# 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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Exercise (Currently Open)

**Heat Transfer Exercise HC Verma Capacities of Gases Questions**

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

**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 cm^{2 }= 10^{-2 }m^{2
}Thickness of the slab, Δx = 1 cm = 10 ^{-2 }m

**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 m^{2} 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 :-**

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 m^{2}

Temperature difference , ΔT = T_{1} – T_{2} = 300 – 80 = 220K

**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 m^{2}, conductivity of the cloth = 0.04 J s^{−1} m^{−1}°C^{−1}, thickness of the cloth = 0.5 cm.

**Answer-3 :-**

Temperature of the body, T_{1} = 97°F= 36. 11°C

Temperature of the surroundings, T_{2} = 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 m^{2}

Difference in the temperature ΔT = T_{1} – T_{2} = 36.11 – 8.33 = 27.78°C

Rate at which heat is flowing out is given by

**Question-4 :-**

Water is boiled in a container having a bottom of surface area 25 cm^{2}, 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 × 10^{6} J kg^{−1}.

**Answer-4 :-**

Area of the bottom of the container, A = 25 cm^{2 }= 25 × 10^{-4} m^{2}

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

⇒ ( 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 cm^{2}. Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 × 10^{5} J kg^{−1}.

**Answer-5 :-**

**Given :**

^{−1}m

^{−1}°C

^{−1}

**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 cm^{2}, 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 × 10^{5} J kg^{−1}.

**Answer-6 :-**

$Area of the walls of the box, A = 2400 cm2 = 2400 × 10−4 m2 Thickness of the ice box, l = 2 mm = 2 × 10−3 m Thermal conductivity of the material of the box, K = 0.06 W m−1 °C−1 Temperature of the water outside the box, T1 = 20°C Temperature of ice, T2 = 0°C$

**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 cm^{2}. The room temperature = 42°C, latent heat of vaporization = 2.27 × 10^{6 }J 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

⇒ 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 cm^{2}, 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 cm^{2} = 0.2 × 10^{−4} m^{2}

Initial temperature*, **T*_{1} = 40°C

Final temperature*, **T*_{2} = 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 cm^{2}. 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^{−1}g °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 :-**

$since heat goes out from both surface. Hence net heat coming out.$

**Question-10 :-**

The left end of a copper rod (length = 20 cm, area of cross section = 0.20 cm^{2}) 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 :-**

$t=20cm=20×1_{−2}m$

$A=0.2m_{2}=0.2×1_{−4}m_{2}$

$θ_{1}=8_{0}C$

$θ_{2}=2_{0}C$ $K=385$

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

**Heat Transfer Exercise HC Verma Capacities of Gases Questions**

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

**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 25∘C**

That means

$QAB = QBC$

⇒ 100 x = 2500

x = 25 cm

From the end with 0°C

**Question-12 :-**

A cubical box of volume 216 cm^{3} 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 m^{2}. 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 water*, **T*_{1} = 1°C

Temperature if ice bath, *T*_{2} = 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} m^{2}

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

**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 × 10^{5} J kg^{−1}and 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, ρ_{ω} = 10^{2} kg/m^{3}

Latent heat of fusion of ice, L_{ice} = 3.36 × 10^{5} 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 ρ_{ω}

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 cm^{2} are joined to form a triangle ABC. The conductivities of the rods are K_{AB}_{ }= 50 J s^{−1} m^{−1}°C^{−1}, K_{BC} = 200 J s^{−1}m^{−1}°C^{−1} and K_{AC} = 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, *K*_{AB} = 50 J/m-s-°C

Temperature of junction at A, *T*_{A} = 40°C

Thermal conductivity of rod BC*, K*_{BC} = 200 J/m-s-°C

Temperature of junction at B*, T*_{B}_{ }= 80°C

Thermal conductivity of rod BC*, **K*_{CA} = 400 J/m-s-°C

temperature of junction at C*, **T*_{C} = 80°C

*l* = 20 cm = 20 × 10^{−2} m

A = 1 cm^{2} = 10^{−4} m^{2}

**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 *q*_{1} be the rate of flow of heat through a semicircular rod.

Rate of flow of heat is given by

Let *q*_{2 }be the rate of flow of heat through a straight rod.

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 cm^{2} 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.

*Q*

_{A}/dt = KA(dθ/dt)

*Q*

_{B}/dt = KA(dθB/dt)

^{-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 :-**

Given K_{ rubber} = 0.15J/m-s-°C

T_{ 2}– T_{ 1} = 90°C

We know for radial conduction in a Cylinder

** **

**= 232.5 ≈ 233 j/s**

**Question-20 :-**

A hole of radius *r*_{1} is made centrally in a uniform circular disc of thickness *d* and radius *r*_{2}. The inner surface (a cylinder a length *d* and radius *r*_{1}) is maintained at a temperature θ_{1} and the outer surface (a cylinder of length *d* and radius *r*_{2}) is maintained at a temperature θ_{2} (θ_{1} > θ_{2}). The thermal conductivity of the material of the disc is K. Calculate the heat flowing per unit time through the disc.

**Answer-20 :-**

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

Consider an annular ring of radius *r *and thickness *dr*.

Rate of flow of heat is given by

dθ/dt = K(2πrd)

Rate of flow of heat is constant.

**Question-21 :-**

A hollow tube has a length *l*, inner radius R_{1} and outer radius R_{2}. 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 T_{1} and T_{2} (T_{2} > T_{1}) (b) the inside of the tube is maintained at temperature T_{1} and the outside is maintained at T_{2}.

**Answer-21 :-**

(a) When the flat ends are maintained at temperatures T_{1} and T_{2} (where T_{2} > T_{1}):

Area of cross section through which heat is flowing, =

Rate of flow of heat = dθ/dt

( b )

When the inside of the tube is maintained at temperature T_{1} and the outside is maintained at T_{2}_{:}

Let us consider a cylindrical shell of radius *r* and thickness *dr*.

Rate of flow of heat,

dr/r = 2πKl dT

**Question-22 :-**

A composite slab is prepared by pasting two plates of thickness L_{1} and L_{2} and thermal conductivity K_{1} and K_{2}. 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.

∴ *R*_{S} = *R*_{1} +*R*_{2}

Resistance of a conducting slab, **R=l/KA**

**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}.

**Answer-23 :-**

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.

= 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 cm^{2} 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 :-**

** **

q_{1} and q_{2} 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 = q _{1} + q_{2}**

Rate of heat flow, **q =KAΔT/l**

As **q = q _{1} + q_{2}**

Kp =K_{1} + k_{2}

= 390 + 200 = 590 W/m°C

q= 2.36 W

**Heat Transfer Exercise HC Verma Capacities of Gases Questions**

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

**Question-25 :-**

Figure (Figure 28-E4) shows an aluminium rod joined to a copper rod. Each of the rods has a length of 20 cm and area of cross section 0.20 cm^{2}. The junction is maintained at a constant temperature 40°C and the two ends are maintained at 80°C. Calculate the amount of heat taken out from the cold junction in one minute after the steady state is reached. The conductivity are K_{At} = 200 W m^{−1}°C^{−1} and K_{Cu} = 400 W m^{−1}°C^{−1}.

**Answer-25 :-**

Area of cross section*, A* = 0.20 cm^{2} = 0.2 × 10^{–4} m^{2}

Thermal conductivity of aluminium*, **K*_{Al} = 200 W/m °C

Thermal conductivity of copper, K_{Cu} = 400 W/m°C

Total heat flowing per second = *q*_{Al} + *q*_{Cu}

= 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.

(Figure 28-E5)

**Answer-26 :-**

*q* = *q*_{1} +* **q*_{2} ……………(1)

R_{1} and R_{2} are in parallel, so total heat across R_{1} and R_{2} will be same.

From equation (1) and (2),

q = q_1 + {7q_1}/6`

**q _{1} = 60 J/sec**

**Question-27 :- (Heat Transfer Exercise HC Verma )**

Suppose the bent part of the frame of the previous problem has a thermal conductivity of 780 J s^{−1} m^{−1} °C^{−1} whereas it is 390 J s^{−1} m^{−}1°C^{−1} for the straight part. Calculate the ratio of the rate of heat flow through the bent part to the rate of heat flow through the straight part.

**Answer-27 :-**

Resistance of any branch, R = l/KA

Here, *K* is the thermal conductivity, *A* is the area of cross section and *l* is the length of the conductor.

since R_{1}_{ }and R_{2}_{ }are in parallel, the amount of heat flowing through them will be same.

**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)

Length, l = 2 mm = 0.0002 m

(b)

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

=381W

**Question-29 :- (Heat Transfer Exercise HC Verma )**

The two rods shown in following figure have identical geometrical dimensions. They are in contact with two heat baths at temperatures 100°C and 0°C. The temperature of the junction is 70°C. Find the temperature of the junction if the rods are interchanged.

**Answer-29 :-**

As the rods are connected in series, the rate of flow of heat will be same in both the cases.

**Case 1:**

Rate of flow of heat is given by **dQ/dt =KAΔT/l**

Rate of heat flow in rod P will be same as that in rod Q.

**Case 2:**

Again, the rate of flow of heat will be same in rod P and Q.

**Question-30 :-**

The three rods shown in figure (28 E-7) have identical geometrical dimensions. Heat flows from the hot end at a rate of 40 W in the arrangement (a). Find the rates of heat flow when the rods are joined as in arrangement (b) and in (c). Thermal conductivities of aluminium and copper are 200 W m^{−1}°C^{−1} and 400 W m^{−1}°C^{−1} respectively.

**Answer-30 :-**

For arrangement (a),

Temperature of the hot end ,*T*_{1} = 100°C

Temperature of the cold end ,*T*_{2} = 0°C

Rate of flow of heat is given by

For arrangement (c),

**Question-31 :- (Heat Transfer Exercise HC Verma )**

Four identical rods AB, CD, CF and DE are joined as shown in figure (Figure 28-E8) . The length, cross-sectional area and thermal conductivity of each rod are l, A and K respectively. The ends A, E and F are maintained at temperature T_{1}, T_{2} and T_{3} respectively. Assuming no loss of heat to the atmosphere, find the temperature at B.

(Figure 28-E8)

**Answer-31 :-**

Let the temperature at junction B be T.

Let *q*_{1}, *q*_{2} and *q*_{3} be the heat currents, i.e. rate of flow of heat per unit time in AB, BCE and BDF, respectively.

From the diagram, we can see that

*q*_{1} = *q*_{2} + *q*_{3}

The rate of flow of heat is given by

Using this tn the above equation, we get

**Question-32 :-**

Seven rods A, B, C, D, E, F and G are joined as shown in the figure (Figure 28-E9). All the rods have equal cross-sectional area A and length *l*. The thermal conductivities of the rods are K_{A} = K_{C} = K_{0}, K_{B} = K_{D} = 2K_{0}, K_{E} = 3K_{0}, K_{F} = 4K_{0} and K_{G} = 5K_{0}. The rod E is kept at a constant temperature T_{1} and the rod G is kept at a constant temperature T_{2} (T_{2} > T_{1}). (a) Show that the rod F has a uniform temperature T = (T_{1} + 2T_{2})/3. (b) Find the rate of heat flowing from the source which maintains the temperature T_{2}.

(Figure 28-E9)

**Answer-32 :-**

Given:

K_{A} = K_{C} = K_{0}

K_{B} = K_{D} = 2K_{0}

K_{E} = 3K_{0}, K_{F} = 4K_{0}

K_{9}= 5K_{0}

Here, K denotes the thermal conductivity of the respective rods.

In steady state, temperature at the ends of rod F will be same.

Rate of flow of heat through rod A + rod C = Rate of flow of heat through rod B + rod D

(b) To find the rate of flow of heat from the source (rod G), which maintains a temperature T_{2} is given by

First, we will find the effective thermal resistance of the circuit.

From the diagram, we can see that it forms a balanced Wheatstone bridge.

Also, as the ends of rod F are maintained at the same temperature, no heat current flows through rod F.

Hence, for simplification, we can remove this branch.

From the diagram, we find that R_{A }and R_{B} are connected in series.

∴ R_{AB} = R_{A} + R_{B}

R_{C} and R_{D }are also connected in series.

∴ R_{CD} = R_{C} + RD

Then, R_{AB}_{ }and R_{CD}_{ }are in parallel connection.

**Question-33 :- (Heat Transfer Exercise HC Verma )**

Find the rate of heat flow through a cross section of the rod shown in figure (28-E10) (θ2 > θ1). Thermal conductivity of the material of the rod is K.

(Figure 28-E10)

**Answer-33 :-**

We can infer from the diagram that ΔPQR is similar to ΔPST.

So, by the property of similar triangles,

Thermal resistance is given by

**Heat Transfer Exercise HC Verma Capacities of Gases Questions**

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

**Question-34 :-**

A rod of negligible heat capacity has length 20 cm, area of cross section 1.0 cm^{2} and thermal conductivity 200 W m^{−1}°C^{−1}. The temperature of one end is maintained at 0°C and that of the other end is slowly and linearly varied from 0°C to 60°C in 10 minutes. Assuming no loss of heat through the sides, find the total heat transmitted through the rod in these 10 minutes.

**Answer-34 :-**

Given:

Length of the rod,* l* = 20 cm = 0.2 m

Area of cross section of the rod, *A* = 1.0 cm^{2} = 1.0 × 10 ^{-4}m^{2}

Thermal conductivity of the material of the rod, k = 200 W m^{-1}° C^{-1}

The temperature of one end of the rod is increased uniformly by 60° C within 10 minutes.

This mean that the rate of increase of the temperature of one end is 0.1° C per second

⇒ 60/10×60 °C/s

So, total heat flow can be found by adding heat flow every second.

Rate of flow of heat = dQ/dt

For each interval,

sum of n terms of an AP is given by

⇒ Q_{net} = 1800J (approximately)

**Question-35 :-**

A hollow metallic sphere of radius 20 cm surrounds a concentric metallic sphere of radius 5 cm. The space between the two spheres is filled with a nonmetallic material. The inner and outer spheres are maintained at 50°C and 10°C respectively and it is found that 100 J of heat passes from the inner sphere to the outer sphere per second. Find the thermal conductivity of the material between the spheres.

**Answer-35 :-**

A = 4π*r*^{2}

Let:

Radius of the inner sphere =* **r*_{1}

Radius of the outer sphere = *r*_{2}

Consider a shell of radii r and thickness dr.For this shell,

Rate of flow of heat , q=-K. (4πr²).dt/dx

Here, the negative sign indicates that the temperature decreases with increasing radius.

⇒ K = 3W /m°c

**Question-36 :- (Heat Transfer Exercise HC Verma )**

Following figure shows two adiabatic vessels, each containing a mass m of water at different temperatures. The ends of a metal rod of length L, area of cross section A and thermal conductivity K, are inserted in the water as shown in the figure. Find the time taken for the difference between the temperatures in the vessels to become half of the original value. The specific heat capacity of water is s. Neglect the heat capacity of the rod and the container and any loss of heat to the atmosphere.

**Answer-36 :-**

Rate of transfer of heat from the rod is given as

ΔQ/Δt = KA(T_{1}-T_{2})/ l

In time t, the temperature difference becomes half.

In time Δt, the heat transfer from the rod will be given by

ΔQ = KA(T_{1 }– T_{2}) /l Δt…………(i)

Heat loss by water at temperature T_{1} is equal to the heat gain by water at temperatureT_{2}.

Therefore, heat loss by water at temperature T_{1} in time Δt is given by

ΔQ = ms(T_{1 }– T_{2})…………(ii)

From equation (i) and (ii),

This gives us the fall in the temperature of water at temperature T_{1}.

Similarly, rise in temperature of water at temperature T_{2} is given by

Change in the temperature is given by

Here, ΔT/Δt is the rate of change of temperature difference.

Taking limit Δ t→ 0,

On integrating within proper limit, we get

**Question-37 :-**

Two bodies of masses *m*_{1} and *m*_{2} and specific heat capacities *s*_{1} and *s*_{2} are connected by a rod of length *l*, cross-sectional area A, thermal conductivity K and negligible heat capacity. The whole system is thermally insulated. At time *t* = 0, the temperature of the first body is T_{1} and the temperature of the second body is T_{2} (T_{2} > T_{1}). Find the temperature difference between the two bodies at time *t*.

**Answer-37 :-**

Rate of transfer of heat from the rod is given by

ΔQ/Δt = KA(T_{2}-T_{1})/l

Heat transfer from the rod in time Δ *t* is given by

ΔQ/Δt = KA(T_{2}-T_{1})/l Δt…………(1)

Heat loss by the body at temperature* **T*_{2} is equal to the heat gain by the body at temperature T_{1}

Therefore, heat loss by the body at temperature t_{2} in time Δt is given by

ΔQ = s_{2}m_{2 }(T_{2}-T_{2})….(2)

from equation (i) and (ii)

This gives us the fall in the temperature of the body at temperature *T*_{2}.

Similarly, rise in temperature of water at temperature *T*_{1} is given by

Change in the temperature is given by

On integrating both the sides, we get

lim Δ t → 0

**Question-38 :- (Heat Transfer Exercise HC Verma )**

An amount *n* (in moles) of a monatomic gas at an initial temperature T_{0} is enclosed in a cylindrical vessel fitted with a light piston. The surrounding air has a temperature T* _{s}* (> T

_{0}) and the atmospheric pressure is P

_{α}. Heat may be conducted between the surrounding and the gas through the bottom of the cylinder. The bottom has a surface area A, thickness

*x*and thermal conductivity K. Assuming all changes to be slow, find the distance moved by the piston in time

*t*.

**Answer-38 :-**

In time *dt**,* heat transfer through the bottom of the cylinder is given by

For a monoatomic gas, pressure remains constant.

For a monoatomic gas,

Integrating both the sides,

From the gas equation,

**Question-39 :-**

Assume that the total surface area of a human body is 1.6 m^{2} and that it radiates like an ideal radiator. Calculate the amount of energy radiated per second by the body if the body temperature is 37°C. Stefan constant σ is 6.0 × 10^{−8} W m^{−2} K^{−4}.

**Answer-39 :-**

Given:

Area of the body, *A* = 1.6 m^{2}^{ }

Temperature of the body,* T *= 310 K

From Stefan-Boltzmann law,

Here, *A* is the area of the body and σ is the Stefan-Boltzmann constant.

Energy radiated per second = 1.6 × 6 × 10^{−8} × (310)^{4}

= 886.58 ≈ 887 J

**Question-40 :-**

Calculate the amount of heat radiated per second by a body of surface area 12 cm^{2} kept in thermal equilibrium in a room at temperature 20°C. The emissivity of the surface = 0.80 and σ = 6.0 × 10^{−8} W m^{−2} K^{−4}.

**Answer-40 :-**

Given:

Area of the body, *A* = 12 × 10^{−4} m^{2} Temperature of the body,* T* = (273 + 20) = 293 K

Emissivity, *e* = 0.80

Rate of emission of heat, *R* = *AeσT*^{4}

*R* = 12 × 10^{−4} × 0.80 × 6.0 × 10^{−8 }× (293)^{4}

*R *= 0.42 J (approximately)

**Question-41 :-**

A solid aluminium sphere and a solid copper sphere of twice the radius are heated to the same temperature and are allowed to cool under identically surrounding temperatures. Assume that the emissivity of both the spheres in the same. Find the ratio of (a) the rate of heat loss from the aluminium sphere to the rate of heat loss from the copper sphere and (b) the rate of fall of temperature of the aluminium sphere to the rate of fall of temperature of the copper sphere. The specific heat capacity of aluminium = 900 J kg^{−1}°C^{−1} and that of copper = 390 J kg^{−1}°C^{−1}. The density of copper = 3.4 times the density of aluminium.

**Answer-41 :-**

(a) Rate of loss of heat = *eAσ**T*^{4}

^{}

(b) Relation between the amount of heat loss by both the spheres in a small time Δ*t* is given by

**Question-42 :-**

A 100 W bulb has tungsten filament of total length 1.0 m and radius 4 × 10^{−5} m. The emissivity of the filament is 0.8 and σ = 6.0 × 10^{−8} W m^{−2}K^{4}. Calculate the temperature of the filament when the bulb is operating at correct wattage.

**Answer-42 :-**

Given:

Power of the bulb, *P* = 100 W

Length of the filament,* l* = 1.0 m

Radius of the filament, *r *= 4 × 10^{−5} m

According to Stefan’s law,

Here, *e* is the emissivity of the tungsten and σ is Stefan’s constant.

**Question-43 :- (Heat Transfer Exercise HC Verma )**

A spherical ball of surface area 20 cm^{2} absorbs any radiation that falls on it. It is suspended in a closed box maintained at 57°C. (a) Find the amount of radiation falling on the ball per second. (b) Find the net rate of heat flow to or from the ball at an instant when its temperature is 200°C. Stefan constant = 6.0 × 10^{−8} W m^{−2} K^{−4}.

**Answer-43 :-**

(a)

Area of the ball,* A* = 20 × 10^{−4} m^{2}

Temperature of the ball, *T* = 57°C = 57 + 273 = 330 K

Amount of heat radiated per second = *AσT*^{4}

= 20 × 10^{−4} × 6 × 10^{−8} × (330)^{4}

= 1.42 J

(b)

Net rate of heat flow from the ball when its

temperature is 200 °C is given by

eAσ (T_{1}^{4} – T_{2}^{4})

= 20 × 10^{-4 }× 6 × 10^{-8} × 1 ((473)^{4} – (330)^{4} [∴ e = 1]
= 4.58 W

**Question-44 :-**

A spherical tungsten piece of radius 1.0 cm is suspended in an evacuated chamber maintained at 300 K. The piece is maintained at 1000 K by heating it electrically. Find the rate at which the electrical energy must be supplied. The emissivity of tungsten is 0.30 and the Stefan constant σ is 6.0 × 10^{−8} W m^{−2} K^{−4}.

**Answer-44 :-**

Given:

Radius of the spherical tungsten, *r* = 10^{−2} m

Emissivity of the tungsten, *e* = 0.3

Stefan’s constant, σ = 6.0 × 10^{−8} W m^{−2} K^{−4}

Surface area of the spherical tungsten, *A* = 4π*r*^{2}

*A* = 4 π (10^{−2})^{2}

A = 4 π × 10^{−4} m^{2}

Rate at which energy must be supplied is given by

eσA (T_{2}^{4} – T_{1}^{4})

= (0.3)× 6 × 10^{-8} × 4π × 10^{-4}×((1000)^{4} – (300)^{4})

= 22.42 ≈ watt = 22 watt

**Question-45 :-**

A cubical block of mass 1.0 kg and edge 5.0 cm is heated to 227°C. It is kept in an evacuated chamber maintained at 27°C. Assuming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decrease. Specific heat capacity of the material of the block is 400 J Kg^{-1 }K-^{1.}

**Answer-45 :-**

It is given that a cube behaves like a black body.

∴ Emissivity, *e* = 1

Stefan’s constant, σ = 6 × 10^{−8} W/(m^{2} K^{4})

Surface area of the cube,* A* = 6 × 25 × 10^{−4}

Mass of the cube, *m* = 1 kg

Specific heat capacity of the material of the cube, *s* = 400 J/(kg-K)

Temperature of the cube,* **T*_{1} = 227 + 273 = 500 K

Temperature of the surrounding,* **T*_{0} = 27 + 273 = 300 K

Rate of flow of heat is given by

**ΔQ/Δ = eAσ ( T ^{4 }-T_{0} ^{4})**

⇒ **ms . ΔT/Δt = 1× 6 × 10 × 25 × 10 ^{-4 }(500^{4} -300^{4})**

**Question-46 :-**

A copper sphere is suspended in an evacuated chamber maintained at 300 K. The sphere is maintained at a constant temperature of 500 K by heating it electrically. A total of 210 W of electric power is needed to do it. When the surface of the copper sphere is completely blackened, 700 W is needed to maintain the same temperature of the sphere. Calculate the emissivity of copper.

**Answer-46 :-**

According to Stefan,s law,

**power = eAσ ( T ^{4 }-T_{0} ^{4})**

Temperature difference = 200 K

Let the emissivity of copper be* e.*

210 = *eAσ *(500^{4} − 300^{4}) …(1)

When the surface of the copper sphere is completely blackened, 700 W is needed to maintain the temperature of the sphere.

For a black body,

*e *= 1

700 = 1·*Aσ*(5004 − 3004) …(2)

On dividing equation (1) by equation (2) we have,

200/700 = e/1

⇒ e = 0.3

**Question-47 :-**

A spherical ball A of surface area 20 cm^{2} is kept at the centre of a hollow spherical shell B of area 80 cm^{2}. The surface of A and the inner surface of B emit as blackbodies. Both A and B are at 300 K. (a) How much is the radiation energy emitted per second by the ball A? (b) How much is the radiation energy emitted per second by the inner surface of B? (c) How much of the energy emitted by the inner surface of B falls back on this surface itself?

**Answer-47 :-**

Given:

Surface area of the spherical ball, *S*_{A} = 20 cm^{2} = 20 × 10 ^{-4} m^{-2}

Surface area of the spherical shell, *S*_{B} = 80 cm^{2} = 80 × 10^{-4} m2

Temperature of the spherical ball, *T*_{A} = 300 K

Temperature of the spherical shell, *T*_{B }= 300 K

Radiation energy emitted per second by the spherical ball A is given by

E_{A} = σS_{A}T_{A}^{4}

⇒E_{A} 6.0 ×10^{-8}× 20 × 10^{-4 }× (300)^{4}

⇒E_{A} = 0.97 J

Radiation energy emitted per second by the inner surface of the spherical shell B is given by

E_{B} = σ S_{B}T_{B}^{4}

⇒ E_{B} = 6.0 ×10 ^{-8 }× 80 × 10^{-4 }× (300)^{4}

⇒ E_{B} = 3.76 J ≈ 3.8 J

Energy emitted by the inner surface of B that falls back on its surface is given by

E = E_{B} – E_{A }= 3.76 – 0.94

⇒ E = 2.82 J

**Question-48 :-(Heat Transfer Exercise HC Verma )**

A cylindrical rod of length 50 cm and cross sectional area 1 cm^{2} is fitted between a large ice chamber at 0°C and an evacuated chamber maintained at 27°C as shown in the figure. Only small portions of the rod are inside the chamber and the rest is thermally insulated from the surrounding. The cross section going into the evacuated chamber is blackened so that it completely absorbs any radiation falling on it. The temperature of the blackened end is 17°C when steady state is reached. Stefan constant σ = 6 × 10^{−8} W m^{−2} K^{−4}. Find the thermal conductivity of the material of the rod.

(Figure 28-E12)

**Answer-48 :-**

Length, *l* = 50 cm

Cross sectional area, *A* =1 cm^{2}

Stefan’s constant*, σ* = 6 × 10^{−8} W m^{−2} K^{−4}

Temperature of the blackened end = 17°C

Temperature of the chamber = 27°C

Temperature of one end of the rod = 0°C

According to Stefan’s law,

= 6 × 10.3 …. (1)

Also , (Delta Q)/(Delta t)=(KA(T_1 – T_2))/l……..(2) From (1) and (2),6×10.3 = (K × 17)/0.5

k = 1.8 w/ °C

**Heat Transfer Exercise HC Verma Capacities of Gases Questions**

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

**Question-49 :-**

One end of a rod of length 20 cm is inserted in a furnace at 800 K. The sides of the rod are covered with an insulating material and the other end emits radiation like a blackbody. The temperature of this end is 750 K in the steady state. The temperature of the surrounding air is 300 K. Assuming radiation to be the only important mode of energy transfer between the surrounding and the open end of the rod, find the thermal conductivity of the rod. Stefan constant σ = 6.0 × 10^{−8} W m^{−2} K^{−4}.

**Answer-49 :-**

Stefan’s constant, σ = 6 × 10^{−8} W m^{−2} K^{−4}

*l *= 0.2 m

T_{1} = 300 K

T_{2} = 750 K

From (1) and (2),

K = 74 W/m °C

**Question-50 :-**

A calorimeter of negligible heat capacity contains 100 cc of water at 40°C. The water cools to 35°C in 5 minutes. The water is now replaced by K-oil of equal volume at 40°C. Find the time taken for the temperature to become 35°C under similar conditions. Specific heat capacities of water and K-oil are 4200 J kg^{−1} K^{−1} and 2100 J kg^{−1} K^{−1}respectively. Density of K-oil = 800 kg m^{−3}.

**Answer-50 :-**

Given:

Volume of water, *V* = 100 cc = 100×10^{-3} m^{3}

Change in the temperature of the liquid, ∆*θ* = 5°C

Time, *T** *= 5 min

For water,

For K-oil,

From (i) and (ii),

⇒ t = 2 min

**Question-51 :-**

A body cools down from 50°C to 45°C in 5 mintues and to 40°C in another 8 minutes. Find the temperature of the surrounding.

**Answer-51 :-**

Let the temperature of the surroundings be T_{0}°C.

Case 1:

Initial temperature of the body = 50°C

Final temperature of the body = 45°C

Average temperature = 47.5 °C

Difference in the temperatures of the body and its surrounding = (47.5 − T)°C

Rate of fall of temperature = Δt/t = 5/5 = 1°C/min

By Newton’s law of cooling,

1 = -K [47.5 – t_{0}]……….. (i)

Case 2:

Initial temperature of the body = 45°C

Final temperature of the body = 40°C

Average temperature = 42.5 °C

Difference in the temperatures of the body and its surrounding = (42.5 − T_{0})°C

Rate of fall of temperature = ΔT/t= 5 8 = 5/8°C/min

From Newton^{,}s law of cooling,

0.625 = -K [42.5 – T_{0}] …….. (2)

Dividing (1) by (2),

⇒ 42.5 – T_0 = 29.68 – 0.625T

⇒ T_{0} = 34° C

**Question-52 :-**

A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. Find the water equivalent of the calorimeter.

**Answer-52 :-**

Let water equivalent to calorimeter be *w.*

Change in temperature *= *5°C

Specific heat of water = 4200 J/Kg °C

Rate of flow of heat is given by

*q *= Energy per unit time = msΔT/t

Case 1:

From calorimeter theory, these two rates of flow of heat should be equal to each other.

⇒ q_{1} = q_{2}

⇒ W = 12.5 ×10^{-3} kg

⇒ W = 12.5 g

**Question-53 :-**

A metal ball of mass 1 kg is heated by means of a 20 W heater in a room at 20°C. The temperature of the ball becomes steady at 50°C. (a) Find the rate of loss of heat to the surrounding when the ball is at 50°C. (b) Assuming Newton’s law of cooling, calculate the rate of loss of heat to the surrounding when the ball rises 30°C. (c) Assume that the temperature of the ball rises uniformly from 20°C to 30°C in 5 minutes. Find the total loss of heat to the surrounding during this period. (d) Calculate the specific heat capacity of the metal.

**Answer-53 :-**

In steady state, the body has reached equilibrium. So, no more heat will be exchanged between the body and the surrounding.

This implies that at steady state,

Rate of loss of heat = Rate at which heat is supplied

Given:

Mass, *m* = 1 kg

Power of the heater = 20 W

Room temperature = 20°C

(a)At steady state,

Rate of loss of heat = Rate at which heat is supplied

And, rate of loss/gain of heat = Power

∴ dQ/dt = p = 20W

(b) By Newton’s law of cooling, rate of cooling is directly proportional to the difference in temperature.

So, when the body is in steady state, then its rate of cooling is given as

When the temperature of the body is 30°C, then its rate of cooling is given as

The initial rate of cooling when the body^{,}s temperature is 20°C is given as

t = 5min = 300s

Heat liberated = 10/3 × 300 = 1000J

Net heat absorbed = Heat supplied − Heat Radiated

= 6000 − 1000 = 5000 J

d) Net heat absorbed is used for raising the temperature of the body by 10°C.

*∴ m *S ∆T = 5000

= 500 J kg^{-1} C^{-1}

**Question-54 :-**

A metal block of heat capacity 80 J°C^{−1} placed in a room at 20°C is heated electrically. The heater is switched off when the temperature reaches 30°C. The temperature of the block rises at the rate of 2°C s^{−1} just after the heater is switched on and falls at the rte of 0.2 °C s^{−1} just after the heater is switched off. Assume Newton’s law of cooling to hold. (a) Find the power of the heater. (b) Find the power radiated by the block just after the heater is switched off. (c) Find the power radiated by the block when the temperature of the block is 25°C. (d) Assuming that the power radiated at 25°C represents the average value in the heating process, find the time for which the heater was kept on.

**Answer-54 :-**

Given:

Heat capacity of the metal block,* s* = 80 J°C^{−1}

Heat absorb by the metal block is,

H = s × ΔT = 80 × (30-10) = 800J

Rate of rise in temperature of the block =(dθ/dt) = 2°C/s

Rate of fall in temperature of the block = (dθ/dt)=-0.2° C/s

The negative sign indicates that the temperature is falling with time.

(a) Energy = *s* (Δθ)*
*(Power = Energy per unit time

∴ Power of the heater = Heat capacity × dθ/dt

P = 80 × 2

P = 160 w

(b) Power Radiated = Energy lost per unit time.

∴ P = heat capacity × (dθ/dt)

Here, dθ/dt represents the rate of decrease in temperature.

p = 80×0.2

p = 16 W

(c) 16 W of power is radiated when the temperature of the block decreases from 30°C to 20°C.

⇒ 16 = *K* (30 − 20)

*K* = 1.6

From newton’s law of cooling,

**Question-55 :- (Heat Transfer Exercise HC Verma )**

A hot body placed in a surrounding of temperature θ_{0} obeys Newton’s law of cooling dθ/dt = -K (θ- θ_{0}) . Its temperature at *t* = 0 is θ_{1}. The specific heat capacity of the body is *s*and its mass is *m*. Find (a) the maximum heat that the body can lose and (b) the time starting from *t* = 0 in which it will lose 90% of this maximum heat.

**Answer-55 :-**

According to Newton’s law of cooling,

dθ/dt = -K (θ- θ_{0})

(a) Maximum heat that the body can lose, ΔQ_{max} = ms (θ_{1} – θ_{0})

(b) If the body loses 90% of the maximum heat, then the fall in temperature will be θ_{. }

From Newton’s law of cooling,

dθ/dt = -K (θ- θ_{0})

Integrating this equation within the proper limit, we get

At time *t* = 0,

θ = θ_{1}

At time *t*,

θ = θ

From (i) and (ii),

—: End of **Heat Transfer** Exercise **HC Verma** **Solutions** Concept of Physics:–

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