Tiêu đề 14. Transmission of Heat Med Ans.pdf ✅

1. (a) Temperature gradient 50 2 1 125 − 25 =  − =   = x x    C cm o = 2 / . 2. (a) According to problem, rate of heat loss in both rods are equal i.e. 1 2        =      dt dQ dt dQ  2 2 2 2 1 1 1 1 l K A l K A   =   K1 A1 = K2 A2 [As  1 =  2 = (T1 – T2) and 1 2 l = l given] 3. (a) l KA dt dQ  = , For both rods K, A and  are same  dt l dQ 1  So   2 2 ( / ) ( / ) = = = r r l l dQ dt dQ dt semicircular straight straight semi circular . 4. (d) Cooking utensil should conduct maximum and absorb minimum heat so it should possess high conductivity and low specific heat. 5. (c) From l KA dt dQ  =  dt dQ K A l    = 4000 400 (100 10 ) 0.1 4    = − = 100oC 6. (c)         = x K A dQ / dt  Rate of flow of heat per unit area = Thermal conductivity  Temperature gradient Temperature gradient (X) Thermal conductivi ty(K) 1  As KC > Km > Kg therefore XC  Xm  Xg . 7. (a) As the temperature of room remain constant therefore the rate of heat generation from the heater should be equal to the rate of flow of heat through a glass window t l t KA R V J . 1 2  =          2 2 0.2 10 0.2 1 (20 ) 20 (200 ) 4.2 1 −    −  =    = 15.24oC [where  = temperature of outside] 8. (b) Rate of flow of heat or power (P) x KA   =  x K R T  = 2 4  Thickness of shell P R KT x 2 4  = . 9. (b) Temperature of interface 1 2 1 1 2 2 K K K K + + =    Substituting R A l K 1 1 = and R A l K 2 2 = we get 1 2 1 2 2 1 R R R R + + =    . 10. (c) Let the thermal resistance of each rod is R Effective thermal resistance between B and D = 2R Temperature of interface 1 2 1 2 2 1 R R R R + + =    140 . 3 420 2 20 2 200 C R R R R o = = +  +   = 11. (b) KA = 2KB = 2 K Heat current through both layers will be same 36 – T d       KAA = T O– d       KBA (36 – T) 2 K = 72 3 T K T = = 24 T = temp diff = 36 – 24 = 12 A B C D R R R R R R A B R D 2R 200oC 20oC 
12. (c) dQ T KA dt  =  4000 = 4 400 100 10 0.1 T −      T = 100°C 13. (d)  = 1 –  thRth1 = 1 – 2 1 1 1 2 Th Th Th R R R    −      − = 2 1 1 1 1 2 1 1 2 K A K A K A d d d        − −      +     = 2 1 1 1 1 2 1 1 2 K K K d d d        − −      +     = 1 1 2 2 2 1 1 2 2 1 K d K d K d K d   + + 14. (d)Utensil should have low thermal resistance R KA     =   and low specific heat so that heat loss is less 15. (c) 1 2 1 1 1 2 2 2 2 2 (2 ) 9 2 8 (3 ) R K A K r R K A K r   = = =   = T R     1 R So, 1 2 2 1 8 9 R R  = =  16. (b) T = 100 – R1 = 100 – 1 1 2 100 20 R R R   −     + = 100 – 80 5 5 3 KA KA KA         +   = 70°C 17. (b)  th = T 100 0 R KA  − = = 100 × 100 × 100 × 10–4 = 100 W/sec = 60 × 100 W/min = 6 × 103 W/min. 18. (d) K depends on material of metal only 19. (b) 1 1 Q t = H1 i = 100 0 2R − = 50 R i H2 = 100 R /2 = 200 R = 2 2 Q t Q1 = Q2 = 10 cal. 50 (2) R  = 2 200 t R  t 2 = 1 2 min. 20. (d) Thermal resistance should be equal, so K A K A 1 1 2 2 =  K1A1 = K2A2 21. (b) Equivalent thermal circuit Req = R1 + R2 = 2 KA = K A K A 1 2 +  K = 1 2 1 2 2K K K K+
22. (c) Equivalent thermal circuit 1 2 1 1 1 R R R eq = +  K A eq 2 KA KA 2 = +  3 2 K K eq = 23. (d) For a rod of length L and area of cross-section A whose faces are maintained at temperatrure T1 and T2 respectively. Then in steady state the rate of heat flowing from one face to the other face in time t is given by The curved surface of rod is kept insulated from surrounding to avoid leakge of heat dQ KA T T ( 1 2 – ) dt L = 24. (a) Square is made of four rod of same material, so the temperature difference of second diagonal will also be100°C. 25. (b)Let q be the temperature of the junction (say B). Thermal resistance of all the three rods is equal. Rate of heat flow through AB + Rate of heat flow through CB = Rate of heat flow through BD  90 R − + 90 R − = 0 R  − Here R = Thermal Resistance  3  = 180 or  = 60oC 26. (d) 1 Q R  In the first case rods are in parallel and thermal resistance is 2 R while in second case rods are in series and thermal resistance is 2R. 27. (a)Net heat given/sec = 1000 – 160 = 840 J/S if it takes a time t then 840 t = 2000 × 4.2 × (77 – 27) t = 500 sec = 8 min 20 sec. 28. (a) Thermal conductivity depends on the material, not on temperature difference. 29. (b) Q1 = 2 T k r         , Q2 = 2 2 (2 ) T k r          2 1 2 Q Q = 30. (c) Let temperature of the interface = T 1 1 1 T T– L AK       = 2 2 2 T T– L K A        T 1 2 1 2 L L K K     +   = 1 2 2 T L K + 2 1 1 TL K  T = 1 1 2 2 1 2 1 2 2 1 T K L T L K L K L K + + 31. (b) TAB = 200 – RA = 200 – (200 18) A A B C R RRR −  + + = 200 – 182 2 2 3 KA KA KA KA  + + = 116°C 32. (c) TBC = 200 – RA – RB = 200 – 182 2 2 2 3 KA KA KA KA KA    +     + + = 74°C 33. (a) 1 2 2 1 R R  =  = 2 2 2 K A KA  = 2   2 = 1 2  = 4 2 = 2 34. (c) dQ dT KA dt dx =  1 dQ dT K A dt dx =
1 dQ A dt is same so dT dx is smaller for higher K. 35. (b) T = 100 – R1 = 100 – 1 1 2 ( ) T R R R   + = 100 – 1 1 2 1 1 2 100 K A K A K A  + After substituting the values T = 60°C 36. (c)   for point source. '  = 2 2 ' 2 d d d d         =     = 1 4  ' = 4  37. (a) Copper has higher thermal conductivity, so lower resistance. 38. (a) In 1st case   = 1 2 20 4 T R R  = + cal/min  2 T R  = 5 cal/min. (since R1 = R2 = R)  T R  = 10 cal/min. In second case  2 = 1 2 1 2 T RR R R  +   2 = 2 T R   20 t =  2 = 2 × 10  t = 1 min. 39. (b)It`s a parallel Combination R1 = 1 d K A R2 = 2 d K A 1 Req = 1 1 R + 2 1 R + .................upto nth 1 Req = 1 2 n R + 2 2 n R = 2 n 1 2 1 2 R R R R   +     Req = 1 2 1 2 2( ) ( ) R R n R R +  ( / ) ( ) eq d n K A = 1 2 1 2 2 1 1 d d K A K A d n A K K           +   1 Keq = 1 2 2 K K+  Keq = 1 2 2 K K+ 40. (a) For balanced wheatstone bridge 1 2 3 4 R R R R =  3 4 1 2 K K K K = K1K4 = K2K3 41. (d) Req = R1 + R2 + R3 where R1 = (2 ) k A , R2 = kA , R3 = 2 kA Thermal current through rods 100 0 Req − = 1 1 100 T R − = 2 1 2 100 T R R − + = 2 3 T 0 R −  100 0 7 2kA − = 1 100 2 T kA − = 2 100 3 2 T kA − = 2 0 2 T kA − after solving T1 = 86°C, T2 = 57°C 42. (c) Thermal Resistance R = 2 1 2 .4 r r dr k r   = 1 4 k 1 2 1 1 – r r       2 1 1 2 ( – ) 4 . r r  k r r Rate of heat flow = T T 2 1 – R = 1 2 2 1 ( – ) ( – ) T T r r . 4k r1 r 2