THERMAL PROPERTIES OF MATTER QUIZ-23
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Q221. Statement 1: A hollow metallic closed container maintained at a uniform temperature can act as a source of black body radiation
Statement 2: All metals act as a black body
Solution
221(c) Hollow metallic closed container maintained at a uniform temperature can act as source of black body. It is also well-known that all metals cannot act as black body because if we take a highly metallic polished surface. It will not behave as a perfect black body
221(c) Hollow metallic closed container maintained at a uniform temperature can act as source of black body. It is also well-known that all metals cannot act as black body because if we take a highly metallic polished surface. It will not behave as a perfect black body
Q222. 
Solution
222(a)
222(a)
Q223. 
Solution
223(a) Initially temperature greatest at A will be more and also area is more at A. So the rate of flow of heat is maximum at A At steady state, rate of flow of heat is constant at all sections as all section ate connected in series As (dQ/dt)=KA(-dT/dx) |dT/dx|∝1/A At steady state, the temperature at each at every section is constant. Hence dT/dt=0 at each section
223(a) Initially temperature greatest at A will be more and also area is more at A. So the rate of flow of heat is maximum at A At steady state, rate of flow of heat is constant at all sections as all section ate connected in series As (dQ/dt)=KA(-dT/dx) |dT/dx|∝1/A At steady state, the temperature at each at every section is constant. Hence dT/dt=0 at each section
Q224.
Solution
214 (c) We have for the sections, AB,BC,CD and DE with (dQ/dt) as the steady state thermal, energy transmitted per second (A being the area of cross section) dQ/dt=(KA(100-T_C))/L=(A(0.8)K(T_C-T_D))/(1.2)L =((1.2)KA(T_D-T_E))/(1.5)L=((1.5)KA T_E)/(0.6)L These give (100-T_c)=(0.8/1.2)(T_C-T_D) =(1.2/1.5)(T_D-T_E )=(1.5/0.6) T_E 6(100-T_C )=4(T_C-T_D )=(4.8)(T_D-T_E )=15 T_E Solving for the differences (100-T_C ),(T_C-T_D ),(T_D-T_E ) and T_E remaining that the sum of these differences is 100, we obtain (T_A-T_C )=24.1,(T_C-T_D )=36.2
214 (c) We have for the sections, AB,BC,CD and DE with (dQ/dt) as the steady state thermal, energy transmitted per second (A being the area of cross section) dQ/dt=(KA(100-T_C))/L=(A(0.8)K(T_C-T_D))/(1.2)L =((1.2)KA(T_D-T_E))/(1.5)L=((1.5)KA T_E)/(0.6)L These give (100-T_c)=(0.8/1.2)(T_C-T_D) =(1.2/1.5)(T_D-T_E )=(1.5/0.6) T_E 6(100-T_C )=4(T_C-T_D )=(4.8)(T_D-T_E )=15 T_E Solving for the differences (100-T_C ),(T_C-T_D ),(T_D-T_E ) and T_E remaining that the sum of these differences is 100, we obtain (T_A-T_C )=24.1,(T_C-T_D )=36.2
Q225. 
Solution
225 (c) i. Rate at which heat is radiated from the body =Q_r J/s =eσAT_14=0.55×5.67×10(-8)×1.5×(323)4 J/s=509 W ii. Rate at which heat radiation is absorbed by the body= =Q_a J/s =eσAT_24=0.55×5.67×10(-8)×1.5×(295)4 J/s =354 W iii. Rate at which net radiation is emitted by the body =Q_n J/s=Q_r-Q_a=(509-354)W=155 W
225 (c) i. Rate at which heat is radiated from the body =Q_r J/s =eσAT_14=0.55×5.67×10(-8)×1.5×(323)4 J/s=509 W ii. Rate at which heat radiation is absorbed by the body= =Q_a J/s =eσAT_24=0.55×5.67×10(-8)×1.5×(295)4 J/s =354 W iii. Rate at which net radiation is emitted by the body =Q_n J/s=Q_r-Q_a=(509-354)W=155 W
Q226. 
Solution
226 (b) Work done by the system can be non-zero in any of the process The relation dU=nC_v=n(R/(r-1))dT is valid for all the process In isothermal process dT=0 In adiabatic process dQ=0 and non-zero for any other process
226 (b) Work done by the system can be non-zero in any of the process The relation dU=nC_v=n(R/(r-1))dT is valid for all the process In isothermal process dT=0 In adiabatic process dQ=0 and non-zero for any other process
Q227. 
Solution
227 (d) Let Q be the heat required to convert 100 g of water at 20°C to 100°C Then mc∆θ=(100)(1)(100-20) Q=8000 cal Now suppose m_0 mass of steam converts into water to librate this much amount of heat. Then m_0=Q/L=8000cal/(540cal/g)=14.8 g Since it is less than m=20 g, the temperature of the mixture is 100°C Mass of steam in the mixture=(20-14.8)=5.2 g Mass of water in the mixture =(100+14.8)=114.8 g If m=10 g, the amount of heat liberated by steam = mL=10×540=5400 Let θ be the final temperature of the mixture _(m_(H_2 O) S_(H_2 O) ) (θ-20)=m_steam L+ms_(H_2 O) (100-θ) 100×1(θ-20)=10×540+10×1(100-θ) 110θ=5400+1000+2000 θ=76.4°C
227 (d) Let Q be the heat required to convert 100 g of water at 20°C to 100°C Then mc∆θ=(100)(1)(100-20) Q=8000 cal Now suppose m_0 mass of steam converts into water to librate this much amount of heat. Then m_0=Q/L=8000cal/(540cal/g)=14.8 g Since it is less than m=20 g, the temperature of the mixture is 100°C Mass of steam in the mixture=(20-14.8)=5.2 g Mass of water in the mixture =(100+14.8)=114.8 g If m=10 g, the amount of heat liberated by steam = mL=10×540=5400 Let θ be the final temperature of the mixture _(m_(H_2 O) S_(H_2 O) ) (θ-20)=m_steam L+ms_(H_2 O) (100-θ) 100×1(θ-20)=10×540+10×1(100-θ) 110θ=5400+1000+2000 θ=76.4°C
Q228. 
Solution
228(b) Solar constant =1.35 kW/m^2 Thermal conductivity of earth’s crust =0.75 J/s mK Heat transferred per second is dQ/dt=-K(4πr2)dT/dr r=R_e=6400 km -dT/dr=-(1°C)/(30 m) Heat lost by the earth per second due to conduction from the core dQ/dt=((0.75 J×4π)/msK)×(6400×103 m)2 (1°C)/30m P_1=[(0.75×4π×(6400×103 )2)/30]J/s =1.286×1013 J/s≈1.3×1013 J/s Heat absorbed from sun =SπR_e2 P_2=(1.35×103 W/m2 )π(6400×103 m)2 =1.7×1017 W Heat lost by the earth by radiation if e=1 P_2=eσA' T4 (A'=4πR_e(2_ )) 1.7×1017=1×5.67×20(-8)×4π×(6400×103 )2 T4 T4=(1.7×1017)/(5.67×10(-8)×4π×(6400×103 )2 ) =5.8×109=58×108 Surface temperature T of earth =(58)(1/4)×102=(7.6)(1/2)×102 =2.76×100=276 K P_1/P_2 =(1.3×1013)/(1.7×1017 )=7.5×10(-5)
228(b) Solar constant =1.35 kW/m^2 Thermal conductivity of earth’s crust =0.75 J/s mK Heat transferred per second is dQ/dt=-K(4πr2)dT/dr r=R_e=6400 km -dT/dr=-(1°C)/(30 m) Heat lost by the earth per second due to conduction from the core dQ/dt=((0.75 J×4π)/msK)×(6400×103 m)2 (1°C)/30m P_1=[(0.75×4π×(6400×103 )2)/30]J/s =1.286×1013 J/s≈1.3×1013 J/s Heat absorbed from sun =SπR_e2 P_2=(1.35×103 W/m2 )π(6400×103 m)2 =1.7×1017 W Heat lost by the earth by radiation if e=1 P_2=eσA' T4 (A'=4πR_e(2_ )) 1.7×1017=1×5.67×20(-8)×4π×(6400×103 )2 T4 T4=(1.7×1017)/(5.67×10(-8)×4π×(6400×103 )2 ) =5.8×109=58×108 Surface temperature T of earth =(58)(1/4)×102=(7.6)(1/2)×102 =2.76×100=276 K P_1/P_2 =(1.3×1013)/(1.7×1017 )=7.5×10(-5)
Q229. 
Solution
229 (a) Fraction of volume submerged f=V_i/V=ρ_1/ρ_2 After increasing the temperature f'=(V_i')/V' =(ρ_1 (1-γ_1 ∆T))/(ρ_2 (1-γ_2 ∆T))>f(because γ_2>γ_1) If γ_1, then f'=f If γ_2<γ_1,then f'
229 (a) Fraction of volume submerged f=V_i/V=ρ_1/ρ_2 After increasing the temperature f'=(V_i')/V' =(ρ_1 (1-γ_1 ∆T))/(ρ_2 (1-γ_2 ∆T))>f(because γ_2>γ_1) If γ_1, then f'=f If γ_2<γ_1,then f'
Q230. A body cools in a surrounding of constant temperature 30°C. Its heat capacity is 2 J/°C. Initial temperature of the body is 40°C. Assume Newton’s law of cooling is valid. The body cools to 38°C in 10 min
In further 10 min it will cool from 38°C to__________________:
Solution
230 (b) We have θ-θ_s=(θ_0-θ_s ) e(-kt), Where θ_0-initial temperature of body =40°C θ= temperature of body after time t Since body cools from 40 to 38 in 10 min, we have 38-30=(40-30) e(-10k) (i) Let after 10 min, the body temperature be θ θ-30°=(38-30) e(-10k) (ii) (Eq.(i))/(Eq.(ii)) gives 8/(θ-30)=10/8,θ-30=6.4 θ=36.4°C
230 (b) We have θ-θ_s=(θ_0-θ_s ) e(-kt), Where θ_0-initial temperature of body =40°C θ= temperature of body after time t Since body cools from 40 to 38 in 10 min, we have 38-30=(40-30) e(-10k) (i) Let after 10 min, the body temperature be θ θ-30°=(38-30) e(-10k) (ii) (Eq.(i))/(Eq.(ii)) gives 8/(θ-30)=10/8,θ-30=6.4 θ=36.4°C
