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11th
Physics
Thermal Properties of Matter
Conduction
$Conduction Questions and An...$

Important questions on Conduction

Q: Two rods of the same length and diameter having thermal conductivities {K_1}\,{K_2} are joined in parallel. The equivalent thermal conductivity of the combination is:

\dfrac{1}{{{K_{eq}}}} = \dfrac{1}{{{K_1}}} + \dfrac{1}{{{K_2}}} \boxed{{K_{eq}} = \dfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}}

Q: State and explain the laws of thermal conductivity.

Consider a metal slab of the area of cross-section A and thickness d. Let the end faces (ABCD & EFGH) be maintained at temperatures \theta_1 and \theta_2 ( \theta_1> \theta_2 ).The amount of heat conducted (Q) from the hotter to colder face is,1. directly proportional to the area of cross-section (A)2. directly proportional to the time for which heat flows (t)3. directly proportional to the temperature difference between the faces ( \theta_1 - \theta_2 )4. inversely proportional to the distance between the faces (d)This is called the law of thermal conductivity.According to the law Q \propto (\frac{\mathrm{At}\left(\theta_{1}-\theta_{2}\right)}{\mathrm{d}}) (\mathrm{Q}=\frac{\mathrm{KAt}\left(\theta_{1}-\theta_{2}\right)}{\mathrm{d}}) K is a constant of proportionality called coefficient of thermal conductivity.

Q: Define the term 'conductivity' of a metallic wire. write its SI unit.

It is defined as the ratio of the current density in the wire to the electric field which brings above the flow of electron in the wire. It is reciprocal of the resistivity of the metalic wire. SI unit is siemens per meter \dfrac{S}{m}

Q: On the average, the temperature of the earth' s crust increases 1^{o} C for every 30 m of depth. The average thermal conductivity of the earth's crust is 0.75 J/m s K. Solar constant is 1.35 kW/m^{2} . Match the items in Column I with that in Column II:

Solar constant = 1.35 kW/m^{2} Thermal conductivity of earth's crust = 0.75 J/s m KHeat transferred per second is \dfrac{dQ}{dt} = -K (4 \pi r^{2}) \dfrac{dT}{dr} r = R_e = 6400 km - \dfrac{dT}{dr} = - \dfrac{1^{o}C}{30 m} Heat lost by the earth per second due to conduction from the core \dfrac{dQ}{dt} = \left(\dfrac{0.75 J \times 4 \pi}{msK} \right) \times (6400 \times 10^{3} m)^{2} \dfrac{1^{o} C }{30 m} P_1 = \left[ \dfrac{0.75 \times 4 \pi \times (6400 \times 10^{3} )^{2}}{30} \right] J/s = 1.286 \times 10^{13} J/s \approx 1.3 \times 10^{13} J/s Heat absorbed from the sun = S \pi R_e^{2} P_2 = \left( 1.35 \times 10^{3} \dfrac{W}{m^{2}} \right) \pi (6400 \times 10^{3} m)^{2} = 1.7 \times 10^{17} W Heat lost by the earth by radiation if e = 1 P_2 = e \sigma A ' T^{4} (A ' = 4\pi R_e^{2}) 1.7 \times 10^{17} = 1 \times 5.67 \times 20^{-8} \times 4 \pi \times (6400 \times 10^{3})^{2} T^{4} T ^{4} = \dfrac{1.7 \times 10^{17}}{5.67 \times 10^{-8} \times 4 \pi \times (6400 \times 10^{3})^{2}} = 5.8 \times 10^{9} = 58 \times 10^{8} Surface temperature T of earth (58)^{1/4} \times 10^{2} = (7.6)^{1/2} \times 10^{2} = 2.76 \times 100 = 276 K \dfrac{P_1}{P_2} = \dfrac{1.3 \times 10^{13}}{1.7 \times 10^{17}} =7.5 \times 10^{-5}

Q: A slab of stone area 3500{cm}^{2} and thickness 10cm is exposed on the lower surface to steam at {100}^{o}C . A block of ice at {0}^{o}C rests on upper surface of the slab. In one hour 4.8kg of ice of melted. The thermal conductivity of the stone is J{s}^{-1} {m}^{-1} {k}_{-1} is(Latent heat of ice =3.36\times { 10 }^{5 }J/kg )

Given : A = 3500 \ cm^2 = 0.35 \ m^2 l = 10 \ cm = 0.1 \ m \Delta T =100-0 = 100^o C Mass of ice melted m = 4.8 \ kg Time taken t = 1 \ hr = 3600 \ s Latent heat of ice L = 3.36\times 10^{5} \ J/kg Heat absorbed by ice = Heat conducted by slab \therefore mL = \dfrac{KA t\Delta T}{l} Or 4.8\times 3.36\times 10^{5} = \dfrac{K(0.35) (3600)(100)}{0.1} \implies \ K = 1.24 \ Js^{-1} m^{-2} k^{-1}

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easy38 Questionsmedium58 Questionshard11 Questions

HARD

Two rods of the same length and diameter having thermal conductivities $K_{1} K_{2}$ are joined in parallel. The equivalent thermal conductivity of the combination is:

$\frac{K _{1} K _{2}}{K _{1} + K _{2}}$

$K_{1} + K_{2}$

$\frac{K _{1} + K _{2}}{2}$

$K_{1} K_{2}$

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HARD

The heat capacity of a metal is 4200 J/k. Its water equivalent is-

$0.5 k g$

$1 k g$

$1.5 k$

$2 k g$

View Answer

HARD

State and explain the laws of thermal conductivity.

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MEDIUM

A cylinder of radius $R$ made of a material of thermal conductivity $K_{1}$ is surrounded by a cylindrical shell of inner radius $R$ and outer radius $2 R$ made of a material of thermal conductivity $K_{2}$ . The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is?

This question has multiple correct options

$K_{1} + K_{2}$

$\frac{K _{1} + 3 K _{2}}{4}$

$\frac{K _{1} + 8 K _{2}}{9}$

$\frac{8 K _{1} + K _{2}}{9}$

View Answer

HARD

Define the term 'conductivity' of a metallic wire. write its SI unit.

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HARD

On the average, the temperature of the earth' s crust increases $1^{o} C$ for every 30 m of depth. The average thermal conductivity of the earth's crust is 0.75 J/m s K. Solar constant is $1.35 k W / m^{2}$ . Match the items in Column I with that in Column II:

List 1i. Heat lost by the core of the earth per second due to conduction (in W)ii. Heat received by the earth per second from the sun (in W)ii. If e = 1, average surface temperature of the earth in equilibrium (in K)iv. Ratio of heat lost by the earth to the heat received from the sun

List 2276$$7.5 \times 10^{-5}$$$$1.3 \times 10^{13}$$$$1.7 \times 10^{17}$$

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HARD

A cylindrical rod of length 50 cm and cross-sectional area $1 c m^{2}$ is fitted between a large ice chamber at $0^{0} C$ and an evacuated chamber maintained at $2 7^{0} c$ `as shown in figure only small portions of the road are inside the chambers 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 $1 7^{0} C$ when steady state is reached.seftan constant $σ = 6 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ find the thermal conductivity of the material of the rod.

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HARD

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 sphere are maintained at $5 0^{0} C$ and $1 0^{0} C$ 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.

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HARD

Two rods A and B of same length and radius are joined together. the thermal conductivity of A and B are $2 K$ and $K$ . Under steady state conditions, if the temperature difference between the open ends of A and B is $3 6^{\circ} C$ , the temperature difference across 'A' is:

$1 2^{\circ} C$

$1 8^{\circ} C$

$2 4^{\circ} C$

$9^{\circ} C$

View Answer

HARD

A slab of stone area $3500 c m^{2}$ and thickness $10 c m$ is exposed on the lower surface to steam at $100^{o} C$ . A block of ice at $0^{o} C$ rests on upper surface of the slab. In one hour $4.8 k g$ of ice of melted. The thermal conductivity of the stone is $J s^{- 1}$ $m^{- 1}$ $k_{- 1}$ is
(Latent heat of ice $= 3.36 \times 10^{5} J / k g$ )

$12.0$

$10.5$

$1.02$

$1.24$

View Answer

Important questions on Conduction

Two rods of the same length and diameter having thermal conductivities K1​K2​ are joined in parallel. The equivalent thermal conductivity of the combination is:

K1​+K2​K1​K2​​

K1​+K2​

2K1​+K2​​

K1​K2​​

The heat capacity of a metal is 4200 J/k. Its water equivalent is-

0.5kg

1kg

1.5k

2kg

State and explain the laws of thermal conductivity.

K1​+K2​

4K1​+3K2​​

9K1​+8K2​​

98K1​+K2​​

Define the term 'conductivity' of a metallic wire. write its SI unit.

On the average, the temperature of the earth' s crust increases 1oC for every 30 m of depth. The average thermal conductivity of the earth's crust is 0.75 J/m s K. Solar constant is 1.35kW/m2. Match the items in Column I with that in Column II:

Two rods A and B of same length and radius are joined together. the thermal conductivity of A and B are 2K and K. Under steady state conditions, if the temperature difference between the open ends of A and B is 36∘C, the temperature difference across 'A' is:

12∘C

18∘C

24∘C

9∘C

A slab of stone area 3500cm2 and thickness 10cm is exposed on the lower surface to steam at 100oC. A block of ice at 0oC rests on upper surface of the slab. In one hour 4.8kg of ice of melted. The thermal conductivity of the stone is Js−1 m−1 k−1​ is (Latent heat of ice =3.36×105J/kg)

12.0

10.5

1.02

1.24

More important questions

Two rods of the same length and diameter having thermal conductivities $K_{1} K_{2}$ are joined in parallel. The equivalent thermal conductivity of the combination is:

$\frac{K _{1} K _{2}}{K _{1} + K _{2}}$

$K_{1} + K_{2}$

$\frac{K _{1} + K _{2}}{2}$

$K_{1} K_{2}$

$0.5 k g$

$1 k g$

$1.5 k$

$2 k g$

$K_{1} + K_{2}$

$\frac{K _{1} + 3 K _{2}}{4}$

$\frac{K _{1} + 8 K _{2}}{9}$

$\frac{8 K _{1} + K _{2}}{9}$

On the average, the temperature of the earth' s crust increases $1^{o} C$ for every 30 m of depth. The average thermal conductivity of the earth's crust is 0.75 J/m s K. Solar constant is $1.35 k W / m^{2}$ . Match the items in Column I with that in Column II:

Two rods A and B of same length and radius are joined together. the thermal conductivity of A and B are $2 K$ and $K$ . Under steady state conditions, if the temperature difference between the open ends of A and B is $3 6^{\circ} C$ , the temperature difference across 'A' is:

$1 2^{\circ} C$

$1 8^{\circ} C$

$2 4^{\circ} C$

$9^{\circ} C$

$12.0$

$10.5$

$1.02$

$1.24$