Title 11.THERMAL PROPERTIES OF MATTER - Explanations.pdf ✅

1 (d) Let θ be the temperature of the mixture. Heat gained by water at 0°C = Heat lost by water at 10°C c m1 (θ − 0) = c m2 (10 − θ) θ = 400 60 = 6.66°C 2 (b) The surface temperature of the stars is determined using Wien’s displacement law. According to this (law)λmT = b where b is Wien’s constant whose value is 2.898 × 10−3 mK. 3 (b) From Stefan’s law, the rate at which energy is radiated by sun at its surface isP = σ × 4πr 2T 4 [Sun is a perfectly black body as it emits radiations of all wavelengths and so for it e=1.] The intensity of this power at earth’s surface(under the assumption r>>r0) is I = P 4πR2 = σ×4πr 2T 4 4πR2 = σR 2σT 4 r 2 The area of earth which receives this energy is only one-half of total surface area of earth, whose projection would be πr0 2 . ∴ Total radiant power as received by earth = πr0 2 × I = πr0 2×σR 2T 4 r 2 = πr0 2R 2σT 4 r 2 4 (c) Let m gram of water, whose temperature is θ(> 30°C), be added to 20 g of water at 30°C. If m × 1(θ − θ0 ) = 20 × 1(θ0 − 30) (m+20)θ0 = 60 + mθ θ0 = 600+mθ 20+m For θ0 to be maximum m should be small and θ should be large 5 (c) Both the cylinders are in parallel, for the heat flow from one end as shown Hence Keq = K1A1+K2A2 A1+A2 Where A1 = Area of cross-section of inner cylinder ∝ πR 2 and A2 = Area of cross-section of cylindrical shell = π{(2R) 2 − (R) 2 } = 3πR 2 ⇒ Keq = K1(πR 2 ) + K2(3πR 2 ) πR2 + 3πR2 = K1 + 3K2 4 6 (a) Anomalous density of water is given by (a). It has maximum density at 4°C. 10 (d) W = JQ ⇒ W = 4.2 × 200 = 840 J 11 (a) dQ dt = K(πr 2)dθ dt ⇒ ( dQ dt ) s ( dQ dt ) l = Ks × rs 2 × l1 Kl × r1 2 × ls = 1 2 × 1 4 × 2 1 ⇒ ( dQ dt ) s = ( dQ dt ) l 4 = 4 4 = 1 12 (a) Here, ρ0 = 10g/cc ρ100 = 9.7g/cc, α = ? From ρ0 = ρ100(1 + γ × 100) γ = ρ0 − ρ100 ρ100 × 100 = 10 − 9.7 9.7 × 100 = 3.09 × 10−4 α = γ 3 = 3.09×10−4 3 =1.03× 10−4°C−1 13 (a) If mass of the bullet is m gm, Then total heat required for bullet to just melt down Q1 = m c ∆θ + m L = m × 0.03(327 − 27) + m × 6 = 15 m cal = (15m × 4.2)J Now when bullet is stopped by the obstacle, the loss in its mechanical energy = 1 2 (m × 10−3)v 2 J (As m g = m × 10−3kg) As 25% of this energy is absorbed by the obstacle, The energy absorbed by the bullet Q2 = 75 100 × 1 2 mv 2 × 10−3 = 3 8 × 10−3 J Now the bullet will melt if Q2 ≥ Q1 i. e. , 3 8 mv 2 × 10−3 ≥ 15m × 4.2 ⇒ vmin = 410m/s 14 (c) At absolute zero (i. e. , 0 K)vrms becomes zero 15 (a) θ = ∆L L0∆α = (1 − 0.9997) 0.9997 × 12 × 10−6 = 25°C 16 (d) T1 = 277°C = 277 + 273 = 550 K T2 = 67°C = 67 + 273 = 340 K Temperature of surrounding T = 27°C = 27 + 273 = 300 K Ratio of loss of heat=T1 4−T 4 T2 4−T4 R 2R K2 K1
= ( T1 T ) 4 −1 ( T2 T ) 4 −1 = ( 550 300) 4 −1 ( 340 300) 4 −1 = 9.5 0.5 = 19 1 17 (a) According to the question, 1 2 × 1 2 mv 2 = m × s × ∆T 1 4 m × 4 × 104 = 125 × m × ∆T ∆T = 4×104 500 = 80°C 18 (a) In series both walls have same rate of heat flow. Therefore dQ dt = K1A(T1 − θ) d1 = K2A(θ − T2) d2 ⇒ K1d2 (T1 − θ) = K2d1(θ − T2) ⇒ θ = K1d2T1 + K2d1T2 K1d2 + K2d1 19 (a) As α = β 2 = γ 3 ⇒ α ∶ β ∶ γ = 1 ∶ 2 ∶ 3 20 (d) If the temperature of a body on Celsius and Fahrenheit scales are recorded as C and F respectively, then C−0 100−0 = F−32 212−32 or C 5 = F−32 9 Here, C = 95°C ∴ 95 5 = F−32 9 Or 5F = 1015 ∴ F = 1015 5 = 203°F 21 (c) E = σT 4 ⇒ 5.6 × 10−8 × T 4 = 1 ⇒ T = [ 1 5.6 × 10−8 ] 1/4 = 65 K 22 (d) Thermal capacity = mc = 40 × 0.2 = 8 cal/°C 23 (c) Thermoelectric thermometer is used for finding rapidly varying temperature 24 (c) From ideal gas equation PV = μ RT ⇒ P = μ RT V Given PT 2 = K ⇒ μ RT V . T 2 = K ⇒ μ RT 3 = KV ... (i) Differentiating both sides, we get 3μ RT 2dT = K dV ... (ii) Dividing equation (ii) by (i), we get 3 T dT = dV V Coefficient of volume expansion = dV V dT = 3 T 26 (a) According to Wien’s displacement law λm = b T ⇒ T = b λm = 2.93×10−3 4000×10−10 = 7325 K 27 (b) For θt plot, rate of cooling = dθ dt = slope of the curve At P, dθ dt = tanφ2 = k(θ2 − θ0 ), where k = constant At Q, dθ dt = tanφ1 = k(θ1 − θ0 ) ⇒ tan φ2 tan φ1 = θ2−θ0 θ1−θ0 28 (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 29 (b) An ideal black body absorbs all the radiations incident upon it and has an emissivity equal to 1. If a black body and an identical another body are kept the same temperature, then the black body will radiate maximum power. Hence, the black object at a temperature of 2000°C will glow brightest. 30 (d) Let the temperature of common interface be T°C. Rate of heat flow H = Q t = KA∆T l ∴ H1 = ( Q t ) 1 = 2KA(T−T1) 4x And H2 = ( Q t ) 2 = KA(T2−T) x In steady state, the rate of heat flow should be same in whole system ie, H1 = H2 ⇒ 2KA(T−T1) 4x = KA(T2−T) x ⇒ T−T1 2 = T2 − T ⇒ T − T1 = 2T2 − 2T ⇒ T = 2T2+T1 3 ...(i) Hence, heat flow from composite slab is H = KA(T2−T) x = KA x (T2 − 2T2+T1 3 ) = KA 3x (T2 − T1) ...(ii) [from Eq. (i)] Accordingly, H = [ A(T2−T1 )K x ] f ... (iii) By comparing Eqs. (ii) and (iii), we get ⇒ f = 1 3 d1 d2 K1 T1  K2 T2
31 (b) For a black body rate of energy Q t = P = AσT 4 ⇒ P ∝ T 4 ⇒ P1 P2 = ( T1 T2 ) 4 = { (273 + 7) (273 + 287) } 4 = 1 16 32 (b) When length of the liquid column remains constant, then the level of liquid moves down with respect to the container, thus γ must be less than 3α Now we can write V = V0(1 + γ∆T) Since V = Al0 = [A0 (1 + 2α∆T)]l0 = V0 (1 + 2α∆T) Hence V0 (1 + γ∆T) = V0 (1 + 2α∆T) ⇒ γ = 2α 33 (d) Q t = KA∆θ 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 34 (b) E1 E2 = ( T1 T2 ) 4 = ( 727 + 273 127 + 273) 4 = (1000) 4 (400) 4 = 104 4 4 = 625 16 35 (a) An opaque body does not transmit any radiation, hence transmission coefficient of an opaque body is zero. 36 (a) Convection is not possible in weightlessness. So the liquid will be heated through conduction 37 (c) Total energy radiated from a body Q = AεσT 4 t Or Q t ∝ AT 4 Q t ∝ r 2T 4 (∵ A = 4πr 2 ) Q1 Q2 = ( r1 r2 ) 2 ( T1 T2 ) 4 j = ( 8 2 ) 2 [ 273 + 127 273 + 527] 4 = 1 38 (c) Stress = Yα∆θ; hence it is independent of length 39 (b) Thermoelectric thermometer is based on Seebeck Effect 40 (d) Rate of cooling RC = dθ dt = Aεσ(T 4−T0 4 ) mc ⇒ dθ dt ∝ A V ∝ r 2 r 3 ⇒ dθ dt ∝ 1 r 41 (b) ∆L = L0 α∆θ Rod A: 0.075 = 20 × αA × 100 ⇒ αA = 75 2 × 10−6 /°C rod B: 0.045 = 20 × αB × 100 ⇒ αB = 45 2 × 10−6 /°C For composite rod : x cm of A and (20 − x)cm of B we have 0.060 = x αA × 100 + (20 − x)αB × 100 = x [ 75 2 × 10−6 × 100 + (20 − x) × 45 2 × 10−6 × 100] On solving we get x = 10 cm 42 (b) In first case 50−40 5 = K [ 50+40 2 − θ0 ] ...(i) In second case 40−33.33 5 = K [ 40+33.33 2 − θ0 ] ...(ii) By solving θ0 = 20°C 43 (d) From Stefan’s law, the rate at which energy is radiated by sun at its surface is P = σ × 4πr 2T 4 (Sun is a perfectly black body as it emits radiations of all wavelengths and so for it e=1) The intensity of this power at earth’s surface(under the assumption R>>r0) is I = P 4πR2 = σ×4πr 2T 4 4πR2 = σr 2T 4 R2 = σr 2(t+273) 4 R2 44 (a) The contraction in the length of the wire due to change in temperature=αLT = 1.2 × 10−5 × 3 × (−170 − 30) = −7.2 × 10−3m The expansion in the length of wire due to stretching force = F L A Y = (10×10)×3 (0.75×10−6)(2×1011) = 2 × 10−3 m Resultant change in length = −7.2 × 10−3 + 2 × 10−3 = −5.2 × 10−3m= −5.2mm Negative sign shows a contradiction. 45 (b) (20 – x) A A B B 20cm x
Q2 Q1 = ( T2 T1 ) 4 ⇒ 2 1 = ( T2 T1 ) 4 ⇒ T2 4 = 2 × T1 4 = 2 × (273 + 727) 4 ⇒ T2 = 1190K 46 (c) When state is not changing ∆Q = mc∆θ 47 (d) Black and rough surfaces are good absorber that’s why they emit well. (Kirchhoff’s law) 48 (a) From Stefan’s law E = σT 4 T 4 = E σ = 6.3 × 107 5.7 × 10−8 = 1.105 × 1015 = 0.1105 × 1016 T = 0.58 × 104K = 5.8 × 103K 50 (d) Thermostat is used in electric apparatus like refrigerator, iron etc for automatic cut off. Therefore for metallic strips to bend on heating their coefficient if linear expansion should be different 51 (c) Let F = K − X As F−32 9 = K−273 5 ∴ x − 32 9 = x − 273 5 9x − 2457 = 5x − 160 4x − 2457 + 160 = 0 x = 2297 4 = 574.25° 52 (b) Wien’s displacement law is given by λmT = constant (say b) Given, b=Wien’s constant=2.93 × 10−3 m − K λm = 2.93 × 10−10m Substituting the values, we obtain T = b λm = 2.93 × 10−3 2.93 × 10−10 = 107 K 53 (c) Star emits thermal radiations these radiations are a mixture of wavelengths and bear the following relation, with temperature (T) as λmT= constant Where λm is maximum wavelength. This is Wien’s displacement law and is used in determining the temperature of stars. 54 (a) Heat current H = KA∆θ l ∴ ∆θ = Hl KA = 30 × 1 × 10−2 0.76 × 100 × 10−4 = 39.47 = 40°C(approx. ) 55 (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) 56 (a) Suppose m′ kg ice melts out of m kg. Then by using W = JQ ⇒ mgh = J(m′L). Hence fraction of ice melts = m′ m = gh JL = 9.8 × 1000 4.18 × 80 = 1 33 57 (b) According to Newton’s law θ1−θ2 t = k [ θ1+θ2 2 − θ0 ] Initially, (80 − 64) 5 = K ( 80 + 64 2 − θ0) ⇒ 3.2 = K(72 − θ0) . . . (i) Finally (64 − 52) 10 = K [ 64 + 52 2 − θ0 ] ⇒ 1.2 = K[58 − θ0] . . . (ii) On solving equation (i) and (ii), θ0 = 49°C 58 (b) Liquid having more specific heat has slow rate of cooling because for equal masses rate of cooling dθ dt ∝ 1 c 59 (c) Equivalent thermal conductivity of the compound, slab, Keq = l1+l2 l1 K1 + l2 K2 = l+l l K + l 2K = 2l 3l 2K = 4 3 K 61 (b) For resistance thermometers t = Rt−R0 R100−R0 × 100°C Here Rt=5.5Ω, Ro=5Ω, R100 = 5.25Ω ∴ t = 5.5−5 5.25−5 × 100 = 0.5 0.25 × 100 = 200°C 63 (c) F − 32 9 = K − 273 5 ⇒ x − 32 9 = x − 273 5 ⇒ x = 574.25 64 (a) Energy gained by water (in 1 s) = Energy supplied−energy lost = (1000 J-160 J)=840 J Total heat required to raise the temperature of water from 27°C to 77°C is ms∆θ. Hence, the required time,