Q: The specific heat capacity of a substance is temperature dependent and is given by the formula C=k T, where k is a constant of suitable dimensions in SI units, and T is the absolute temperature. If th

aligned & T _ i =-73^ C =200 ~K \\ & ~T _ f =27^ C =300 ~K aligned aligned & Q = msdT \\ & = 1 kTdT \\ & = kTdT = K _ 200 ^ 300 TdT \\ & = K 2 [ ~T ^2 ]_ 200 ^ 300 = K 2 [300^2-200^2 ] \\ & Q =25000 ~K aligned Hence n=25000

Q: Match the temperature of a black body given in List- I with an appropriate statement in List- II , and choose the correct option. [Given: Wien’s constant as 2 . 9 × 10 - 3 m -

Maximum wavelength for each case: P 2000 K λ m T = b ⇒ λ m = b T = 2 . 9 × 10 - 3 2000 = 1 . 45 × 10 - 6 m = 1450 nm ( Q ) 3000 K λ m T = b ⇒ λ m = 2 . 9 × 10 - 3 3000 = 966 . 66 nm ( R ) 5000 K λ m T = b ⇒ λ m = 2 . 9 × 10 - 3 5000 = 580 nm ( S ) 10000 K λ m T = b ⇒ λ m = 2 . 9 × 10 - 3 10 , 000 = 290 nm Now for List-

Q: A small object is placed at the centre of a large evacuated hollow spherical container. Assume the container is maintained at 0 K . At time t = 0 , the temperature of the object is&

Using Stefan-Boltzmann law, P = d Q d t = σ e A T B 4 − T S 4 , ∵ T S = 0 K So, d Q d t = e σ A T 4 , consider , T = T B also, d Q d t = - m s d T d t = e σ A T 4 - ∫ 200 T d T T 4 = ∫ 0 t k d t , where k = σ A m s ⇒ 1 T 3 − 1 200 3 = 3 k t Now, For case 1 1 100 3 − 1 200 3 = 3 k t 1

Q: A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the su

P i = 700 × 0 . 05 = 35 watts P i = P r 35 = e σ A 4 T 0 3 ( Δ T ) . . . 1 ∴ K = 4 e σ A T 0 3 m C = 0 . 0001 4 e σ A T 0 3 = 0 . 001 × m C = 4 . 2 from equation (1) Δ T = 35 4 . 2 = 50 6 = 25 3 = 8 . 33

Q: The filament of a light bulb has surface area 64 mm 2 . The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far.

P = σ A e T 4 P = 5 . 6 × 10 - 8 × 64 × 10 - 6 × 1 × ( 2500 ) 4 P = 14175 × 10 - 14 × 10 8 × 10 4 (a) P = 141 . 75 W (b) σ A e T 4 4 π ( 100 ) 2 × π 3 × 10 - 3 2 = 141 . 75 × 9 × 10 - 6 4 × 10 4 318 . 937 × 10 - 10 3 . 18937 × 10 8 W (c) λ T = b λ = 2 . 93 × 10 - 6 2500 = 1160 nm (d) 3 . 18937 × 10 - 8 = n sec h c λ 3 . 18937 ×

Question 1

As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition (P_1) and a freely movable but thermally insulated piston (P_2). The partition P_1 with the

Accepted Answer

Let T_1 and T_2 be the temperatures of the gases in sections S_1 and S_2 respectively. Assume T_1 > T_2, so heat flows from S_1 to S_2. The rate of heat transfer through the partition P_1 is given by Fourier's law of heat conduction: dQ dt = KA x (T_1 - T_2) For section S_1, the volume is constant (since P_1 is immovable). Thus, the process is isochoric. The heat lost by S_1 is: dQ = -n_1 C_V dT_1 Since the gas is monoatomic and n_1 = 1 mole, C_V = 3 2 R. dT_1 = - 2 dQ 3R For section S_2, the pr

Question 2

Two identical plates P and Q , radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures $ T _ P $ and $ T _ Q $, respectively, with \( T _ Q

Accepted Answer

Initially : W _0= ( T _ P ^4- T _ Q ^4 ) Finally : Putting heat currents equal in steady state : aligned & ( T _ P ^4- T _1^4 )= ( T _1^4- T _2^4 ) \ & ( T _1^4- T _2^4 )= ( T _2^4- T _ Q ^4 ) aligned Adding : T _ P ^4- T _1^4= T _2^4- T _ Q ^4 T _1^4+ T _2^4= T _ P ^4+ T _ Q ^4 and T _1^4- T _2^4= T _ P ^4- T _1^4 Adding : T _1^4= 2 ~T _ P ^4+ T _ Q ^4 3 So W _ S = ( T _ P ^4- T _1^4 ) = ( T _ P ^4- ( 2 ~T _ P ^4+ T _ Q ^4 3 ) )= ( T _ P ^4- T _ Q ^4 3 ) hence W _ S W _0 =3

Question 3

The specific heat capacity of a substance is temperature dependent and is given by the formula C=k T, where k is a constant of suitable dimensions in SI units, and T is the absolute temperature. If th

Accepted Answer

aligned & T _ i =-73^ C =200 ~K \ & ~T _ f =27^ C =300 ~K aligned aligned & Q = msdT \ & = 1 kTdT \ & = kTdT = K _ 200 ^ 300 TdT \ & = K 2 [ ~T ^2 ]_ 200 ^ 300 = K 2 [300^2-200^2 ] \ & Q =25000 ~K aligned Hence n=25000

Question 4

Match the temperature of a black body given in List- I with an appropriate statement in List- II , and choose the correct option. [Given: Wien&rsquo;s constant as  2 . 9 &#215; 10 - 3 &#160; m -

Accepted Answer

Maximum wavelength for each case: P &#160; 2000 &#160; K &#955; m T = b &#8658; &#955; m = b T = 2 . 9 &#215; 10 - 3 2000 = 1 . 45 &#215; 10 - 6 &#160; m = 1450 &#160; nm ( Q ) &#160; 3000 &#160; K &#955; m T = b &#8658; &#955; m = 2 . 9 &#215; 10 - 3 3000 = 966 . 66 &#160; nm ( R ) &#160; 5000 &#160; K &#955; m T = b &#8658; &#955; m = 2 . 9 &#215; 10 - 3 5000 = 580 &#160; nm ( S ) &#160; 10000 &#160; K &#955; m T = b &#8658; &#955; m = 2 . 9 &#215; 10 - 3 10 , 000 = 290 &#160; nm Now for List-

Question 5

A small object is placed at the centre of a large evacuated hollow spherical container. Assume the container is maintained at  0 &#160; K . At time  t = 0 , the temperature of the object is&

Accepted Answer

Using Stefan-Boltzmann law, P = d Q d t = &#963; e A T B 4 &#8722; T S 4 , &#160; &#160; &#160; &#160; &#160; &#160; &#8757; T S = 0 &#160; K So, d Q d t = e &#963; A T 4 , &#160; &#160; &#160; &#160; consider , &#160; T = T B also,     d Q d t = - m s d T d t = e &#963; A T 4 - &#8747; 200 T d T T 4 = &#8747; 0 t k d t , &#160; where &#160; k = &#963; A m s &#8658; 1 T 3 &#8722; 1 200 3 = 3 k t Now,  For case  1 1 100 3 &#8722; 1 200 3 = 3 k t 1 &#160; &#160; &#160;

Question 6

A container with  1 &#160; kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the su

Accepted Answer

P i = 700 &#215; 0 . 05 = 35 watts P i = P r 35 = e &#963; A 4 T 0 3 ( &#916; T ) &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . 1 &#8756; K = 4 e &#963; A T 0 3 m C = 0 . 0001 4 e &#963; A T 0 3 = 0 . 001 &#215; m C = 4 . 2 from equation (1) &#916; T = 35 4 . 2 = 50 6 = 25 3 = 8 . 33

Question 7

The filament of a light bulb has surface area 64 &#160; mm 2 . The filament can be considered as a black body at temperature 2500 &#160; K emitting radiation like a point source when viewed from far.

Accepted Answer

P = &#963; A e T 4 P = 5 . 6 &#215; 10 - 8 &#215; 64 &#215; 10 - 6 &#215; 1 &#215; ( 2500 ) 4 P = 14175 &#215; 10 - 14 &#215; 10 8 &#215; 10 4 (a)  P = 141 . 75 &#160; W (b)  &#963; A e T 4 4 &#960; ( 100 ) 2 &#215; &#960; 3 &#215; 10 - 3 2 = 141 . 75 &#215; 9 &#215; 10 - 6 4 &#215; 10 4 318 . 937 &#215; 10 - 10 3 . 18937 &#215; 10 8 &#160; W (c)  &#955; T = b &#955; = 2 . 93 &#215; 10 - 6 2500 = 1160 &#160; nm (d)  3 . 18937 &#215; 10 - 8 = n sec h c &#955; 3 . 18937 &#215;

Thermal Properties of Matter — Previous Year Questions & Solutions