HC Verma Questions and Solutions: Chapter 28: Heat Transfer- 3 | HC Verma Solutions - JEE PDF Download

Q.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.

Let :
`R_{BE} = 60/{KA} = R_2

q = q₁ + q₂ .......(1)
R₁ and R₂ are in parallel, so total heat across R₁ and R₂ will be same.

From equation (1) and (2),

q₁ = 60 J/sec

Q.27. 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.

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



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

Q.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 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 .

(a)

Length, l = 2 mm = 0.0002 m
Rate of flow of heat

= 8000 J / sec
(b)


Resistance of glass,
Resistance of air,
From the circuit diagram, we can find that all the resistors are connected

Rate of flow of heat ,=

=381W

Q.29. 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.

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
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.

Q.30. The three rods shown in figure  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 condcutivities of aluminium and copper are 200 W m^?1°C^?1 and 400 W m^?1°C^?1 respectively.

For arrangement (a),

Temperature of the hot end , T₁ = 100°C
Temperature of the cold end , T₂ = 0°C

Given :
q = 40W

For arrangement (b),


Rate of flow of heat is given by

For arrangement (c),


rate of heat flow =

Q.31. Four identical rods AB, CD, CF and DE are joined as shown in following figure . 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 T₁, T₂ and T₃ respectively. Assuming no loss of heat to the atmosphere, find the temperature at B.

Let the temperature at junction B be T.
Let q₁, q₂ and q₃ 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
q₁ = q₂ + q₃
The rate of flow of heat is given by

Using this tn the above equation, we get

Q.32. Seven rods A, B, C, D, E, F and G are joined as shown in the figure. All the rods have equal cross-sectional area A and length l. The thermal conductivities of the rods are K_A = K_C = K₀, K_B = K_D = 2K₀, K_E = 3K₀, K_F = 4K₀ and K_G = 5K₀. The rod E is kept at a constant temperature T₁ and the rod G is kept at a constant temperature T₂ (T₂ > T₁). (a) Show that the rod F has a uniform temperature T = (T₁ + 2T₂)/3. (b) Find the rate of heat flowing from the source which maintains the temperature T₂.

Given:
K_A = K_C = K₀
K_B = K_D = 2K₀
K_E = 3K₀, K_F = 4K₀
K₉= 5K₀
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 T₂ is given by
Rate of flow of heat, q =
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 RA and RB are connected in series.
? R_AB = R_A + R_B
R_C and R_D are also connected in series.
? R_CD = R_C + R_D
Then, R_AB and R_CD are in parallel connection.

Q.33. Find the rate of heat flow through a cross section of the rod shown in figure (28-E10) (?₂ > ?₁). Thermal conductivity of the material of the rod is K.

We can infer from the diagram that ?PQR is similar to ?PST.
So, by the property of similar triangles,

Thermal resistance is given by


Rate of flow of heat = ?Q/R

Q.34. A rod of negligible heat capacity has length 20 cm, area of cross section 1.0 cm² 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.

Given:
Length of the rod, l = 20 cm = 0.2 m
Area of cross section of the rod, A = 1.0 cm² = 1.0 × 10 ^-4m²
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

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)

Q.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.

A = 4?r²
Let:
Radius of the inner sphere = r₁
Radius of the outer sphere = r₂
Consider a shell of radii r and thickness dr. For this shell,
Rate of flow of heat,
Here, the negative sign indicates that the temperature decreases with increasing radius.

? K = 3W /m°c

Q.36. 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.

Rate of transfer of heat from the rod is given as

In time t, the temperature difference becomes half.
In time ?t, the heat transfer from the rod will be given by  

Heat loss by water at temperature T₁ is equal to the heat gain by water at temperatureT₂.
Therefore, heat loss by water at temperature T₁ in time ?t is  given by

From equation (i) and (ii),

This gives us the fall in the temperature of water at temperature T₁.
Similarly, rise in temperature of water at temperature T₂ is given by

On integrating within proper limit, we get

Q.37. Two bodies of masses m₁ and m₂ and specific heat capacities s₁ and s₂ 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 T₁ and the temperature of the second body is T₂ (T₂ > T₁). Find the temperature difference between the two bodies at time t.

Rate of transfer of heat from the rod is given by

Heat transfer from the rod in time ?? t is given by

Heat loss by the body at temperature T₂ is equal to the heat gain by the body at temperature T₁
Therefore, heat loss by the body at temperature t₂  in time ?t is given by

This gives us the fall in the temperature of the body at temperature T₂.
Similarly, rise in temperature of water at temperature T₁ is given by

Change in the temperature is given by

On integrating both the sides, we get
lim ? t ? 0

Q.38. An amount n (in moles) of a monatomic gas at an initial temperature T₀ is enclosed in a cylindrical vessel fitted with a light piston. The surrounding air has a temperature T_s (> T₀) and the atmospheric pressure is P_?. 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.

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,

Q.39. Assume that the total surface area of a human body is 1.6 m² 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 ? is 6.0 × 10^?8 W m^?2 K^?4.

Given:
Area of the body, A = 1.6 m² 
Temperature of the body, T = 310 K
From Stefan-Boltzmann law,

Here, A is the area of the body and ? is the Stefan-Boltzmann constant.
Energy radiated per second = 1.6 × 6 × 10^?8 × (310)⁴
=  886.58 ? 887 J

Q.40. Calculate the amount of heat radiated per second by a body of surface area 12 cm² kept in thermal equilibrium in a room at temperature 20°C. The emissivity of the surface = 0.80 and ? = 6.0 × 10^?8 W m^?2 K^?4.

Given:
Area of the body, A = 12 × 10^?4 m² Temperature of the body, T = (273 + 20) = 293 K
Emissivity, e = 0.80
Rate of emission of heat, R = Ae?T⁴
R = 12 × 10^?4 × 0.80 × 6.0 × 10^?8 × (293)⁴
R = 0.42 J (approximately)

Q.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.

(a) Rate of loss of heat = eA?T⁴


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

Q.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 ? = 6.0 × 10^?8 W m^?2K⁴. Calculate the temperature of the filament when the bulb is operating at correct wattage.

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 ? is Stefan's constant.

Q.43. A spherical ball of surface area 20 cm² 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.

(a) Area of the ball, A = 20 × 10^?4 m²
Temperature of the ball, T = 57°C = 57 + 273 = 330 K
Amount of heat radiated per second = A?T⁴
= 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
eA? (T₁⁴ - T₂⁴)
= 20 × 10^-4 × 6 × 10^-8 × 1 ((473)⁴ - (330)⁴  [? e = 1]
= 4.58 W

Q.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 ? is 6.0 × 10^?8 W m^?2 K^?4.

Given:
Radius of the spherical tungsten, r = 10^?2 m
Emissivity of the tungsten, e = 0.3
Stefan's constant, ? = 6.0 × 10^?8 W m^?2 K^?4
Surface area of the spherical tungsten, A = 4?r²
A = 4 ? (10^?2)²
A = 4 ? × 10^?4 m²
Rate at which energy must be supplied is given by
e?A (T₂⁴ - T₁⁴)
= (0.3)× 6 × 10^-8 × 4? × 10^-4×((1000)⁴ - (300)⁴)
= 22.42 ? watt = 22 watt

Q.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.

It is given that a cube behaves like a black body.
? Emissivity, e = 1
Stefan's constant, ? = 6 × 10^?8 W/(m² K⁴)
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, T₁ = 227 + 273 = 500 K
Temperature of the surrounding, T₀ = 27 + 273 = 300 K
Rate of flow of heat is given by
 

Q.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.

According to Stefan,s law,

Temperature difference = 200 K
Let the emissivity of copper be e.
210 = eA?(500⁴ ? 300⁴)   ...(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·A?(5004 ? 3004)   ...(2)
On dividing equation (1) by equation (2) we have,
200/700 = e/1
? e = 0.3  

Q.47. A spherical ball A of surface area 20 cm² is kept at the centre of a hollow spherical shell B of area 80 cm². 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?

Given:
Surface area of the spherical ball, S_A = 20 cm² = 20 × 10 ^-4 m^-2
Surface area of the spherical shell, S_B = 80 cm² = 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 = ?S_AT_A⁴
? 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 = ? S_BT_B⁴
? 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

Q.48. A cylindrical rod of length 50 cm and cross sectional area 1 cm² 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 ? = 6 × 10^?8 W m^?2 K^?4. Find the thermal conductivity of the material of the rod.

Length, l = 50 cm
Cross sectional area, A =1 cm²
Stefan's constant, ? = 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,

Also , (Delta Q)/(Delta t) = (KA(T_1 - T_2))/l ...
k = 1.8 w/ °C

Q.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 ? = 6.0 × 10^?8 W m^?2 K^?4.

Stefan's constant, ? = 6 × 10^?8 W m^?2 K^?4
l = 0.2 m
T₁ = 300 K
T₂ = 750 K

From (1) and (2),

Q.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^?1respectively. Density of K-oil = 800 kg m^?3.

Given:
Volume of water, V = 100 cc = 100×10^-3 m³
Change in the temperature of the liquid, ?? = 5°C
Time, T = 5 min
For water,

For K-oil,

From (i) and (ii),

? t = 2 min 

Q.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.

Let the temperature of the surroundings be T0°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 =
By Newton's law of cooling,

1 = -K [47.5 - t₀]........... (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 ? T₀)°C
Rate of fall of temperature =
From Newton,s law of cooling,

? 42.5 - T_0 = 29.68 - 0.625T
? T₀ = 34° C.

Q.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.

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 =
Case 1:

From calorimeter theory, these two rates of flow of heat should be equal to each other.
? q₁ = q₂

? W = 12.5 ×10^-3 kg
? W = 12.5 g

Q.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.

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

(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 =
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

Q.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.

Given:
Heat capacity of the metal block, s = 80 J°C^?1
Heat absorb by the metal block is,

Rate of rise in tempreature of the block
Rate of fall in temperature of the block =
The negative sign indicates that the temperature is falling with time.
(a) Energy = s(??)
(Power = Energy per unit time
? Power of the heater = Heat capacity × (d?/dt)
P = 80 × 2
P = 160 w
(b) Power Radiated = Energy lost per unit time.
? P = heat capacity × (d?/dt)
Here, (d?/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.

From newton's law of cooling,

Q.55. A hot body placed in a surrounding of temperature ?₀ obeys Newton's law of cooling  Its temperature at t = 0 is ?₁. The specific heat capacity of the body is sand 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.

According to Newton's law of cooling,

(a) Maximum heat that the body can lose, ?Q_max = ms (?₁ - ?₀)
(b) If the body loses 90% of the maximum heat, then the fall in temperature will be ?.
 
From Newton's law of cooling,

Integrating this equation within the proper limit, we get
At time t = 0,
? = ?₁
At time t,
? = ?