C Also 9 32 5 − = C F  9 32 5 4 − = F  F = 39.2F 11. (c) Production and measurement of temperature close to 0 K is done in cryogenics 12. (c) 13. (c) At absolute zero (i.e. 0 K) vrms becomes zero. 14. (c) 15. (c) We know that (1 ) 0 P = P +t and (1 ) 0 V = V +t and  = (1 / 273 )/ C for t = −273 C , we have P = 0 and V = 0 Hence, at absolute zero, the volume and pressure of the gas become zero. 16. (d) Zero kelvin = −273 C (absolute temperature). As no matter can attain this temperature, hence temperature can never be negative on Kelvin scale. 17. (b) C 5 = F−32 9  25 5 = F−32 9  F = 77°F. 18. (c) Thermoelectric thermometer is used for finding rapidly varying temperature. 19. (c) Due to evaporation cooling is caused which lowers the temperature of bulb wrapped in wet hanky. 20. (c) 5 273 9 32 − = F − K  5 273 9 32 − = x − x  x = 574 .25 21. (c) 9 32 5 − = C F  9 (140 32) 5 − = C  C = 60 22. (a) 9 32 5 − = C F  9 32 5 − = t t  t = − 40 23. (d) Standardisation of thermometers is done with gas thermometer. 24. (a) For gases  is more. 25. (c) The boiling point of mercury is 400°C. Therefore, the mercury thermometer can be used to measure the temperature upto 360°C. 26. (a) C C P P P P t t  =  − −   = − − = 100 25 (90 50) (60 50) 100 ( ) ( ) 100 0 0 27. (b) By filling nitrogen gas at high pressure, the boiling point of mercury is increased which extend the range upto 500°C. 28. (a) Pyrometer is used to measure very high temperature. 29. (c) 5 273 9 32 − = F − K  5 0 273 9 32 − = F −  F = −459 .4F  −460 F 30. (c) Initial volume V1 = 47.5 units Temperature of ice cold water T1 = 0C = 273 K Final volume of V2 = 67 units Applying Charle’s law, we have 2 2 1 1 T V T V = (where temperature T2 is the boiling point)
or T K C V V T = =   =  = 385 112 47.5 67 273 1 1 2 2 31. (a) Temperature on any scale can be converted into other scale by UFP LFP x LFP − − = Constant for all scales 100 60 150 20 20 = − x −  x = 98C 32. (d) 9 32 5 − = C F  9 140 32 5 − = C  C = 60C 33. (a) Rapidly changing temperature is measured by thermocouple thermometers. 34. (d) Difference of 100°C = difference of 180°F  Difference of 30° =  30 = 54 100 180 35. (a) 36. (c) When a copper ball is heated, it’s size increases. As Volume (radius)3 and Area  ((radius)2 , so percentage increase will be largest in it’s volume. Density will decrease with rise in temperature. 37. (a) (1 ) (1 ) 2 1 2 1 2 1       + + = = h h       + = (1 ) 0     1 100 1 50 60 50 +  +  =     = 0.005 / C 38. (b) γr = γa + γv; where γr =coefficient of real expansion, γa =coefficient of apparent expansion and γv =coefficient of expansion of vessel. For copper γr = C + 3αCu = C + 3A For silver γr = S + 3αAg  C + 3A = S + 3αAg  αAg = C−S+3A 3 39. (d) Fractional change in period ΔT T = 1 2 αΔθ = 1 2 × 2 × 10 −6 × 10 = 10 −5 % change= ΔT T × 100 = 10 −5 × 100 = 10 −3% 40. (c) (1 ) L = L0 +  2 1 2 1 1 ( ) 1 ( )     +  +  = L L  1 11 10 19 10 1 11 10 20 6 6 2 +   +   = − − L  L2 = 9.99989  Length is shorten by cm 5 10 9.99989 0.00011 11 10 − − = =  41. (c) Stress = Y ; hence it is independent of length. 42. (c) Solids, liquids and gases all expand on being heated as result density (= mass/volume) decreases. 43. (c) As coefficient of cubical expansion of liquid equals coefficient of cubical expansion of vessel, the level of liquid will not change on heating. 44. (b) Loss in time per second ( 0) 2 1 2 1 =  = −  t T T     loss in time per day 86400 2 1 (24 60 60) 2 1 2 1  =    =       t = t t t t 45. (c) A bimetallic strip on being heated bends in the form of an arc with more expandable metal (A) outside (as shown) correct. 46. (a) When the ball is heated, expansion of ball and cavity both occurs, hence volume of cavity increases. 47. (b) In summer alcohol expands, density decreases, so 1 litre of alcohol will weigh less in summer than in winter. 48. (b) Similar to previous question, benzene contracts in winter. So 5 litre of benzene will weigh more in winter than in summer. 49. (d) Water has maximum density at 4°C. 50. (a) Since coefficient of expansion of steel is greater than that of bronze. Hence with small increase in it's temperature the hole expand sufficiently. 51. (d) 2 A  L  L L A A  =   2  = 2 2 = 4%  A A . 52. (d) 2 1 2 1 1 1 t t V V   + + =  1 100 1 20 125 100 +  +  =     = 0.0033/°C 53. (d) α = β 2 = 2×10 −5 2 = 10 −5/°C 54. (d) Coefficient of volume expansion 1 2 4 3 10 10 (100 0) (10 9.7) .( ) ( ) . − =   − − =  − =   =        T Hence, coefficient of linear expansion = = C − 10 / 3  4  A B A B A B A >B B A
55. (a) ρ = ρ0 (1 − γ. Δθ) = 13.6[1 − 0.18 × 10 −3 (473 − 273)] = 13.6[1 − 0.036] = 13.11gm/cc. 56. (b) As we know γreal = γapp. + γvessel  γapp. = γglycerine − γglass = 0.000597 − 0.000027 = 0.00057/°C 57. (c) Water has maximum density at 4°C, so if the water is heated above 4°C or cooled below 4°C density decreases i.e. volume increases. In other words, it expands so it overflows in both the cases. 58. (a) γ = ΔV V.ΔT = 0.24 100×40 = 6 × 10 −5 /°C  α = γ 3 = 2 × 10 −5 /°C 59. (a) As 2 3    = =   :  :  = 1 : 2 : 3 60. (a) γapp. = Mass expelled Mass remained× ΔT = x/100 x × 80 = 1 8000 = 1.25 × 10 −4/°C 61. (b) In anomalous expansion, water contracts on heating and expands on cooling in the range 0°C to 4°C. Therefore water pipes sometimes burst, in cold countries. 62. (c) On heating the system; x, r, d all increases, since the expansion of isotropic solids is similar to true photographic enlargement. 63. (d) α = ΔL L0×Δθ = 0.01 5×100 = 2 × 10 −5/°C 64. (a) α = ΔL L0(Δθ) = 0.19 100(100−0) = 1.9 × 10 −5 /°C Now  = 3 = 3  1.9  10–5 /°C = 5.7  10–5 /°C 65. (d) Since, the coefficient of linear expansion of brass is greater than that of steel. On cooling, the brass contracts more, so, it get loosened. 66. (b) Increase in length L = L0   = 10  10  10–6  (100 – 0) = 10–2 m = 1 cm 67. (a) α = ΔL L0Δθ = (1−0.9997) 0.9997×12×10−6 = 25°C 68. (c) The densest layer of water will be at bottom. The density of water is maximum at 4°C. So the temperature of bottom of lake will be 4°C. 69. (c) Given Δl1 = Δl2 or l1αat = l2αs t ∴ l1 l2 = αs αa or l1 l1+l2 = αs αa+αs . 70. (b) In vapor to liquid phase transition, heat liberates. 71. (b) Pressure inside the mines is greater than that of normal. Pressure. Also we know that boiling point increases with increase in pressure. 72. (c) Q = m.c.   = m. Q c ; when  = 0  c =  73. (c) Mass and volume of the gas will remain same, so density will also remain same. 74. (d) 75. (a) The latent heat of vaporization is always greater than latent heat of fusion because in liquid to vapour phase change there is a large increase in volume. Hence more heat is required as compared to solid to liquid phase change. 76. (c) When state is not changing Q = mc. 77. (a) Heat taken by ice to melt at 0°C is Q1 = mL = 540 × 80 = 43200cal Heat given by water to cool up to 0°C is Q2 = msΔθ = 540 × 1 × (80 − 0) = 43200cal Hence heat given by water is just sufficient to melt the whole ice and final temperature of mixture is 0°C. 78. (d) Due to large specific heat of water, it releases large heat with very small temperature change. 79. (a) Q = m. c. Δθ = 5 × (1000 × 4.2) × (100 − 20) = 1680 × 10 3 J = 1680kJ 80. (b) Melting point of ice decreases with increase in pressure (as ice expands on solidification). 81. (c) Conversion of ice (0°C) into steam (100°C) is as follows Density 0°C 4°C Temp. (Q1 = mLi) Water at 0°C ice 0°C Water at100°C (Q3 = mLV) (Q2 = mcW) Steam at 100°C
Heat required in the given process = Q1 + Q2 + Q3 = 180 +11(100 − 0)+1536 = 716 cal 82. (a) If m gm ice melts then Heat lost = Heat gain 80 1(30 − 0) = m  80  m = 30 gm 83. (c) At boiling point saturation vapour pressure becomes equal to atmospheric pressure. Therefore, at 100°C for water. S.V.P. = 760 mm of Hg (atm pressure). 84. (b) Thermal capacity = Mass  Specific heat Due to same material both spheres will have same specific heat. Also mass = Volume (V)  Density ()  Ratio of thermal capacity = m1 m2 = V1ρ V2ρ = 4 3 πr1 3 4 3 πr2 3 = ( r1 r2 ) 3 = ( 1 2 ) 3 = 1: 8 85. (a) Ice (–10°C) converts into steam as follows (ci = Specific heat of ice, cW = Specific heat of water) Total heat required Q = Q1 + Q2 + Q3 + Q4  Q = 1 0.5(10) + 1 80 + 11(100 − 0) + 1 540 = 725 cal Hence work done W = JQ = 4.2725 = 3045 J 86. (b) When water is cooled at 0°C to form ice then 80 calorie/gm (latent heat) energy is released. Because potential energy of the molecules decreases. Mass will remain constant in the process of freezing of water. 87. (a) Steam at 100°C contains extra 540 calorie/gm energy as compare to water at 100°C. So it’s more dangerous to burn with steam then water. 88. (a) Same amount of heat is supplied to copper and water so mcccΔθc = mWcWΔθW  ΔθW = mccc(Δθ)c mWcW = 50×10 −3×420×10 10×10−3×4200 = 5°C 89. (c) Temperature of mixture A B A A B B mix c c c c + + =     A B A B c c c c +  +  = 32 24 28  A B A B 28 c + 28c = 32 c + 24 c  1 1 = B A c c 90. (b) Heat lost by hot water = Heat gained by cold water in beaker + Heat absorbed by beaker  440 (92 −) = 200 ( − 20) + 20 ( − 20)   = 68C 91. (b) Q = m.c. ; if  = 1 K then Q = mc = Thermal capacity. 92. (a) Latent heat is independent of configuration. Ordered energy spent in stretching the spring will not contribute to heat which is disordered kinetic energy of molecules of substance. 93. (d) Temperature of mixture θmix = m1c1θ1+m2c2θ2 m1c1+m2c2 = m×c×2T+ m 2 (2c)T m.c+ m 2 (2c) = 3 2 T 94. (a) 95. (a) θmix = θW− Li cW 2 = 100− 80 1 2 = 10°C 96. (b) When pressure decreases, boiling point also decreases. 97. (a) Boiling occurs when the vapour pressure of liquid becomes equal to the atmospheric pressure. At the surface of moon, atmospheric pressure is zero, hence boiling point decreases and water begins to boil at 30°C. 98. (d) Thermal capacity = mc = 40 0.2 = 8 cal/ C . 99. (b) Q = m.c.   = m. Q c In temperature measurement scale F  C so F C c c    ( ) ( ) . 100.(a) Increasing pressure lowers melting point of ice. 101.(b) Work done changes into heat energy, when the temperature of palm becomes above the atmosphere so it starts losing heat to the surroundings. 102.(b) Firstly the temperature of bullet rises up to melting point, then it melts. Hence according to W = JH.  1 2 mv 2 = J.[m. c. Δθ + mL] = J[mS(475 − 25) + mL]  mS(475 − 25) + mL = mv 2 2J (Q1 = mci) Water at 100°C ice –10°C Water at 0°C (Q4 = mLV) (Q2 = mLf) Steam at 100°C ice 0°C (Q3 = mcW)