Tuesday, August 11, 2020

H C Verma solutions, HEAT TRANSFER, EXERCISES, Q11-Q20, Chapter-28, Concepts of Physics, Part-II

Heat Transfer

EXERCISES, Q11 - Q20


   11. The ends of a meter stick are maintained at 100°C and 0°C. One end of a rod is maintained at 25°C. Where should its other end be touched on the meter stick so that there is no heat current in the rod in steady-state?  


Answer: There will be no heat current in the rod if there is no temperature difference between the ends. It means the other end of the rod should touch the meter stick at that point where the temperature is 25°C.
Diagram for Q-11

   In the steady-state, the temperature gradient is ΔT/Δx =(100°C-0°C)/(100 cm)

= 1°C/cm.

If the temperature at a distance  x cm from the cold end is 25°C, then

x * ΔT/Δx = 25°C

→x * 1°C/cm = 25°C

→x = 25 cm from the cold end.    




 

    12. A cubical box of volume 216 cm³ is made up of 0.1 cm thick wood. The inside is heated electrically by a 100 W heater. It is found that the temperature difference between the inside and outside surface is 5°C in steady-state. Assuming that the entire electrical energy spent appears as heat, find the thermal conductivity of the material of the box.  


Answer: In the steady-state, 100 W of heat current is flowing out of the box because the temperature inside will be constant. So the heat current through one face of the cube = 100/6 W. Length of an edge of the cube = Cubic root of 216 cm³ = 6 cm. Area of one face of the cube, A = 6² cm² = 36 cm² =0.0036 m². The thickness of the wooden wall, x = 0.1 cm =0.001 m. The temperature difference between outer and inner faces, T -T' = 5°C. If the thermal conductivity of the material = K, then we have

ΔQ/Δt = KA(T -T')/x

→100/6 = K*0.0036*5/0.001

→100/6 = K*3.6*5

→K = 100/(6*3.6*5) =0.92 W/m-°C.    




 

    13. Figure (28-E1) shows water in a container having 2.0 mm thick walls made of material of thermal conductivity 0.50 W/m-°C. The container is kept in a melting-ice bath at 0°C. The total surface area in contact with water is 0.05 m². A wheel is clamped inside the water and is coupled to a block of mass M as shown in the figure. As the block goes down, the wheel rotates. It is found that after some time a steady-state is reached in which the block goes down with a constant speed of 10 cm/s and the temperature of the water remains constant at 1.0°C. Find the mass M of the block. Assume that the heat flows out of the water only through the walls in contact. Take g = 10 m/s².    
Figure for Q-13 

 

Answer: Here, T -T' = 1°C, Thickness  of the wall, x = 2 mm = 0.002 m, Area of contact, A = 0.05 m², K =0.50 W/m-°C.

Constant speed of the block under the gravitational force, v =10 cm/s =0.1 m/s. The gravitational force on the block =weight of the block, W = Mg. Work-done per unit time (power) by the gravitational force on the block 

= force*speed

=Mgv

This work-done/unit time by the gravitational force is converted into heat energy in the water through the pully system. In the steady-state, this heat energy per unit time is dissipated through the area in contact as heat current. So the heat current ΔQ/Δt = Mgv.

But, ΔQ/Δt = KA(T-T')/x

→Mgv = KA(T-T')/x

→M = KA(T-T')/(xgv)

   =0.50*0.05*1/(0.002*10*0.10)

   =12.5 kg.     




 

    14. On a winter day when the atmospheric temperature drops to -10°C, ice forms on the surface of a lake. (a) Calculate the rate of increase of thickness of the ice when 10 cm of ice is already formed. (b) Calculate the total time taken in forming 10 cm of ice. Assume that the temperature of the entire water reaches 0°C before the ice starts forming. The density of water = 1000 kg/m³, latent heat of fusion of ice = 3.36x10⁵ J/kg and thermal conductivity of ice = 1.7 W/m-°C. Neglect the expansion of water on freezing.    


Answer: (a) Given, K = 1.7 W/m-°C. The thickness of ice, x = 10 cm =0.10 m. The underside of the ice layer the temperature will be 0°C. The temperature difference between both the surfaces = 0°C -(-10°C) = 10°C. In the steady-state, suppose the latent heat of fusion of the mass of the ice layer takes time  Δt second to cross the layer. Let us take area = A m². The volume of ice =A*x m³. Mass of this volume, (neglecting the expansion on freezing) =Ax*1000 kg

     Latent heat of fusion for this mass, 

ΔQ = mL 

    = 1000Ax*3.36x10⁵ J

We have, ΔQ/Δt =KA(T -T')/x

x/Δt =KA(T -T')/ΔQ

  =KA(T-T')/(1000Ax*3.36x10⁵)

         {A will cancel out} 

   =1.7*10/(0.10*3.36x10⁸)

   = 5.0x10⁻⁷ m/s

   = Rate of increase of thickness of the ice layer.



(b) Suppose dx thickness of ice is formed in dt time. In this time the amount of heat going out is equal to the latent heat of fusion for the mass of ice formed i.e. dQ =A.dx.d.L, where d = density of watr. Now,

dQ/dt = KA(T -T')/x

→A.dx.d.L/dt = KA(T -T')/x

→dt = x.dx.d.L/K(T -T')

→∫dt = {d.L/K(T -T')}∫x.dx

→t = d.L.(x²/2)/K(T -T')

   = 1000*3.36x10⁵*(0.10²/2)/(1.7*10) s

   = 98823/3600

    ≈ 27.5 hours. 

             




    15. Consider the situation of the previous problem. Assume that the temperature of the water at the bottom of the lake remains constant at 4°C as the ice forms on the surface (the heat required to maintain the temperature of the bottom layer may come from the bed of the lake). The depth of the lake is 1.0 m. Show that the thickness of the ice formed attains a steady-state maximum value. Find this value. The thermal conductivity of water = 0.50 W/m-°C. Take other relevant data from the previous problem.   


Answer: Suppose in the steady-state the thickness of ice formed is x meter. Now the temperature outside is -10°C, just below the ice at a depth x = 0°C and at the bottom is 4°C. Now in the steady-state, the heat current through the water = heat current through the ice. So, considering area A,

 dQ(ice)/dt =dQ(water)/dt

→KA(T-T')/x = K'A(T"-T)/(1-x)

→1.7*10/x = 0.50*4/(1-x)

17(1-x) = 2x

→17 -17x =2x

→19x =17

→x =17/19 m 

→x = 1700/19 cm = 89 cm.     




 

    16. Three rods of lengths 20 cm each and area of cross-section 1 cm² are joined to form a triangle ABC. The conductivities of the rods are KAB = 50 J/m-s-°C, KBC = 200 J/m-s-°C and KAC =400 J/m-s-°C. The junctions A, B and C are maintained at 40°C, 80°C and 80°C respectively. Find the rate of heat flowing through the rods AB, AC and BC.    


Answer: Given KAB = K =50 J/m-s-°C, KBC =K' = 200 J/m-s-°C and KAC = K" =400 J/m-s-°C. 

 The temperature of the junction A = 40°C

of B = 80°C and

of C = 80°C.

The temperature difference between A and B =80°C -40°C =40°C.

The rate of heat flow = KA(T -T')/x

=50*(1/10000)*40/0.20 W

= 1 W.  

   The rate of heat flow through AC, 

  = 400*(1/10000)*40/0.20 W

   = 8 W.

 Since the temperature difference between the junctions B and C is zero, hence the rate of heat flow through the rod BC = zero. 




 

    17. A semicircular rod is joined at its end to a straight rod of the same material and the same cross-sectional area. The straight rod forms a diameter of the other rod. The junctions are maintained at different temperatures. Find the ratio of the heat transferred through a cross-section of the semicircular rod to the heat transferred through a cross-section of the straight rod in a given time.


Answer: Let the thermal conductivity of the material and the area of cross-section be K and A respectively. If the length of the diameter i.e. straight rod =X then the length of the semi-circular rod =πX/2. The temperature difference between the ends of both rods is the same, sat T-T'. If ΔQ and ΔQ' are the heat transferred through the semi-circular and the straight rods respectively during a time interval Δt then ΔQ/ΔQ' = ?.

Now, ΔQ/Δt =KA(T-T')/(πX/2) .... (i) 

and ΔQ'/Δt =KA(T-T')/X .... (ii)

Dividing (i) by (ii) we get,

ΔQ/ΔQ' = X/(πX/2) =2/π =2:π

 




    18. A metal rod of a cross-sectional area 1.0 cm² is being heated at one end. At one time, the temperature gradient is 5.0°C/cm at cross-section A and is 2.5°C/cm at cross-section B. Calculate the rate at which the temperature is increasing in the part AB of the rod. The heat capacity of the part AB = 0.40 J/°C, the thermal conductivity of the material of the rod = 200 W/m-°C. Neglect any loss of heat to the atmosphere.   


Answer: The temperature gradient at A, dθ'/dx = 5°C/cm =500°C/m and at B, dθ"/dx = 2.5°C/cm =250°C/m. 

The rate of heat flow through A, 

dQ'/dt =KA*dθ'/dx, and through B,

dQ"/dt =KA*dθ"/dx

Hence the net heat flow into the rod AB,

dQ/dt =dQ'/dt -dQ"/dt 

=KA(dθ'/dx -dθ"/dx) ----- (i)


Heat capacity, ms =0.40 J/°C

If the temperature increase of part AB =θ, then the rate of heat increase of the rod, dQ/dt =ms.dθ/dt 

    →dQ/dt = 0.40*dθ/dt ----- (ii)

Equating (i) and (ii)

0.40*dθ/dt = KA(dθ'/dx-dθ"/dx)

→dθ/dt = 200*(0.0001)(500-250)/0.40

→dθ/dt = 12.5°C. 




 

    19. Steam at 120°C is continuously passed through a 50 cm long rubber tube of inner and outer radii 1.0 cm and 1.2 cm. The room temperature is 30°C. Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber = 0.15 J/m-s-°C.  


Answer: This problem can be solved by two methods.

First:-

The temperature difference, 

T-T' = 120°C -30°C =90°C,

Average radius, r =(1.2+1.0)/2 =1.1 cm.

Average area, A = 2πrl 

=2π*1.1*50/10000 m² =0.035 m²

Thickness of the wall, x =1.2 - 1.0 

=0.2 cm =0.002 m.

K =0.15 J/m-s-°C,

Hence the rate of heat flow through the walls =KA(T -T')/x

=0.15*0.035*90/0.002 W

= 236 W

----

Second:-

Consider dx thickness of the wall at a distance of x  from the inside layer. The rate of heat flow, q=KA*dT/dx

=K*2πxl*dT/dx

→dT ={q/(2πlK)}dx/x

→∫dT ={q/(2πlK)}*∫dx/x

→T -T' ={q/(2πlK)}*ln(r/r')

→q =2πlK(T-T')/ln(r/r')

    =2π*0.50*0.15*90/ln(1.2/1.0)

    =233 J/s. 

    


 


    20. A hole of radius r₁ is made centrally in a uniform circular disc of thickness d and radius r₂. The inner surface (a cylinder of length d and radius r₁ is maintained at a temperature θ₁ and the outer surface (a cylinder of length d and radius r₂) is maintained at a temperature θ₂ (θ₁ > θ₂). The thermal conductivity of the material of the disc is K. Calculate the heat flowing per unit time through the disc.  


Answer: Let the rate of heat flow through dx thickness of the cylinder at a radius x = -q. Negative sign is for the increase in radius there is decrease in temperature. If the temperature difference in this thickness = dT, then

-q = KA*dT/dx

→-q =K*(2πxd)*dT/dx

→dT ={-q/(2πKd)}*dx/x

→∫dT ={-q/(2πKd)}*.dx/x

→[T] = {-q/(2πKd)}[ln(x)]

The limits of integration on the left side is θ₁ to θ₂ and on the right side is r₁ to r₂. 

→θ₂-θ₁ = {-q/(2πKd)}{lnr₂-lnr₁}

→q =2πKd(θ₁-θ₂)/ln(r₂/r₁). 


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Links to the Chapters











CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

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CHAPTER- 7 - Circular Motion

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CHAPTER- 5 - Newton's Laws of Motion


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CHAPTER- 4 - The Forces

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CHAPTER- 3 - Kinematics - Rest and Motion

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