Heat Transfer Exercise HC Verma Solutions Vol-2 Class-12 Ch-28

Heat Transfer Exercise HC Verma Solutions Vol-2 Class-12 Ch-28. Step by Step Solutions of Exercise Questions of Chapter-28 Heat Transfer (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Heat Transfer Exercise HC Verma Solutions Vol-2 Class-12 Ch-28

Board ISC and other board
Publications Bharti Bhawan Publishers
Chapter-28 Heat Transfer
Class 12
Vol  2nd
writer HC Verma
Book Name Concept of Physics
Topics Solution of Exercise Questions
Page-Number 98, 99, 100, 101, 102

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Objective-II

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Heat Transfer Exercise HC Verma Capacities of Gases Questions

HC Verma Solutions of Ch-28  Vol-2 Concept of Physics for Class-12

(Page-98)

Question-1 :-

A uniform slab of dimension 10 cm × 10 cm × 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−1 °C−1. Find the amount of heat flowing through the slab per minute.

Answer-1 :-

Given:
Thermal conductivity of the material, k = 0.80 W m-1  °c-1
Area of the cross section of the slab, A = 100 cm2  = 10-2 m2
Thickness of the slab, Δx = 1 cm = 10 -2 m

hc verma heat transfer exercise img 1

Question-2 :-

A liquid-nitrogen container is made of a 1 cm thick styrofoam sheet having thermal conductivity 0.025 J s−1 m−1 °C−1. 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 us 300 K. Calculate the rate of heat flow from the atmosphere to the liquid nitrogen.

Answer-2 :-

hc verma heat transfer exercise img 2

Thickness of the container, l = 1 cm = 10 -2 m
Thermal conductivity of the styrofoam sheet, k = 0.025 J s-1 m-1 °C -1
Area, A= 0.80 m2

hc verma heat transfer exercise img 3

Temperature difference , ΔT = T1 – T2 = 300 – 80 = 220K

hc verma heat transfer exercise img 4

Question-3 :- (Heat Transfer Exercise HC Verma )

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, surface of the body under clothes = 1.6 m2, conductivity of the cloth = 0.04 J s−1 m−1°C−1, thickness of the cloth = 0.5 cm.

Answer-3 :-

hc verma heat transfer exercise img 2

Temperature of the body, T1 = 97°F= 36. 11°C

Temperature of the surroundings, T2 = 47°F= 8.33°C`

Conductivity of the cloth, K = 0.04  J s-1 m-1  °c

Thickness of the cloth l = 0.5 cm = 0.005 m

Area of the cloth, A=1.6 m2

Difference in the temperature ΔT = T1 – T2 = 36.11 – 8.33 = 27.78°C

hc verma heat transfer exercise img 5

Rate at which heat is flowing out is given by

hc verma heat transfer exercise img 6

Question-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−1°C−1. 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 vaporisation of water = 2.26 × 106 J kg−1.

Answer-4 :-

Area of the bottom of the container, A = 25 cm= 25 × 10-4 m2

Thickness of the vaporisation of water , l = 1 mm = 10-3 m.

Latent heat of vaporisation of water, L = 2.26 × 106  j Kg -1

Thermal conductivity of the container , K =50W m- 1 °C-1 mass = 100 g = 0.1 Kg

mass = 100g = 0.11 kg

Rate of heat transfer from the base of the container is given by

hc verma heat transfer exercise img 7

⇒ ( T – 100 ) = 3.008 × 10
⇒ T = 130° C

Question-5 :-

One end of a steel rod (K = 46 J s−1 m−1°C−1) of length 1.0 m is kept in ice at 0°C and the other end is kept in 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 × 105 J kg−1.

Answer-5 :-

Given :

K = 46 J s−1 m−1°C−1
We know that

Question-6 :-

A 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−1°C−1. Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice = 3.4 × 105 J kg−1.

Answer-6 :-

Question-7 :-

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

Answer-7 :-

Thickness of porous walls, l = 1mm = 10-3 m

mass, m =10 kg

Latent heat of vapourisation, Lv = 2.27 × 106 J/kg

Thermal conductivity, K =   0.80 J/m s °C

ΔQ = 2.27 × 106 × 10 J

0.1 g of water evaporate in 1 sec, so 10 kg water will evaporate in 105 s

hc verma heat transfer exercise img 11

⇒ T = 27.8° C

⇒ T = 28° C

Question-8 :-

A steel frame (K = 45 W m−1°C−1) of total length 60 cm and 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-8 :-

Thermal conductivity, K = 45 W m–1 °C–1

Length, l = 60 cm = 0.6 m

Area of cross section, A = 0.2 cm2 = 0.2 × 10−4 m2

Initial temperatureT1 = 40°C

Final temperatureT2 = 20°C

Question-9 :-  (Heat Transfer Exercise HC Verma )

Water at 50°C is filled in a closed cylindrical vessel of height 10 cm and cross sectional area 10 cm2. The walls of the vessel are adiabatic but the flat parts are made of 1-mm thick aluminium (K = 200 J s−1 m−1°C−1). Assume that the outside temperature is 20°C. The density of water is 100 kg m−3, and the specific heat capacity of water = 4200 J k−1g °C−1. Estimate the time taken for the temperature of fall by 1.0 °C. Make any simplifying assumptions you need but specify them.

Answer-9 :-

Water at 50°C is filled in a closed cylindrical vessel of height 10 cm and cross sectional area 10 cm2. The walls of the vessel are adiabatic but the flat parts are made of 1-mm thick aluminium (K = 200 J s−1 m−1°C−1). Assume that the outside temperature is 20°C. The density of water is 100 kg m−3, and the specific heat capacity of water = 4200 J k−1g °C−1. Estimate the time taken for the temperature of fall by 1.0 °C. Make any simplifying assumptions you need but specify them.

Question-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) the 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−1°C−1.

Answer-10 :-




 
(a) Constant heat current follow through the wire, temperature gradient

For 11 cm fro left end,

Heat Transfer Exercise HC Verma Capacities of Gases Questions

HC Verma Solutions of Ch-28  Vol-2 Concept of Physics for Class-12

(Page-99)

Question-11 :-

The ends of a metre 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 metre stick so that there is no heat current in the rod in steady state?

Answer-11 :-

Let the point to be touched be B

No Heat will flow when the temperature at that point is also 25C

That means

 100 x = 2500

x = 25 cm

From the end with C

Question-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 the 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-12 :-

Question-13 :- (Heat Transfer Exercise HC Verma )

Following Figure (Figure 28-E1) shows water in a container having 2.0 mm thick walls made of a material of thermal conductivity 0.50 W m−1°C−1. 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−1 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 s−2.

Answer-13 :-

Temperature of waterT1 = 1°C

Temperature if ice bath, T2 = 0°C

Thermal conductivity, K = 0.5 W/m °C

Length through which heat is lost, l = 2 mm = 2 × 10–3 m

Area of cross section, A = 5 × 10−2 m2

Velocity of the block, v = 10 cm/sec = 0.1 m/s

Let the mass of the block be m.

Power = F · v
= (mg) v          ……(1)

Also,

From equation (1), (2) and (3), we get

hc verma heat transfer exercise img 21

Question-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 the 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. Density of water = 1000 kg m−3, latent heat of fusion of ice = 3.36 × 105 J kg−1and thermal conductivity of ice = 1.7 W m−1°C−1. Neglect the expansion of water of freezing.

Answer-14 :-

Thermal conductivity, K = 1.7 W/m°C

Density of water, ρω = 102 kg/m3

Latent heat of fusion of ice, Lice = 3.36 × 105 J/kg

Length, l = 10 × 10−2 m

(a) Rate of flow of heat is given by

= 5 × 10-7 m/s

(b) To form a thin ice layer of thickness dx, let the required be dt.
Mass of that thin layer, dm = A dx ρω
Heat absorbed by that thin layer, dQ = Ldm

Question-15 :- (Heat Transfer Exercise HC Verma )

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−1°C−1. Take other relevant data from the previous problem.

Answer-15 :-

Let the point up to which ice is formed is at a distance of x m from the top of the lake.

Under steady state, the rate of flow of heat from ice to this point should be equal to the rate flow of heat from water to this point.

Temperature of the top layer of ice = −10°C

Temperature of water at the bottom of the lake = 4°C
Temperature at the point up to which ice is formed =

Question-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 s−1 m−1°C−1, KBC = 200 J s−1m−1°C−1 and KAC = 400 J s−1 m−1°C−1. 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-16 :-

Thermal conductivity of rod AB, KAB = 50 J/m-s-°C

Temperature of junction at A, TA = 40°C

Thermal conductivity of rod BC, KBC = 200 J/m-s-°C

Temperature of junction at B, TB = 80°C

Thermal conductivity of rod BCKCA = 400 J/m-s-°C

temperature of junction at CTC = 80°C

l = 20 cm = 20 × 10−2 m

A = 1 cm2 = 10−4 m2

Question-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-17 :-

Let A be the area of cross section and K be the thermal conductivity of the material of the rod.
Let q1 be the rate of flow of heat through a semicircular rod.

Rate of flow of heat is given by

Let qbe the rate of flow of heat through a straight rod.

hc verma heat transfer exercise img 29

Ratio of the rate of flow of heat through the 2 rods

Question-18 :- (Heat Transfer Exercise HC Verma )

A metal rod of cross sectional area 1.0 cm2 is being heated at one end. At one time, the temperatures gradient is 5.0°C cm−1 at cross section A and is 2.5°C cm−1 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−1, thermal conductivity of the material of the rod = 200 W m−1°C−1. Neglect any loss of heat to the atmosphere

Answer-18 :-

The rate of heat flow per sec.

= dQA/dt = KA(dθ/dt)
The rate of heat flow per sec.
= dQB/dt = KA(dθB/dt)
This part of heat is absorbed by the red.
hc verma heat transfer exercise img 31
hc verma heat transfer exercise img 32
= 1250 x 10-2
= 12.5 °C/m

Question-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 s−1 m−1°C−1.

Answer-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 s−1 m−1°C−1.

Given K rubber = 0.15J/m-s-°C

T 2– T 1 = 90°C

We know for radial conduction in a Cylinder

hc verma heat transfer exercise img 34

= 232.5 ≈ 233 j/s

Question-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 a 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-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 a 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.

Let dθ/dt be the rate of flow of heat.

Consider an annular ring of radius and thickness dr.

Rate of flow of heat is given by

dθ/dt = K(2πrd)

Rate of flow of heat is constant.

hc verma heat transfer exercise img 35

Question-21 :-

A hollow tube has a length l, inner radius R1 and outer radius R2. The material has a thermal conductivity K. Find the heat flowing through the walls of the tube if (a) the flat ends are maintained at temperature T1 and T2 (T2 > T1) (b) the inside of the tube is maintained at temperature T1 and the outside is maintained at T2.

Answer-21 :-

(a)  When the flat ends are maintained at temperatures T1 and T2 (where T2 > T1):

Area of cross section through which heat is flowing, =

hc verma heat transfer exercise img 37

A hollow tube has a length l, inner radius R1 and outer radius R2. The material has a thermal conductivity K. Find the heat flowing through the walls of the tube if (a) the flat ends are maintained at temperature T1 and T2 (T2 > T1) (b) the inside of the tube is maintained at temperature T1 and the outside is maintained at T2.

Rate of flow of heat = dθ/dt

hc verma heat transfer exercise img 38

( b )

When the inside of the tube is maintained at temperature T1 and the outside is maintained at T2:

Let us consider a cylindrical shell of radius r and thickness dr.
Rate of flow of heat, hc verma heat transfer exercise img 39

hc verma heat transfer exercise img 40

dr/r = 2πKl   dT

hc verma heat transfer exercise img 41

Question-22 :-

A composite slab is prepared by pasting two plates of thickness L1 and L2 and thermal conductivity K1 and K2. The slabs have equal cross-sectional area. Find the equivalent conductivity of the composite slab.

Answer-22 :-

It is equivalent to the series combination of 2 resistors.

∴ RS = R1 +R2

Resistance of a conducting slab,  R=l/KA

hc verma heat transfer exercise img 42

Question-23 :-  (Heat Transfer Exercise HC Verma )

Figure (Figure 28-E2) shows a copper rod joined to a steel rod. The rods have equal length and equal cross sectional area. The free end of the copper rod is kept at 0°C and that of the steel rod is kept at 100°C. Find the temperature at the junction of the rods. Conductivity of copper = 390 W m−1°C−1 and that if steel = 46 W m−1°C−1.

Figure (28-E2) shows a copper rod joined to a steel rod. The rods have equal length and equal cross sectional area. The free end of the copper rod is kept at 0°C and that of the steel rod is kept at 100°C. Find the temperature at the junction of the rods. Conductivity of copper = 390 W m−1°C−1 and that if steel = 46 W m−1°C−1.

Answer-23 :-

Figure (Figure 28-E2) shows a copper rod joined to a steel rod. The rods have equal length and equal cross sectional area. The free end of the copper rod is kept at 0°C and that of the steel rod is kept at 100°C. Find the temperature at the junction of the rods. Conductivity of copper = 390 W m−1°C−1 and that if steel = 46 W m−1°C−1.

K Cu = 390w/m-°C  K St = 46w/m-°C Now, Since they are in series connection, So, the heat passed through the cross sections in the same.

hc verma heat transfer exercise img 44

hc verma heat transfer exercise img 45

= 10.55 ≈ 10.6°C

Question-24 :-

An aluminium rod and a copper rod of equal length 1.0 m and cross-sectional area 1 cm2 are welded together as shown in the figure . One end is kept at a temperature of 20°C and the other at 60°C. Calculate the amount of heat taken out per second from the hot end. Thermal conductivity of aluminium = 200 W m−1°C−1 and of copper = 390 W m−1°C−1.

An aluminium rod and a copper rod of equal length 1.0 m and cross-sectional area 1 cm2 are welded together as shown in the figure . One end is kept at a temperature of 20°C and the other at 60°C. Calculate the amount of heat taken out per second from the hot end. Thermal conductivity of aluminium = 200 W m−1°C−1 and of copper = 390 W m−1°C−1.

(Figure 28-E3)

Answer-24 :-

An aluminium rod and a copper rod of equal length 1.0 m and cross-sectional area 1 cm2 are welded together as shown in the figure . One end is kept at a temperature of 20°C and the other at 60°C. Calculate the amount of heat taken out per second from the hot end. Thermal conductivity of aluminium = 200 W m−1°C−1 and of copper = 390 W m−1°C−1.

An aluminium rod and a copper rod of equal length 1.0 m and cross-sectional area 1 cm2 are welded together as shown in the figure . One end is kept at a temperature of 20°C and the other at 60°C. Calculate the amount of heat taken out per second from the hot end. Thermal conductivity of aluminium = 200 W m−1°C−1 and of copper = 390 W m−1°C−1.

q1 and q2 are heat currents. In other words, they are the rates of flow of heat through aluminium and copper rod, respectively.

Applying KVL at the hot junction, we get

q = q1 + q2

Rate of heat flow, q =KAΔT/l

As q = q1 + q2

hc verma heat transfer exercise img 49

Kp =K1 + k2

= 390 + 200 = 590 W/m°C

hc verma heat transfer exercise img 50

q= 2.36 W

Answer-25 :-

Area of cross section, A = 0.20 cm2 = 0.2 × 10–4 m2

Thermal conductivity of aluminiumKAl = 200 W/​m ​°C

Thermal conductivity of copper, KCu = 400 W/m​°C

Total heat flowing per second = qAl + qCu

hc verma heat transfer exercise img 51

= 8 × 10–1+16 × 10–1

= 24 × 10–1

= 2.4 J/s

Heat drawn in 1 minute = 2.4 × 60 = 144 J

Question-26 :-

Consider the situation shown in the figure (Figure 28-E5) . The frame is made of the same material and has a uniform cross-sectional area everywhere. Calculate the amount of heat flowing per second through a cross section of the bent part if the total heat taken out per second from the end at 100°C is 130 J.

Consider the situation shown in the figure . The frame is made of the same material and has a uniform cross-sectional area everywhere. Calculate the amount of heat flowing per second through a cross section of the bent part if the total heat taken out per second from the end at 100°C is 130 J.

(Figure 28-E5)

Answer-26 :-

Consider the situation shown in the figure . The frame is made of the same material and has a uniform cross-sectional area everywhere. Calculate the amount of heat flowing per second through a cross section of the bent part if the total heat taken out per second from the end at 100°C is 130 J.

hc verma heat transfer img 1

Consider the situation shown in the figure . The frame is made of the same material and has a uniform cross-sectional area everywhere. Calculate the amount of heat flowing per second through a cross section of the bent part if the total heat taken out per second from the end at 100°C is 130 J.

q = q1 + q2        ……………(1)

R1 and R2 are in parallel, so total heat across R1 and R2 will be same.

hc verma heat transfer img 2

From equation (1) and (2),

q = q_1 + {7q_1}/6`

hc verma heat transfer img 3

q1 = 60 J/sec

since R1 and R2 are in parallel, the amount of heat flowing through them will be same.

hc verma heat transfer img 6

Question-28 :-

A room has a window fitted with a single 1.0 m × 2.0 m glass of thickness 2 mm. (a) Calculate the rate of heat flow through the closed window when the temperature inside the room is 32°C and the outside is 40°C. (b) The glass is now replaced by two glass panes, each having a thickness of 1 mm and separated by a distance of 1 mm. Calculate the rate of heat flow under the same conditions of temperature. Thermal conductivity of window glass = 1.0 J s−1 m−1°C−1 and that of air = 0.025 m-1°C-1 .

Answer-28 :-

(a) A room has a window fitted with a single 1.0 m × 2.0 m glass of thickness 2 mm. (a) Calculate the rate of heat flow through the closed window when the temperature inside the room is 32°C and the outside is 40°C. (b) The glass is now replaced by two glasspanes, each having a thickness of 1 mm and separated by a distance of 1 mm. Calculate the rate of heat flow under the same conditions of temperature. Thermal conductivity of window glass = 1.0 J s−1 m−1°C−1 and that of air = 0.025 m-1°C-1 .

Length, l = 2 mm = 0.0002 m

hc verma heat transfer img 8

(b) A room has a window fitted with a single 1.0 m × 2.0 m glass of thickness 2 mm. (a) Calculate the rate of heat flow through the closed window when the temperature inside the room is 32°C and the outside is 40°C. (b) The glass is now replaced by two glasspanes, each having a thickness of 1 mm and separated by a distance of 1 mm. Calculate the rate of heat flow under the same conditions of temperature. Thermal conductivity of window glass = 1.0 J s−1 m−1°C−1 and that of air = 0.025 m-1°C-1 .

hc verma heat transfer img 10

From the circuit diagram, we can find that all the resistors are connected

hc verma heat transfer img 11

=381W

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