Friday, July 31, 2020

H C Verma solutions, HEAT TRANSFER, EXERCISES, Q1-Q10, Chapter-28, Concepts of Physics, Part-II

Heat Transfer

EXERCISES, Q1 - Q10


   1. A uniform slab of dimension 10 cm x 10 cm x 1 cm is kept between two heat reservoirs at temperatures 10°C and 90°C. The larger surface areas touch the reservoirs. The thermal conductivity of the material is 0.80 W/m-°C. Find the amount of heat flowing through the slab per minute. 

Diagram for Q-1

Answer: The heat flows from the hotter face towards the colder face. The area of the cross-section perpendicular to the heat flow is 

A = 10 cm x 10 cm = 100 cm² =0.01 m². 

Given that 

K = 0.80 W/m-°C, 

T - T' =90°C -10°C =80 °C,

Thickness of the wall, 

x = 1 cm =0.01 m

The amount of heat flowing per second is

ΔQ/Δt = KA(T-T')/x

=0.80*0.01*80/0.01 J

=64 J

Hence the amount of heat flowing per minute = 64*60 J

=3840 J.


 


  

   2. A liquid nitrogen container is made of a 1 cm thick thermocoal sheet having thermal conductivity 0.025 J/m-s-°C. Liquid nitrogen at 80 K is kept in it. A total area of 0.80 m² is in contact with the liquid nitrogen. The atmospheric temperature is 300 K. Calculate the rate of heat flow from the atmosphere to the liquid nitrogen. 


Answer: Difference in temperature,

T - T' = 300 K- 80K =220 K.

Thermal conductivity K =0.025 J/m-s-°C,

Area of contact, A = 0.80 m²

Thickness of the container, x =1 cm =0.01 m.

Hence the heat flowing per second,

ΔQ/Δt = KA(T-T')/x

=0.025*0.80*220/0.01 J/s

=20*22 W =440 W    




 

 

    3. The normal body temperature of a person is 97°F. Calculate the rate at which heat is flowing out of his body through the clothes assuming the following values. Room temperature = 47°F, the surface of the body under clothes = 1.6 m², conductivity of the cloth = 0.04 J/m-s-°C, the thickness of the cloth = 0.5 cm.


Answer: The difference of the temperature,

T - T' = 97°F - 47°F =50°F 

=50*5/9°C

=27.78°C.

Thermal conductivity K =0.04 J/m-s-°C,

Area of contact, A = 1.60 m²

Thickness of clothes, x =0.5 cm =0.005 m.

Hence the heat flowing out of the body through clothes per second,

ΔQ/Δt = KA(T-T')/x

=0.04*1.60*27.78/0.005 J/s

=356 W  

 





 

   4. Water is boiled in a container having a bottom of surface area 25 cm², thickness 1.0 mm and thermal conductivity 50 W/m-°C. 100 g of water is converted into steam per minute in the steady-state after the boiling starts. Assuming that no heat is lost to the atmosphere, calculate the temperature of the lower surface of the bottom. Latent heat of vaporization of water = 2.26 x 10⁶ J/kg. 


Answer: Temperature of the boiling water, T' =100°C = temperature of the upper surface of the bottom, 

Let the temperature of the lower surface of the bottom = T. So the temperature difference = (T - 100)

Surface area, A = 25 cm² =0.0025 m²

Thickness of the bottom, x = 1 mm =0.001 m.

Thermal conductivity, K = 50 W/m-°C

In the steady-state 100 g of water is converted into steam, hence the heat flowing per second through the bottom

= (100/1000)*2.26x10⁶/60 J

= 3767 J

But the heat flowing per second is given as

KA(T-T')/x, equating the two with substituting data

50*0.0025*(T-100)/0.001 =3767

→T-100 =3.767/(50*0.0025)

→T-100 = 30

→T = 130°C.    

 




 

  5. One end of a steel rod (K = 46 J/m-s-°C) of length 1.0 m is kept in ice at 0°C and the other end is kept boiling water at 100°C. The area of cross-section of the rod is 0.04 cm². Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 x 10⁵ J/kg. 


Answer: K = 46 J/m-s-°C, 

A =0.04 cm² =4x10⁻⁶ m²

Temperature difference, T-T' =100°C

x = 1.0 m

Hence heat flowing per second,

= KA(T-T')/x

= 46*4x10⁻⁶*100/1.0

= 0.0184 J

Latent heat of fusion of ice, L = 3.36x10⁵ J/kg

Hence the mass of ice melting per second = 0.0184/3.36x10⁵ kg

= 0.0055x10⁻⁵ kg    

= 5.5x10⁻⁵ g 





 

   6. An icebox almost completely filled with ice at 0°C is dipped into a large volume of water at 20°C. The box has walls of surface area 2400 cm², thickness 2.0 mm and thermal conductivity 0.06 W/m-°C. Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice = 3.4 x10⁵ J/kg. 


Answer:  The temperature difference, 

T-T' = 20-0 =20°C

A = 2400 cm² =0.24 m²
K = 0.06 W/m-°C
x = 2 mm = 0.002 m
Hence heat flowing per second 
= KA(T-T')/x 
= 0.06*0.24*20/0.002 J/s 
= 144 J/s
Hence the mass of ice melting per second
= 144/3.4x10⁵ kg
And the mass of ice melting per hour
= 144*3600/3.4x10⁵ kg/h
= 1.5 kg/h 

 




 


    7. A pitcher with 1 mm thick porous walls contains 10 kg of water. Water comes to its outer surface and evaporates at a rate of 0.1 g/s. The surface area of the pitcher (one side) = 200 cm². The room temperature = 42°C, latent heat of vaporization = 2.27x10⁶ J/kg, and the thermal conductivity of the porous walls = 0.80 J/m-s-°C. Calculate the temperature of water in the pitcher when it attains a constant value. 


Answer:  When the water vaporizes it takes the heat from the water. Per-second evaporation of water = 0.1 g =1x10⁻⁴ kg.  

 

Per second heat loss from the water

= 1x10⁻⁴*2.27x10⁶ J = 227 J

Let the temperature of water in steady-state = T,

Temperature difference = 42 - T

A = 200 cm² = 0.02 m²

x = 1 mm = 0.001 m 

Writing the equation for heat flow,

KA(T-T')/x =227

→0.80*0.02*(42-T)/0.001 =227

→42-T =227*0.001/(0.8*0.02)

→42-T = 14

→T =42 - 14 = 28°C.   

 





 

    8. A steel frame (K = 45 W/m-°C) of a total length of 60 cm and a cross-sectional area 0.20 cm², forms three sides of a square. The free ends are maintained at 20°C and 40°C. Find the rate of heat flow through a cross-section of the frame. 


Answer: Here, x = 60 cm = 0.60 m.

A = 0.20 cm² = 2x10⁻⁵ m²

K = 45 W/m-°C

Temperature difference. T -T' =20°C, 

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

=45*2x10⁻⁵*20/0.60

= 0.03 W.  

 





 

    9. Water at 50°C is filled in a closed cylindrical vessel of height 10 cm and a cross-sectional area 10 cm². The walls of the vessel are adiabatic but the flat parts are made of 1 mm thick aluminum (K = 200 J/m-s-°C). Assume that the outside temperature is 20°C. The density of water is 1000 kg/m³, and the specific heat capacity of water = 4200 J/kg-°C. Estimate the time taken for the temperature to fall by 1.0°C. Make any simplifying assumptions you need but specify them.


Answer: Assuming that during the fall of temperature by 1°C the rate of heat flow outside is constant. Also that the temperature of the water remains the same everywhere in its volume. So when the water temperature is dropped by 1°C, the heat lost by it is = m*s

= (100/1000)*4200 J

=420 J
Heat lost through each of the flat sides = 210 J. If the time taken during the fall of temperature by 1°C = t, then the rate of heat transfer through one flat side = 210/t J/s. Hence
210/t = KA(T-T')/x
→t =210x/KA(T-T')
→t = 210*0.001/(200*0.001*29.5)
{we take average temperature=(50+49)/2=49.5°C}
→t = 0.035 s.  

 





 

    10. The left end of a copper rod (length = 20 cm, area of cross-section = 0.20 cm²) is maintained at 20°C and the right end is maintained at 80°C. Neglecting any loss of heat through radiation, find (a) temperature at a point 11 cm from the left end and (b) the heat current through the rod. Thermal conductivity of copper = 385 W/m-°C. 


Answer: (a) The temperature gradient of the rod =(80-20)/20 =3°C/cm. Hence the temperature at a point 11 cm from the left end = 20°C +(11 cm)*3°C/cm

= 20+33 =53°C.


(b) Heat current flowing through the rod =KA(T-T')/x 

= 385*(0.00002)*(80-20)/0.2

= 2.31 J/s.

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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

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


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Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

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