28.HEAT TRANSFER by Aditya Bhardwaj

CHAPTER 28

HEAT TRANSFER 1.

t1 = 90°C, t2 = 10°C –3 l = 1 cm = 1 × 10 m 2 –2 2 A = 10 cm × 10 cm = 0.1 × 0.1 m = 1 × 10 m K = 0.80 w/m-°C

10 cm

KA (1   2 ) 8  10 1  1 10 2  80 Q = = = 64 J/s = 64 × 60 3840 J. t l 1  10  2 2 t = 1 cm = 0.01 m, A = 0.8 m 1 = 300, 2 = 80 K = 0.025, KA(1   2 ) Q 0.025  0.8  (30030 ) = = = 440 watt. t l 0.01 2 K = 0.04 J/m-5°C, A = 1.6 m t1 = 97°F = 36.1°C t2 = 47°F = 8.33°C l = 0.5 cm = 0.005 m

1 cm

KA (1   2 ) 4  10 2  1.6  27.78 Q = = = 356 J/s t l 5  10  3 2 –4 2 A = 25 cm = 25 × 10 m –3 l = 1 mm = 10 m K = 50 w/m-°C Q = Rate of conversion of water into steam t 10 1  2.26  10 6 100  10 3  2.26  10 6 4 = = = 0.376 × 10 1 min 60 KA (1   2 ) 50  25  10 4  (  100 ) Q 4 =  0.376 ×10 = t l 10  3 =

10 3  0.376  10 4 50  25  10

4

10 5  0.376 = 30.1 ≈ 30 50  25

K = 46 w/m-s°C l=1m 2 –6 2 A = 0.04 cm = 4 × 10 m 5 Lfussion ice = 3.36 × 10 j/Kg

46  4  10 6  100 Q –8 –5 = = 5.4 × 10 kg ≈ 5.4 × 10 g. t 1 A = 2400 cm2 = 2400 × 10–4 m2 –3 ℓ = 2 mm = 2 × 10 m K = 0.06 w/m-°C 1 = 20°C 2 = 0°C

0°C

KA(1   2 ) Q 0.06  2400  10 4  20 –1 = = = 24 × 6 × 10 × 10 = 24 × 6 = 144 J/sec 3 t  2  10 144 144  3600 m Q = = Kg/h = Kg/s = 1.52 kg/s. Rate in which ice melts = t t L 3.4  10 5 3.4  10 5 –3 ℓ = 1 mm = 10 m m = 10 kg 2 –2 2 A = 200 cm = 2 × 10 m 6 Lvap = 2.27 × 10 J/kg K = 0.80 J/m-s-°C 28.1

100°C

Heat Transfer 6

dQ = 2.27 × 10 × 10,

2.27  10 7 dQ 2 = = 2.27 × 10 J/s dt 10 5 Again we know

dQ 0.80  2  10 2  ( 42  T ) = dt 1 10  3 So,

3

= 2.27 × 10

10  16 × 42 – 16T = 227  T = 27.8 ≈ 28°C K = 45 w/m-°C –2 ℓ = 60 cm = 60 × 10 m 2 –4 2 A = 0.2 cm = 0.2 × 10 m Rate of heat flow, =

8  2  10 3 ( 42  T )

Q2 = 20°

Q1 = 40°

KA (1   2 ) 45  0.2  10 4  20 –3 = = 30 × 10 0.03 w  60  10  2 2

A = 10 cm ,

h = 10 cm

KA(1   2 ) Q 200  10 3  30 = = = 6000 t  1  10  3 Since heat goes out from both surfaces. Hence net heat coming out. Q Q  = = 6000 × 2 = 12000, = MS t t t   –3 –1  6000 × 2 = 10 × 10 × 1000 × 4200 × t  72000  = = 28.57 t 420 So, in 1 Sec. 28.57°C is dropped 1 Hence for drop of 1°C sec. = 0.035 sec. is required 28.57 –2 10. ℓ = 20 cm = 20× 10 m 2 –4 2 A = 0.2 cm = 0.2 × 10 m 1 = 80°C, 2 = 20°C, K = 385 KA(1   2 ) Q 385  0.2  10 4 (80  20) –4 –3 = = = 385 × 6 × 10 ×10 = 2310 × 10 = 2.31 (a) t  20  10  2 (b) Let the temp of the 11 cm point be   Q = 20°C l tKA  2.31 11 cm  = l 385  0.2  10  4   20 2.31  = 2 11  10 385  0.2  10  4

2.31  10 4  11 10  2 = 33 385  0.2   = 33 + 20 = 53 11. Let the point to be touched be ‘B’ No heat will flow when, the temp at that point is also 25°C i.e. QAB = QBC C KA(100  25) KA(25  0) So, = 100  x x  75 x = 2500 – 25 x  100 x = 2500  x = 25 cm from the end with 0°C

80°C

  – 20 =

28.2

100 cm B x

A 100–x

Heat Transfer 12. V = 216 cm a = 6 cm,

3 2

Surface area = 6 a = 6 × 36 m Q = 100 W, t

t = 0.1 cm

KA(1   2 ) Q = t   100 = K=

K  6  36  10 4  5 0.1 10  2 100

= 0.9259 W/m°C ≈ 0.92 W/m°C 6  36  5  10 1 13. Given 1 = 1°C, 2 = 0°C –3 K = 0.50 w/m-°C, d = 2 mm = 2 × 10 m –2 2 v = 10 cm/s = 0.1 m/s A = 5 × 10 m , Power = Force × Velocity = Mg × v KA(1   2 ) dQ = Again Power = dt d KA(1   2 ) So, Mgv = d

KA(1   2 ) 5  10 1  5  2 1 = = 12.5 kg. dvg 2  10  3  10 1  10 3 14. K = 1.7 W/m-°C ƒw = 1000 Kg/m 5 –2 Lice = 3.36 × 10 J/kg T = 10 cm = 10 × 10 m KA(1   2 ) KA(1   2 )  KA (1   2 ) Q 10 cm =  = = (a) t  t Q mL KA (1   2 ) 1.7  [0  ( 10)] = = At ƒ w L 10  10  2  1000  3.36  10 5 M=

–0°C

0°C

17 –7 –7  10  7 = 5.059 × 10 ≈ 5 × 10 m/sec 3.36 (b) let us assume that x length of ice has become formed to form a small strip of ice of length dx, dt time is required. dQ KA ( ) dmL KA ( ) Adx ƒL KA ( ) =  =  = dt x dt x dt x xdxƒL dx ƒL K (  )  =  dt = x dt x K (  ) =





dt =

ƒ L t xdx K( ) 0

Putting values



t=

ƒ L l 2 ƒ L  x 2    = K( )  2  K 2





3.36  10 6 1000  3.36  10 5  10  10 2 3.36 =  10 6 sec. = hrs = 27.45 hrs ≈ 27.5 hrs. 1.7  10  2 2  17 2  17  3600 15. let ‘B’ be the maximum level upto which ice is formed. Hence the heat conducted at that point from both the levels is the same. A Let AB = x –10°C K ice  A  10 K water  A  4 Q Q x ice = water  = i.e. 1 cm t t x (1  x ) t=

1.7  10 17 2 5  10 1  4 =  = x 1 x x 1 x 17  17 – 17 x = 2x  19 x = 17  x = = 0.894 ≈ 89 cm 19 

28.3

1–x C

4°C

Heat Transfer 16. KAB = 50 j/m-s-°c A = 40°C B = 80°C KBC = 200 j/m-s-°c KAC = 400 j/m-s-°c C = 80°C –2 Length = 20 cm = 20 × 10 m 2 –4 2 A = 1 cm = 1 × 10 m (a)

Q AB K  A ( B   A ) 50  1 10 4  40 = AB = = 1 W. t l 20  10  2

(b)

Q AC K  A(C   A ) 400  1 10 4  40 –2 = AC = = 800 × 10 = 8 t l 20  10  2

QBC K  A ( B   C ) 200  1 10 4  0 = BC = =0 t l 20  10  2 KA(1   2 ) 17. We know Q = d KA (1   2 ) KA (1   2 ) Q1 = , Q2 = d1 d2 (c)

KA (1  1 ) Q1 2r 2 r = = = [d1 = r, d2 = 2r] KA (1  1 ) Q2 r  2r 18. The rate of heat flow per sec. dQ A d = = KA dt dt The rate of heat flow per sec. dQB d = KA B = dt dt This part of heat is absorbed by the red. Q ms d = where = Rate of net temp. variation t dt dt d d d  msd d  d  = KA A  KA B  ms = KA  A  B  dt  dt dt dt dt  dt

d –4 = 200 × 1 × 10 (5 – 2.5) °C/cm dt d -4  0 .4  = 200 × 10 × 2.5 dt  0 .4 

200  2.5  10 4 d –2 = °C/m = 1250 × 10 = 12.5 °C/m dt 0.4  10  2 19. Given T2 - T1 = 90°C Krubber = 0.15 J/m-s-°C We know for radial conduction in a Cylinder 2Kl(T2  T1 ) Q = ln(R 2 / R1 ) t 

= 20.

2  3.14  15  10 2  50  10 1  90 = 232.5 ≈ 233 j/s. ln(1.2 / 1)

dQ = Rate of flow of heat dt Let us consider a strip at a distance r from the center of thickness dr. dQ K  2rd  d = [d = Temperature diff across the thickness dr] dt dr 28.4

120°C

50 cm

Heat Transfer C=

K  2rd  d dr

d   c  dr   

dr r

dr = K2d d r Integrating  C

C

 r1

dr = K2d r

2



 Clog r r2 = K2d (2 – 1) r

d

1

r   C(log r2 – log r1) = K2d (2 – 1)  C log  2  = K2d (2 – 1)  r1  K 2d( 2  1 ) C= log(r2 / r1 ) 21. T1 > T2 2 2 A = (R2 – R1 ) 2

KA (T2  T1 ) KA (R 2  R1 )(T2  T1 ) = l l Considering a concentric cylindrical shell of radius ‘r’ and thickness ‘dr’. The radial heat flow through the shell dQ d = – KA [(-)ve because as r – increases  H= dt dt decreases] d H = –2rl K A = 2rl dt

So, Q =



2LK dr =  H r

R2 T1

 d

Integrating and simplifying we get 2KL(T2  T1 ) 2KL( T2  T1 ) dQ = =  H= dt Loge(R 2 / R1 ) ln(R 2 / R1 ) 22. Here the thermal conductivities are in series, K 1A(1  2 ) K 2 A(1   2 )  KA (1   2 ) l1 l2  = K 1A(1   2 ) K 2 A(1   2 ) l1  l2  l1 l2

K1 K 2  l l2  1 K1 K 2  l1 l2 

L1 L2

K l1  l 2

K 1K 2 (K 1K 2 )(l1  l2 ) K = K= K 1l 2  K 2l1 l1  l2 K 1l2  K 2l1

23. KCu = 390 w/m-°C KSt = 46 w/m-°C Now, Since they are in series connection, So, the heat passed through the crossections in the same. So, Q1 = Q2 K  A  (  0) K  A  (100  ) Or Cu = St l l  390( – 0) = 46 × 100 – 46  436  = 4600 4600 = = 10.55 ≈ 10.6°C 436 28.5

0°C

Steel °C

100°C

Heat Transfer 24. As the Aluminum rod and Copper rod joined are in parallel

Q Q Q =      t t  1  Al  t  Cu

40°C 80°C

80°C

KA(1   2 ) K A(1   2 ) K 2 A(1   2 ) = 1  l l l  K = K1 + K2 = (390 + 200) = 590 

KA (1   2 ) 590  1 10 4  (60  20) Q –4 = = = 590 × 10 × 40 = 2.36 Watt t l 1 KCu = 400 w/m-°C 25. KAl = 200 w/m-°C 2 –5 2 A = 0.2 cm = 2 × 10 m –1 l = 20 cm = 2 × 10 m Heat drawn per second K Al  A(80  40) K Cu  A(80  40) 2  10 5  40 =  [200  400] = 2.4 J l l 2  10 1 Heat drawn per min = 2.4 × 60 = 144 J 26. (Q/t)AB = (Q/t)BE bent + (Q/t)BE KA (1   2 ) KA (1   2 ) (Q/t)BE bent = (Q/t)BE = 70 60 (Q / t )BE bent 60 6 = = D (Q / t )BE 70 7 = QAl + QCu =

0°C F

(Q/t)BE bent + (Q/t)BE = 130  (Q/t)BE bent + (Q/t)BE 7/6 = 130

7     1 (Q/t)BE bent = 130 6  27.

 (Q/t)BE bent =

20 cm

A 100°C

B 60 cm

20 cm

130  6 = 60 13

Q 780  A  100 bent = t 70 Q 390  A  100 str = t 60 (Q / t ) bent 780  A  100 60 12 =  = (Q / t ) str 70 390  A  100 7

28. (a)

KA(1   2 ) Q 1 2  1( 40  32) = = = 8000 J/sec. t  2  10  3

60 cm 5 cm 20 cm

5 cm 20 cm

1 mm

   (b) Resistance of glass = ak g ak g Resistance of air = Net resistance =

 ak a

     ak g ak g ak a =

1    2   2k a  k g  =     a  kg ka  a  K gk a 

1 10 3  2  0.025  1    2 0.025  

1 10 3  1.05 0.05 8  0.05 = = 380.9 ≈ 381 W 1 10  3  1.05 =

  2 Q = 1 t R

28.6

Heat Transfer 29. Now; Q/t remains same in both cases K  A  (100  70) K  A  (70  0) = B In Case  : A    30 KA = 70 KB K  A  (100  ) K  A  (   0) In Case  : B = A    100KB – KB  = KA  70 KB   100KB – KB  = 30 7 300 = = 30°C  100 =    3 10 30. 1 – 2 = 100   2 Q 0°C = 1 t R    1   2    R = R1 + R2 +R3 = =   = aK Al aK Cu aK Al a  200 400 

100°C

70°C A

0°C

°C B

0°C

100°C

 1   4  1   = a  400  a 80

100 Q a =  40 = 80 × 100 ×  / a 1 / 80 t  a 1 =  200 For (b) 

l l l   R CuR Al R CuR Al AK Al AK Cu AK Al R = R1 + R2 = R1 + = RAl + = l l R Cu  R Al R Cu  R Al  A Cu A Al =

  2 100 Q 100  600 A 100  600 1 = 1 = = =  = 75 l / A 4 / 600 t R 4 l 4 200 For (c) 1 1 1 1 1 1 1 =   =   l l l R R1 R 2 R 3 aK Al aK Cu aK Al



T  T3 T  T2 T1  T =  l 3l / 2 3l / 2

 – 7T = – 3T1 – 2(T2 + T3)

 3T1 – 3T = 4T – 2(T2 + T3)

3T1  2(T2  T3 ) T= 7 28.7

Al 0°C

a a a (K Al  K Cu  K Al ) = 2  200  400  = 800  l l l l 1 R=  a 800   2 Q 100  800  a  = 1 = t R l 100  800 = = 400 W 200 31. Let the temp. at B be T Q QA Q KA(T1  T ) KA(T  T3 ) KA(T  T2 )  = B  C  = t t t l l  (l / 2) l  (l / 2) =

R 100°C

l l l l 4 l  1 1    =   =  AK Al A K Al  K Cu A  200 200  400  A 600

100°C

Al T2

F QB

QC B

D QA T1

F QB

QC C

D QA T1

Heat Transfer 32. The temp at the both ends of bar F is same Rate of Heat flow to right = Rate of heat flow through left  (Q/t)A + (Q/t)C = (Q/t)B + (Q/t)D K (T  T )A K C (T1  T )A K (T  T2 )A K D (T  T2 )A  A 1 = B   l l l l  2K0(T1 – T) = 2 × 2K0(T – T2)  T1 – T = 2T – 2T2 T  2T2 T= 1 3 y  r1  r2  r1 33. Tan  = = L x  xr2 – xr1 = yL – r1L Differentiating wr to ‘x’ Ldy  r2 – r 1 = 0 dx r r dyL dy  = 2 1  dx = …(1) r2  r1  dx L

(r2 – r1) r2 r1

QLdy (r2  r1 )Ky 2

Integrating both side 2



d =

1

QL r2  r1 k

 r1

dy y r

  1 2 QL  (2 – 1) =   r2  r1 K  y  r 1

1 1 QL     (2 – 1) = r2  r1 K  r1 r2   (2 – 1) =

r  r  QL   2 1 r2  r1 K  r1  r2 

Kr1r2 ( 2  1 )  L d 60 = = 0.1°C/sec dt 10  60 dQ KA 1  2  = dt d KA  0.1 KA  0.2 KA  60   .......  = d d d KA KA 600 = (0.1  0.2  ........  60) =   (2  0.1  599  0.1) d d 2 [ a + 2a +……….+ na = n/2{2a + (n – 1)a}]

Q= 34.

200  1 10 4

 300  (0.2  59.9) =

20  10  2 = 3 × 10 × 60.1 = 1803 w ≈ 1800 w



Ky 2 d Q dx 2 =  = ky d T dx T Ldy 2 = Ky d from(1)  r2r1



200  10 2  300  60.1 20

28.8

d

Now

 d

Heat Transfer 35. a = r1 = 5 cm = 0.05 m b = r2 = 20 cm = 0.2 m 1 = T1 = 50°C 2 = T2 = 10°C Now, considering a small strip of thickness ‘dr’ at a distance ‘r’. 2 A = 4 r d 2 H = – 4 r K [(–)ve because with increase of r,  decreases] dr b dr 4K 2 = On integration, = d 2 a r H 1 4ab(1   2 ) dQ H= = K dt (b  a) Putting the values we get



20 cm



K  4  3.14  5  20  40  10 3 K=

15  10  2 15

= 100

= 2.985 ≈ 3 w/m-°C 4  3.14  4  10 1 KA (T1  T2 ) KA (T1  T2 ) Q 36. = Rise in Temp. in T2  t L Lms KA (T1  T2 ) KA(T1  T2 ) Fall in Temp in T1 = Final Temp. T1  T1  Lms Lms KA (T1  T2 ) Final Temp. T2 = T2  Lms KA(T1  T2 ) KA (T1  T2 ) T Final = T1   T2  dt Lms Lms = T1  T2    Ln 37.

2KA(T1  T2 ) 2KA(T1  T2 ) dT = =   Lms dt Lms

(T1  T2 ) / 2 2KAt = (T1  T2 ) Lms

 ln (1/2) =

KA(T1  T2 ) Q = t L Fall in Temp in T1 

( T1  T2 )



( T1  T2 )

2KAt Lms

Rise in Temp. in T2 

KA (T1  T2 ) Lm 2 s 2

Final Temp. T2 = T2 

Final Temp. T1 = T1 

2KA dt = dt Lms (T1  T2 )

 ln2 =

2KAt Lms

 t = ln2

KA (T1  T2 ) Lm1s1 KA(T1  T2 ) Lm1s1

KA (T1  T2 ) Lm1s1

 KA(T1  T2 ) KA (T1  T2 )  KA(T1  T2 ) KA (T1  T2 ) T  T2  = T1  = T1  T2      dt Lm1s1 Lm 2 s 2 Lm 2 s 2   Lm1s1 

KA T1  T2   1 dT 1  =   dt L  m1s1 m 2 s 2

 lnt = 

  



dT KA  m 2 s 2  m1s1  dt  =  T1  T2  L  m1s1m 2 s 2 

KA  m2 s 2  m1s1  t  C  L  m1s1m2 s 2 

At time t = 0, T = T0,  ln

5 cm

T = T0

T T KA  m 2 s 2  m1s1  t   =  = e T0 T0 L  m1s1m 2 s 2 

 T = T0 e



KA  m1s1  m2s2   t L  m1s1m2s2 

= T2  T1  e



 C = lnT0 KA  m1s1  m2s2  t   L  m1s1m2s2 

KA  m1s1  m2s2   t L  m1s1m2s2  

28.9

Lms 2KA

Heat Transfer 38.

KA(Ts  T0 ) nCP dT KA(Ts  T0 ) Q =  = t x dt x KA(Ts  T0 ) 2LA n(5 / 2)RdT dT  =  = (TS  T0 ) 5nRx dt x dt dT 2KAdt 2KAdt  =   ln(TS  T0 )TT0 =  (TS  T0 ) 5nRx 5nRx

ℓ

2KAt

 ln

 TS  T 2KAdt =   TS – T = (TS  T0 )e 5nRx TS  T0 5nRx

 T = TS  ( TS  T0 )e



2KAt 5nRx

= TS  (TS  T0 )e

 T = T – T0 = (TS  T0 )  (TS  T0 )e



2KAt 5nRx



2KAt 5nRx

2KAt   = (TS  T0 )  1  e 5nRx  

   

2KAt    Pa AL P AL = (TS  T0 )  1  e 5nRx  [padv = nRdt PaAl = nRdt dT = a ]   nR nR   2KAt    nR L= (TS  T0 )  1  e 5nRx     Pa A   2 –8 2 4 39. A = 1.6 m , T = 37°C = 310 K,  = 6.0 × 10 w/m -K Energy radiated per second 4 –8 4 –4 = AT = 1.6 × 6 × 10 × (310) = 8865801 × 10 = 886.58 ≈ 887 J 2 –4 2 40. A = 12 cm = 12 × 10 m T = 20°C = 293 K –8 2 4 e = 0.8  = 6 × 10 w/m -k Q 4 –4 –8 4 12 –13 = Ae T = 12 × 10 0.8 × 6 × 10 (293) = 4.245 × 10 × 10 = 0.4245 ≈ 0.42 t 41. E  Energy radiated per unit area per unit time Rate of heat flow  Energy radiated (a) Per time = E × A



So, EAl =

eT 4  A 4

4r 2 2

4(2r ) eT  A (b) Emissivity of both are same m1S1dT1 =1 = m 2S 2 dT2 

1 4

1:4

dT1 m S s 4r 3  S 2 1   900 = 2 2 = 1 13 = =1:2:9 dT2 m1S1 3 . 4  8  390 s 2 4r2  S1

Q 4 = Ae T t 100  4 4 T = T = teA 0.8  2  3.14  4  10  5  1  6  10  8  T = 1697.0 ≈ 1700 K 2 –4 2 43. (a) A = 20 cm = 20× 10 m , T = 57°C = 330 K 4 –4 –8 4 4 E = A T = 20 × 10 × 6 × 10 × (330) × 10 = 1.42 J E 4 4 2 –4 2 (b) = Ae(T1 – T2 ), A = 20 cm = 20 × 10 m t –8  = 6 × 10 T1 = 473 K, T2 = 330 K –4 –8 4 4 = 20 × 10 × 6 × 10 × 1[(473) – (330) ] 10 10 = 20 × 6 × [5.005 × 10 – 1.185 × 10 ] –2 from the ball. = 20 × 6 × 3.82 × 10 = 4.58 w

42.

28.10

Heat Transfer –3

44. r = 1 cm = 1 × 10 m –2 2 –4 2 A = 4(10 ) = 4 × 10 m –8 E = 0.3,  = 6 × 10 E 4 4 = Ae(T1 – T2 ) t –8 –4 4 4 = 0.3 × 6 × 10 × 4 × 10 × [(100) – (300) ] –12 12 = 0.3 × 6 × 4 × 10 × [1 – 0.0081] × 10 –4 = 0.3 × 6 × 4 × 3.14 × 9919 × 10 –5 = 4 × 18 × 3.14 × 9919 × 10 = 22.4 ≈ 22 W 45. Since the Cube can be assumed as black body e=ℓ –8 2 4  = 6 × 10 w/m -k –4 2 A = 6 × 25 × 10 m m = 1 kg s = 400 J/kg-°K T1 = 227°C = 500 K T2 = 27°C = 300 K d 4 4  ms = eA(T1 – T2 ) dt 



eA T1  T2 d = dt ms



1 6  10 8  6  25  10 4  [(500 ) 4  (300) 4 ] 1 400 36  25  544 –4 =  10  4 = 1224 × 10 = 0.1224°C/s ≈ 0.12°C/s. 400 4 4 46. Q = eA(T2 – T1 ) For any body, 210 = eA[(500)4 – (300)4] For black body, 700 = 1 × A[(500)4 – (300)4] 210 e Dividing =  e = 0.3 700 1 2 2 AB = 80 cm 47. AA = 20 cm , (mS)A = 42 J/°C, (mS)B = 82 J/°C, TA = 100°C, TB = 20°C KB is low thus it is a poor conducter and KA is high. Thus A will absorb no heat and conduct all =

E 4 4   = AA [(373) – (293) ] t  A





 d   mS A   =  dt  A





B A

AA [(373) – (293) ]



A a (373 ) 4  (293 ) 4 6  10 8 (373 ) 4  (293 ) 4  d     = = = 0.03 °C/S (mS ) A 42  dt  A

 d  Similarly   = 0.043 °C/S  dt B 48.

Q 4 4 = eAe(T2 – T1 ) t Q –8 4 4 –8 8 8  = 1 × 6 × 10 [(300) – (290) ] = 6 × 10 (81 × 10 – 70.7 × 10 ) = 6 × 10.3 At KA (1   2 ) Q = t l K(1   2 ) Q K  17 K  17 6  10.3  0.5  = = = 6 × 10.3 = K= = 1.8 tA l 0 .5 0 .5 17 28.11

Heat Transfer –8

49.  = 6 × 10 w/m -k L = 20 cm = 0.2 m, K=? KA(1   2 ) 4 4 E= = A(T1 – T2 ) d s(T1  T2 )  d 6  10 8  (750 4  300 4 )  2  10 1 K= = 1  2 50  K = 73.993 ≈ 74. 50. v = 100 cc  = 5°C t = 5 min For water mS  KA =  dt l

100  10 3  1000  4200 KA = 5 l For Kerosene ms KA = at l 



100  10 3  800  2100 KA = t l

100  10 3  800  2100 100  10 3  1000  4200 = t 5 5  800  2100 T= = 2 min 1000  4200 51. 50°C 45°C 40°C Let the surrounding temperature be ‘T’°C 50  45 Avg. t = = 47.5 2 Avg. temp. diff. from surrounding T = 47.5 – T 50  45 Rate of fall of temp = = 1 °C/mm 5 From Newton’s Law 1°C/mm = bA × t 1 1  bA = = …(1) t 47.5  T In second case, 40  45 Avg, temp = = 42.5 2 Avg. temp. diff. from surrounding t = 42.5 – t 45  40 5 Rate of fall of temp = = °C/mm 8 8 From Newton’s Law 5 = bAt B 1 5 =  ( 42.5  T )  8 ( 47.5  T ) By C & D [Componendo & Dividendo method] We find, T = 34.1°C 

28.12

300 K

750 K

800 K 20 cm

Heat Transfer 52. Let the water eq. of calorimeter = m

(m  50  10 3 )  4200  5 = Rate of heat flow 10 (m  100  10 3 )  4200  5 = Rate of flow 18 (m  50  10 3 )  4200  5 (m  100  10 3 )  4200  5  = 10 18 –3 –3  (m + 50 × 10 )18 = 10m + 1000 × 10 –3 –3  18m + 18 × 50 × 10 = 10m + 1000 × 10 –3  8m = 100 × 10 kg –3  m = 12.5 × 10 kg = 12.5 g 53. In steady state condition as no heat is absorbed, the rate of loss of heat by conduction is equal to that of the supplied. i.e. H = P m = 1Kg, Power of Heater = 20 W, Room Temp. = 20°C d = P = 20 watt (a) H = dt (b) by Newton’s law of cooling d = K( – 0) dt –20 = K(50 – 20)  K = 2/3 d 2 20 = K( – 0) =  (30  20) = w Again, 3 dt 3 20 10  dQ   dQ   dQ  = (c)    = 0,  =   3 3  dt  20  dt  30  dt  avg T = 5 min = 300 

10  300 = 1000 J 3 Net Heat absorbed = Heat supplied – Heat Radiated = 6000 – 1000 = 5000 J Now, m = 5000 5000 5000 –1 –1 S= = = 500 J Kg °C m 1  10 54. Given: Heat capacity = m × s = 80 J/°C Heat liberated =

 d  = 2 °C/s    dt  increase  d  = 0.2 °C/s    dt  decrease

 d  = 80 × 2 = 160 W (a) Power of heater = mS   dt  increa sin g  d  = 80 × 0.2 = 16 W (b) Power radiated = mS   dt  decrea sin g  d  = K(T – T0) (c) Now mS   dt  decrea sin g  16 = K(30 – 20)

K=

16 = 1.6 10

d = K(T – T0) = 1.6 × (30 – 25) = 1.6 × 5 = 8 W dt (d) P.t = H  8 × t

Now,

28.13

30°C

T 20°C

Heat Transfer 55.

d = – K(T – T0) dt Temp. at t = 0 is 1 (a) Max. Heat that the body can loose = Qm = ms(1 – 0) ( as, t = 1 – 0) (b) if the body loses 90% of the max heat the decrease in its temp. will be (  0 )  9 Q m  9 = 1 10ms 10 If it takes time t1, for this process, the temp. at t1 101  91  90   9 0 9 = 1  (1  0 ) = = 1 1 10 10 10 d Now, = – K( – 1) dt Let  = 1 at t = 0; &  be temp. at time t 

 

d   K dt    o

 0

  0 or, ln = – Kt 1  0 –kt

…(2) or,  – 0 = (1 – 0) e Putting value in the Eq (1) and Eq (2) 1  90 –kt  0 (1 – 0) e 10 ln 10  t1 = k

 

28.14