PHYSICS Class
II IIT-JEE Achiever 2016-17 Intensive Revision Program Solution to Worksheet - 13
Topics
Heat Transfer and Thermal Radiation
Date:
19-12-2016
Multiple choice questions with one correct alternative
1. Two bodies A and B having temperature 327 °C and 427 °C are radiating heat to the surroundings. The surroundings temperature is 27 °C. The ratio of heat radiation of A to that of B is (A) 5.2
(B) 0.32
(C) 0.52
(D) 3.2
Ans (C) Ans (C) According to Stefan’s law 4 4 E = σ ( T − T0 )
4
4
2
4
E1 = σ [(600) − (300) ] E2 = 2 [(700) − (300) ] E1 (600) 4 − (300) 4 E2
=
(700) 4 − (300) 4
= 0.52
2. Three rods of same dimensions have thermal conductivities 3K, 2K and K respectively. They are arranged as shown below 100°C
3K
50°C
T
2K
K T 0°C
What will be the temperature “T” of the junction? 200 100 °C °C (A) (B) 3 3
(C) 75°C
(D)
50 3
°C
Ans (A) Ans (A) In equilibrium, Rate of flow of heat through rod 1 = Sum of rate of flow of heat through rods 2 and 3 dQ dQ dQ i.e., = + dt 1 dt 2 dt 3 or (3k) (100 −T) = (2k) (T – 50) + k (T − 0) 200°C or T = or 400 = 6T or 3 3. The ends A of a rod AB of length 1 m is maintained at 100 °C and the end B at 10°C. The temperature at a distance of 60 cm from the end B is (A) 64°C
(B) 36°C
(C) 46°C
Ans (A) Ans (A) Rate of flow of heat
2IIT1617PPWS13S
1
(D) 72°C
θ t
∴
=
KA(θ1 − θ2 )
d KA(100 − θ)
=
40
100 − θ 2
=
KA(θ − 10) 60
θ − 10 3
300 − 3θ = 2θ − 20 5θ = 320
θ = 64°C 4. A wall has two layers A and B, each made of different materials. Both layers are of same thickness. But, the thermal conductivity of material A is twice that of B. If, in the steady state, the temperature difference across the wall is 24 °C, then the temperature difference across the layer B is (A) 8 °C
(B) 12 °C
Ans (C) Ans (C) k A ( θ1 − θ ) t l
=
(C) 16 °C
(D) 20 °C
k B ( θ − θ2 ) t l
That is k A (θ1 − θ) = k B(θ2 − θ) Given that k A= 2k B, therefore 2( θ1 − θ) = (θ − θ2) Also, (θ1 − θ) + (θ − θ2) = 24 Solving these equations, we find (θ − θ2) = 16 °C 5. Heat is flowing through two cylindrical rods of same material. The diameter of the rods are in the ratio of 1 : 2 and lengths are in the ratio of 2 : 1. If the temperature temperature difference across the ends ends of the rods is same, then the ratio of heat conducted per second by them will be (A) 1 : 1
(B) 1 : 2
(C) 1 : 4
(D) 1 : 8
Ans (D) Ans (D)
πD12 k ( T1 − T2 ) t 4 Q1 = l1
πD22 k ( T1 − T2 ) t 4 Q2 = l2
Therefore,
Q1 Q2
=
D12 l2 D
2 2 l1
=
1 4
×
1 2
=
1 8
6. Two spheres of radii R 1 and R2 have densities ρ1 and ρ2 and specific heat C 1 and C2. If they are heated to the same temperature, the ratio of their rates of cooling will be R ρ C Rρ C R ρC (A) 2 2 2 (B) 1 2 2 (C) 2 1 2 R 1ρ1C1 R 2 ρ1C1 R 1ρ 2 C1
(D)
R 2ρ2 C1 R 1ρ1C 2
Ans (A) Ans (A) Rate of emission of radiations per unit area per second will be same for both the spheres. That is
2IIT1617PPWS13S
2
Q1
Q2
=
( 4πR ) ∆t ( 4πR 22 ) ∆t 2 1
or
m1C1∆θ1
m 2C 2 ∆θ 2
=
( 4πR 12 ) ∆t ( 4πR 22 ) ∆t
4 3 4 3 ∆θ 2 πR 1 ρ1C1∆θ1 πR 2ρ 2C 2∆θ 3 3 or = 2 2
( 4πR1 ) ∆t
( 4πR 2 ) ∆t
∆θ1 R ρC Hence ∆t = 2 2 2 ∆θ2 R1ρ1C1 ∆t 7. A body cools in 7 minutes from 60 °C to 40°C. What time (in minutes) does it take to cool from 40 °C to 28°C if the surrounding temperature in 10°C? Assume Newton’s law of cooling is valid (A) 3.5
(B) 10
(C) 7
(D) 10
Ans (C) Ans (C) Rate of cooling ,
dt 60 − 40
st
In 1 case, or K =
dQ
7
∝ temperature difference
= K(60 − 10)
2 35
nd
In 2 case,
40 − 28 t
= K(40 − 10)
Hence t = 7 minutes. 8. The energy spectrum of a black body exhibits a maximum around a wavelength λ0. The temperature of the black body is now around a wavelength 3 λ0 /4. The power radiated by one black body will now increase by a factor of 64 (A) 27
(B)
256 81
(C)
4 3
Ans (B) Ans (B) Let at temperature T, black body emits energy E at wavelength λ0.
⇒ E ∝ (T)4
…(i)
(From Stefan’s law) Also, from Wein’s displacement law, 1 λ′ ∝ T
…(ii)
Thus from (i) and (ii), we get 4
1 E∝ λ0
…(iii)
Hence eqn (iii), we get 4
1 1 E′ ∝ or E′ ∝ ′ λ 3 λ / 4 0 0
…(iv)
From (iii) and (iv), we get 2IIT1617PPWS13S
3
(D)
16 9
256 E 81
E′ =
9. The ends of two rods of different materials with their thermal conductivities, radii of cross sections and lengths all in the ratio 1 : 2 are maintained at the same temperature difference. If the rate of flow of heat in the larger rod is 4 cal/sec, then that in the shorter rod will be (A) 1 cal/sec
(B) 2 cal/sec
(C) 8 cal/sec
(D) 16 cal/sec
Ans (A) Ans (A) The amount of heat transmitted through a conductor is given by KA∆θt Q= l
⇒ Rate of transmission of heat, Q
r=
t
=
KA ∆θ l
Since, temperature difference ∆θ = constant. Therefore, we have KA R∝ l
R1
⇒ or
R2 R1 R2
K1A1 / l1
=
K 2 A 2 / l2
K1 A1 l2 K 2 A2 l1
=
2
k r l = 1 1 2 therefore, R 2 k 2 r2 l1 R1
Substituting,
k1 k2
1 r1
= ;
2 r2
R2 = 4 cal s we get, R1 = 1 cal s
1
l2
2
l1
= ;
2 1
=
−2
−1
10. Five rods of same dimensions dimensions are arranged as shown shown in the figure. They have thermal conductivities k 1, k 2, k 3, k 4 and k 5 when points A and B are maintained at different temperatures. No heat flows through the C
central rod if k 1
(A) k 1k 4 = k 2k 3 (B) k 1 = k 4 and k 2 = k 3 k k (C) 1 = 2 k 4 k 3
k 2
A
k 5 k 3
k 4 D
(D) k 1k 2 = k 3k 4 Ans (A) Ans (A) No heat flows through the central rod, if we have K1 K 3 K2
=
K4
⇒ K1K4 = K2K3
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B
4
11. Three identical thermal conductors are connected as shown in the figure. Considering no heat loss due to radiation, the temperature at the junction is
60 °C
(A) 60°C
Q1
θ
(B) 20°C
Q
(C) 50°C
70 °C
(D) 10°C
20 °C
Q2
Ans (C) Ans (C) Suppose temperature of junction is θ. As no heat loss occurs, so KA KA KA (60° − θ) + (70° − θ) = (θ − 20) L L L
⇒ 60° – θ + 70° − θ = θ = 20° 130° − 2θ = θ – 20° 3θ = 150°
θ = 50° 12. The radiation emitted by a star A is 10,000 times than of the sun. If the surface temperature of the sun and the star are 6000 K and 2000 K respectively, the ratio of the radii of the star A and the sun is (A) 300 : 1
(B) 600 : 1
(C) 900 : 1
(D) 1200 : 1
Ans (C) Ans (C) From Stefan’s law, we have 4
E = A e σ T
⇒ E ∝ A T4 2
4
2
or E ∝ r T
⇒
E1 E2
=
[∵ A = πr ]
r12T14 2
4
r2 T2
⇒
r1 r2
=
E1 T2
2
E2 T1
Substituting E1 = 10,000 E 2; T1 = 6000 K; r 900 T1 = 2000K; we get 1 = r2 1 13. One end of a thermally insulated rod is kept at a temperature T 1 and the other at T 2. The rod is composed of two sections of lengths l1 and l2 and thermal conductivities K1 and K2 respectively. The temperature at the interface of the two sections is (K1l1T1 + K 2l2T2 ) (A) (K1l1 + K 2l2 ) (B)
(K 2l2T1 + K 1l1T2 )
(C)
(K 2l1T1 + K 1l2T2 )
(D)
(K1l2T1 + K 2l1T2 )
(K1l1 + K 2l2 ) (K 2l1 + K 1l2 ) (K1l2 + K 2l1 )
Ans (D) Ans (D) Two sections of thermally insulated rod are in series, therefore rate of flow of heat in them is K1A (T1 − T) K 2A( A (T − T2 ) l1 2IIT1617PPWS13S
=
l2 5
Where T is the temperature of the interface.
∴ K1l2 (T1 − T) = K 2l1 (T −T2) or K1l2T1 − K1l2T = K2l1T − K2l1T2 or K1l2T1 + K2l2T2 = (K1l2 + K2l1) T K l T + K 2l1T2 Hence T = 1 2 1 K1l2 + K 2l1 14. Two identical rods AC and CB made of two different metals having thermal conductivities in the ratio 2 : 3 are kept in contact with each other at the end C as shown in the figure. A is at 100 °C and B is at 25°C. Then the junction C is at (A) 55°C (B) 60°C (C) 75°C (D) 50°C Ans (A) Ans (A) For conduction of heat K (θ − θ2 ) t Q= A 1 d Let temperature of junction θ
∴ K1 (100 − θ) = K2 (θ − 25) K1 θ − 25 = K 2 100 − θ θ − 25 3 100 − θ 3θ − 75 = 200 – 2 θ 5θ = 275 2
=
θ = 55°C 15. Two spheres of radii 8 cm and 2 cm are cooling. Their temperatures are 127 °C and 527°C respectively. Find the ratio of energy radiated by them at the same time. (A) 0.06
(B) 0.5
(C) 1
Ans (C) Ans (C) T1 = 127°C = 273 + 127 K = 400 K T2 = 527°C = 273 + 527 K = 800 K 4
Now, Q = A σ0, T t Q Q ⇒ ∝ AT4 or, ∝ r 2T 4 t t 2
∵ A = 4πr 2
4
r1 T1 8 2 400 4 ∴ = = Q 2 r2 T2 2 800 Q1
4
1 1 = (4) × = 16 × = 1 16 2 2
2IIT1617PPWS13S
6
(D) 2
DPP 16. A spherical shell of inner radius R1 & outer shell R2 is having variable thermal conductivity given by K = a0Tx where a0 is constant, T is temperature in Kelvin & r is the distance from centre. Two surfaces of shell are maintained at temperature T1 (for inner surface) & T 2 (outer surface) respectively. (T 1 > T 2). Heat current flowing through the shell would be 4πa 0 (T12 − T22 ) × R1R 2 (A) R 2 − R1 (C)
4πa 0 (T ( T1 − T2 )R )R 1R 2 R 2 − R1
(B)
4πa 0 R 12 R 22 (T (T12 − T22 )
(D)
4πa 0 (T (T12 − T22 )( )(R 1 + R 2 )2
R 22 − R12 R 2 − R1
Ans (B) Ans (B) Temperature is decreasing as we are going out, let at a distance x from the centre, the temperature dT gradient is − dx dT At this location, K = a 0 Tx from H = −a 0 Tx × 4πx 2 × dx dT ⇒ H = −a 0Tx × 4πx 2 × dx
⇒
R2
∫
R1
Hdx x3
⇒H=
T2
∫
= − 4πa 0 TdT T1
4πa 0 R 12 R 22 (T (T12 − T22 ) R 22 − R 12
17. Three rods of material x and three rods of material y are connected as shown in the figure. All the rods are of identical length and cross-section area. If the end A is maintained at 60°C and the junction E at 10°C, calculate the temperature of the junction D. The thermal conductivities of x and y are 0.92 and 0.46 SI units. (A) 20°C
(B) 30°C
(C) 40°C
(D) 35°C
Ans (A) Use the principle total heat entering at a junction is equal to total heat leaving that junction. If T 1, T2 and T3 are the temperature at B, C and D respectively, write equations at these junctions. Simplify these to obtain. 4T1 − 2T2 − T3 = 60 2T1 − 6T2 + 2T3 = − 20 T1 + 2T2 − 4T3 = −10 The temperatures: at B 30°C and at C, D 20 °C. 18. The figure shows a system of two concentric spheres of radii r 1 and r2 and kept at temperatures T1 and T2 respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to r1r2 (A) (r2 − r1 ) (C)
(r2 − r1 ) r1r2
2IIT1617PPWS13S
(B) (r2 – r1)
r2 r1
(D) log
7
Ans (A) Ans (A)
19. Assuming the sun to be a spherical body of radius R at a temperature of T K, evaluate the total radiant power, incident on the earth, at a distance r from the sun, is 4πr02 R 2σT 4 πr02 R 2σT 4 R 2 σT 4 (A) (B) (C) r2 r2 r2
(D)
r02 R 2 σT 4 4πr 2
Ans (C) Ans (C) Energy radiated per second by the sun: 4
E = σT 4πR
2
This energy falls uniformly on the inner surface of spheres centred around the sun. If r is the distance of the earth from the sun, then energy falling per second on unit area of the sphere of radius r is 2πR 2 σT 4 σR 2T 4 2πr 2
=
r2
It is only from the front hemisphere of the sun that the energy is falling on the earth and it is only on the front half of the earth. 4πR 2 4πr 2 Therefore, and have to be used 2 2 The radiant power incident on the earth is given by σR 2 T 4 πr02 R 2σT 4 2 Q = πr0 × = r2 r2 2
20. An electric heater is placed inside a room of total wall area 137 m and maintained at a temperature 20°C inside, outside temperature – 10°C. The walls are made up of three composite materials. Innermost layer is made up of wood of thickness 2.5 cm, middle layer is of cement of thickness 1 cm and the exterior layer is 25 cm thick. Assuming there is no loss of heat through any other way, the power of electric 2
2
heater is [the thermal conductivity of wood = 0.125 W/m °C, cement = 1.5 W/m °C and 2
brick = 1 W/m °C] (A) 9000 W
(B) 8500 W
(C) 8800 W
Ans (A) Ans (A)
2IIT1617PPWS13S
8
(D) 9400 W
21. A rod of length l with thermally insulated lateral surface is made of a material whose thermal C conductivity K varies as K = , where C is a constant. The ends are at temperatures T 1 and T 2. The heat T flow density is (A) Clog
T2 T1
(B)
C l
T2 T 1
log
(C)
C l
log(TT 1 2)
T2 T1
(D) Cl log
Ans (B) Ans (B)
22. A planet radiates heat at a rate proportional to the fourth power of its surface temperature T. If such a steady temperature of the planet is due to an exactly equal amount of heat received from the sun then which of the following statement is true? (A) The planet’s surface temperature varies inversely as the distance of the sun. (B) The planet’s surface temperature varies directly as the square of its distance from the sun. (C) The planet’s surface temperature varies inversely as the square root of its distance from the sun. (D) The planet’s surface temperature is proportional to the fourth power of its distance from the sun. Ans (C) Ans (C) 4
Rate of loss of energy by unit area of the planet = σT , where σ is the Stefan’s constant. Let Q be the total energy emitted by the sun in every second. If d is the distance of the planet from the sun, then Q falls uniformly over the inner surface of the sphere Q of radius d. Rate of gain of heat by unit area of planet = 4πd 2 For steady temperature of planet Q σT 4 = 4πd 2 4
T =
2IIT1617PPWS13S
1/4
Q 4πσd
2
or
Q T= 2 4πσd
or
9
T∝
1 d
23. What is the rate of flow of heat through a tapering rod of length l tapering from radius r 1 to r2, when the temperature of the ends are θ1 °C and θ2 °C and coefficient of thermal conductivity is K? (A)
πKr1r2 (θ1 − θ 2 ) l
(B)
Kr1r2 (θ1 − θ 2 ) l
(C)
2 2 Kr1 r2 (θ1 − θ2 )
l
(D)
K (r1 − r2 )(θ1 − θ2 ) l
Ans (A) Ans (A) Let O be the apex of the rod. Consider a section at a distance x from O and let x 1 and x2 be the distances to ends from O.
24. A sphere and a cube of same material and same total surface area are placed in the same evacuated space turn by turn after they are heated to the same temperature. Find the ratio of their initial rates of cooling in the enclosure. (A)
π 6
:1
(B)
π 3
:1
(C)
Ans (A) Ans (A)
2IIT1617PPWS13S
10
π 6
:1
(D)
π 3
:1
25. A planet is at an average distance d from the sun and its average surface temperature is T. Assume that the planet receives energy only from the sun and loses energy only through radiation from its surface. –n
Neglect atmospheric effects. If T ∝ d , the value of n is (A) 2
(B) 1
(C)
1 2
(D)
1 4
Ans (C) Ans (C) Let P be the power radiated by the sun and R be the radius of planet. 2
4
Energy radiated by planet = 4πR (σT ) For thermal equilibrium P × πR 2 = 4πR 2 (σT 4 ) 2 4πd 1 ∴ T4 ∝ 2 or T ∝ d −1/2 d 1 Hence, n = 2 26. Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle ABC right angled at B. The points points A and B are maintained at at temperatures temperatures T and
2T
respectively in the steady state. Assuming that only heat conduction takes place, temperature of point C will be 3T (A) 2 +1 T (B) 2 +1 T (C) 3 2 −1
(
)
T
(D)
2 −1
Ans (A) Ans (A) As TB > TA, heat flows from B to A through both paths BA and BCA Rate of heat flow in BC = Rate of heat flow in CA KA
(
2T − Tc l
)
=
KA(Tc − T ) 2l
Solving this, we get, Tc =
2IIT1617PPWS13S
3T 2 +1 11
27. A body cools from 62°C to 50°C in 10 minutes and to 42°C in the next 10 minutes. The temperature of the surrounding is (A) 16°C
(B) 26°C
(C) 36°C
(D) 21°C
Ans (B) Ans (B) Let the temperature of the surrounding be θ0. According to Newton’s law of cooling, 62 − 50 62 + 50 = K − θ0 10 2
112 − θ0 10 2 50 − 42 50 + 42 and = K − θ0 10 2 8 92 or = K − θ0 10 2 or
12
= K
divide (1) by (2), we get 12 10 112 − 2θ0 10
×
8
=
or
92 − 2θ0
or 276 – 6θ0 = 224 – 4 θ0 52 or θ0 = = 26°C 2
or
... (1)
... (2)
3 2
=
112 − 2θ0 92 − 2θ0
276 – 224 = 2θ0
Read the passage given below and answer questions by choosing the correct alternative
A highly conducting solid sphere of radius R, density ρ and specific heat s is kept in an evacuated chamber. A parallel beam of thermal radiation of intensity I is incident on its surface, consider the sphere to be perfectly black body and its temperature at certain instant considered as t = 0 is T 0. Take Stefan’s constant as σ. 28. The equation which gives the temperature T of the sphere as a function of time is t
T
⌠ 3dt (A) ⌠ = 4 ⌡ I − 4 σ T ⌡ 4 Rρ s T0
0
T
(C) ⌠
T
t
T0
0
dT ⌠ 3dt (B) ⌠ = − 4 ⌡ 4σT ⌡ 4Rρs
dT
T
dT
⌡ I − 4σT
4
=
3t 8Rρs
(D) ⌠
3dT
⌡ I − 4σT
T0
4
=
5t 8Rρs
T0
Ans (A) Ans (A) 2
The rate at which energy is absorbed by sphere is, P 1= πR I Let at any time t, the temperature of the sphere is T, then rate at which heat has been radiated is 2
P2 = σ × 4πR T
4 2
2
Net absorption rate, P = πR I − 4πσ R T dT ⇒ ms = πR 2 [1 − 4σT 4 ] dt dT 3dt πR 2 dt = = ⇒ I − 4σT 4 4 4Rρs πR 3ρs 3
2IIT1617PPWS13S
4
12
t
T
⌠ 3dt ⇒ ⌠ = 4 ⌡ I − 4σT ⌡ 4Rρs dT
T0
0
Solving above equation you get T as a function of t. 29. Maximum attainable temperature of the sphere is 1/2
I (A) 4σ
1/3
I (B) 2σ
1/4
I (C) 4σ
Ans (C) Ans (C) 1/4
I For maximum temperature, get Tmax = = 0 we get dt 4σ dT
***
2IIT1617PPWS13S
13
(D) Never occurs
