Heat Transfer

Q: What are the three modes of heat transfer?

Heat is transferred by conduction, convection and radiation. Conduction passes energy between neighbouring particles without bulk motion, convection carries heat by the movement of a fluid, and radiation sends energy as electromagnetic waves, which can cross a vacuum.

Q: What is Fourier's law of heat conduction?

Fourier's law says the rate of heat flow through a layer is proportional to its area and to the temperature gradient across it: H = -kA dT/dx. The constant k is the thermal conductivity, measured in W m⁻¹ K⁻¹.

Q: What is thermal resistance?

Thermal resistance of a slab is R = L/(kA), in K W⁻¹. The heat current is H = ΔT/R, just as electric current is I = V/R.

Q: What is meant by steady state in heat conduction?

In the steady state the temperature at each point of a conductor no longer changes with time. No part stores heat, so the same heat current passes through every cross-section, and in a uniform lagged rod the temperature falls linearly along its length.

Q: Why do metals feel colder than wood at the same temperature?

Metals have a much larger thermal conductivity, mostly because of their free electrons. They draw heat out of your hand much faster than wood does, so the skin cools quickly and the metal feels colder, although both are at the same temperature.

Q: Is heat current a vector quantity?

No. Heat current is a scalar, like electric current. It has a sense of flow along a conductor, from hot to cold, but heat currents meeting at a junction add algebraically, not by the vector law.

Q: How is heat transfer asked in NEET?

NEET asks about the three modes of heat transfer, conductivity values and good or bad conductors, rods in series and parallel, junction temperature, the equivalent conductivity of two rods and everyday examples of convection such as sea breeze.

Q: What kind of conduction problems come in JEE Main?

JEE Main uses thermal resistance networks with series, parallel, junction and bridge rods, spherical and cylindrical shells, tapered rods, ice forming on a pond and heating through a rod, where the heat balance must be integrated over time.

PhysicsThermal Properties Of MatterFor NEET aspirants

Heat transfer is the flow of heat from a hotter region to a colder one by conduction, convection or radiation. For conduction, Fourier's law $H = - k A \frac{d T}{d x}$ and the idea of thermal resistance $R = \frac{L}{k A}$ turn every problem into a circuit: rods in series and parallel, junctions and bridges. This page covers the three modes of heat transfer, conduction in steady and changing states, radial and tapered conductors, and convection, all of which are tested in JEE Main and NEET.

On this page1Three modes2Fourier's law3Steady state4Thermal resistance5Series and parallel6Junctions and bridges7Shells and cones8Time-dependent problems9Convection

Key Formulas - Quick Reference

★ Must learnFourier's law: $H = \frac{d Q}{d t} = - k A \frac{d T}{d x}$ ; uniform rod in steady state: $H = \frac{k A ( T _{1} - T _{2} )}{L}$
★ Must learnThermal resistance $R = \frac{L}{k A}$ ( $K W^{- 1}$ ); $H = \frac{Δ T}{R}$
★ Must learnSeries: $R = R_{1} + R_{2} + \dots$ ; parallel: $\frac{1}{R} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots$
Equal-size rods: series $k_{eq} = \frac{L _{1} + L _{2}}{L _{1} / k _{1} + L _{2} / k _{2}}$ ; parallel $k_{eq} = \frac{k _{1} A _{1} + k _{2} A _{2}}{A _{1} + A _{2}}$
Junction law: $\sum H_{in} = \sum H_{out}$ ; balanced bridge: $\frac{R _{1}}{R _{2}} = \frac{R _{3}}{R _{4}}$
Spherical shell: $R = \frac{r _{2} - r _{1}}{4 π k r _{1} r _{2}}$ ; cylinder: $R = \frac{ln ( r _{2} / r _{1} )}{2 π k L}$ ; cone: $R = \frac{ℓ}{π k r _{1} r _{2}}$
Ice on a pond: $t = \frac{ρ L}{2 k θ} (x_{2}^{2} - x_{1}^{2})$
Heating through a rod: $t = \frac{L m s}{k A} ln \frac{θ _{0} - θ _{1}}{θ _{0} - θ _{2}}$

1. Three Modes of Heat Transfer

Heat is energy in transit, flowing because of a temperature difference from a body at higher temperature to one at lower temperature. It travels by three routes:

Figure 1: Heat travels from hot to cold by conduction (through a medium, without bulk motion), convection (by bulk motion of a fluid) and radiation (by electromagnetic waves, even through vacuum).

Feature	Conduction	Convection	Radiation
Medium	Needed	Needed (fluid)	Not needed
How	Collisions and free electrons pass energy from particle to particle	Hot fluid moves bodily, carrying heat	Electromagnetic waves
Matter moves?	No	Yes	No
Speed	Slow	Faster	Speed of light, $3 \times 1 0^{8} m s^{- 1}$
Main in	Solids	Liquids and gases	Everywhere, including vacuum
Heats the medium in between?	Yes	Yes	No
Example	Handle of a spoon in tea	Boiling water, sea breeze	Sunlight, warmth of a fire

Real situations mix the modes: a room heater warms air by convection and your face by radiation; a hot iron conducts heat into cloth.

2. Conduction and Fourier's Law

Consider a rod whose ends touch a hot reservoir at $T_{1}$ and a cold one at $T_{2}$ , with its sides lagged so heat flows only along it. Molecules at the hot end vibrate more strongly and pass energy to their neighbours by collisions; in metals, free electrons carry energy much faster, which is why metals are good conductors of both heat and electricity. There is no bulk movement of matter.

Figure 2: A lagged rod carries a heat current

H = \frac{d Q}{d t}

from the hot to the cold reservoir. Across any thin slice,

H = - k A \frac{d T}{d x}

(Fourier's law).

★ Must learn

Fourier's law: the rate of heat flow (heat current) through a slice is proportional to the area $A$ and to the temperature gradient:

H = \frac{d Q}{d t} = - k A \frac{d T}{d x}

$k$ is the thermal conductivity of the material. The minus sign shows heat flows towards falling temperature ( $\frac{d T}{d x} < 0$ gives $H > 0$ ). SI unit of $k$ : $W m^{- 1} K^{- 1}$ ; dimensions $[MLT^{- 3} K^{- 1}]$ .

Material	$k$ ( $W m^{- 1} K^{- 1}$ )	Material	$k$ ( $W m^{- 1} K^{- 1}$ )
Silver	406	Glass	0.8
Copper	385	Water	0.8
Aluminium	205	Brick	0.6
Brass	109	Wood	0.12
Steel	50.2	Glass wool	0.04
Lead	34.7	Air	0.024

A metal rod at $5^{\circ} C$ feels colder than wood at $5^{\circ} C$ : the metal conducts heat away from your hand much faster.
Poor conductors (wood, cork, glass wool, air trapped in wool or feathers) are insulators. Cooking pots have copper bottoms and wooden or plastic handles.
Gases are the poorest conductors; trapped still air is why a quilt or double glazing keeps you warm.

3. Steady State and Thermal Resistance

When a rod is first heated, each section absorbs some heat to warm up, so the heat current decreases along the rod. After some time the temperature of every section stops changing: this is the steady state. The temperatures are then constant in time (but different from point to point), no section stores heat, and the same heat current $H$ passes through every section.

Figure 3: Before steady state (dashed, computed from the heat-conduction equation) parts of the rod are still warming up, so the heat current varies along it. In the steady state (solid) no part stores heat,

H

is the same at every section and

T

falls linearly along a uniform lagged rod.

For a uniform lagged rod in steady state, $\frac{d T}{d x}$ is constant, so $T$ falls linearly from $T_{1}$ to $T_{2}$ :

H = \frac{k A ( T _{1} - T _{2} )}{L}, T (x) = T_{1} - (T_{1} - T_{2}) \frac{x}{L}

★ Must learn

Thermal resistance of a rod: $R = \frac{L}{k A}$ , so $H = \frac{T _{1} - T _{2}}{R}$ . SI unit $K W^{- 1}$ . This is exactly Ohm's law with temperature in place of potential.

Heat conduction	Electric conduction
Heat current $H = \frac{d Q}{d t}$	Electric current $I = \frac{d q}{d t}$
Temperature difference $Δ T$	Potential difference $Δ V$
$H = \frac{Δ T}{R}$ , $R = \frac{L}{k A}$	$I = \frac{Δ V}{R}$ , $R = \frac{ℓ}{σ A}$
Thermal conductivity $k$	Electrical conductivity $σ$
Junction law: heat in = heat out (steady state)	Kirchhoff's current law

Key idea

In the steady state a conductor network is a resistor circuit:

Δ T = H R

for every part, and heat currents obey the junction law.

4. Rods in Series and in Parallel

4.1 Series

Slabs placed one after another carry the same heat current. With junction temperature $T$ : $T_{H} - T = H R_{1}$ and $T - T_{C} = H R_{2}$ . Adding,

T_{H} - T_{C} = H (R_{1} + R_{2}) \Rightarrow R_{eq} = R_{1} + R_{2} + \dots

and the junction temperature is $T = \frac{T _{H} R _{2} + T _{C} R _{1}}{R _{1} + R _{2}}$ (for equal areas, $T = \frac{k _{1} T _{H} / L _{1} + k _{2} T _{C} / L _{2}}{k _{1} / L _{1} + k _{2} / L _{2}}$ ).

Figure 4: Slabs in series carry the same heat current. The temperature drop across each slab is

H R

, so the poorer conductor (larger

R

) takes the larger drop and has the steeper profile.

R_{eq} = R_{1} + R_{2}

4.2 Parallel

Slabs placed side by side between the same reservoirs have the same temperature difference; their heat currents add: $H = H_{1} + H_{2} = (T_{H} - T_{C}) (\frac{1}{R _{1}} + \frac{1}{R _{2}})$ , so

\frac{1}{R _{eq}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots

Figure 5: Slabs in parallel have the same temperature difference; the heat currents add:

H = H_{1} + H_{2}

, so

\frac{1}{R _{eq}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}

Exam Trick

Two identical-size rods: in series $k_{eq} = \frac{2 k _{1} k _{2}}{k _{1} + k _{2}}$ (harmonic mean); in parallel $k_{eq} = \frac{k _{1} + k _{2}}{2}$ (arithmetic mean). The larger temperature drop always falls across the poorer conductor, because $Δ T = H R$ with the same $H$ .

Series

Same heat current $H$ . Temperature drops add. $R_{eq} = R_{1} + R_{2}$ . Poor conductor gets the big drop.

Parallel

Same temperature difference. Heat currents add. $\frac{1}{R _{eq}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}$ . Good conductor carries more heat.

5. Junctions and Bridges

In the steady state no heat accumulates at a junction: the total heat current flowing in equals the total flowing out (junction law). Writing every current as flowing out of the junction, $\sum \frac{T _{x} - T _{i}}{R _{i}} = 0$ , so the junction temperature is a resistance-weighted average of the far-end temperatures. Heat current, like electric current, is a scalar: it has a sense along a conductor but does not add as a vector.

Figure 6: Networks of rods are solved like circuits. Left (Solved Example 7): junction law at

x

. Right (Solved Example 10): two paths of resistance

R

and

3 R

in parallel between the same temperatures.

A thermal Wheatstone bridge is balanced when $\frac{R _{1}}{R _{2}} = \frac{R _{3}}{R _{4}}$ : the two middle junctions are at the same temperature and the bridging rod carries no heat, so it can be ignored.

Figure 7: Solved Examples 8 and 9. When

\frac{R _{1}}{R _{2}} = \frac{R _{3}}{R _{4}}

the two middle junctions are at the same temperature, so the bridging rod carries no heat and can be removed.

Quick Recall: tap to check

In a series combination, which quantity is the same for each rod?

The heat current.

Where is the temperature drop largest in a series combination?

Across the rod of largest thermal resistance (poorest conductor).

Is heat current a vector?

No, it is a scalar like electric current; direction along a rod matters but it does not add vectorially.

6. Non-uniform Conductors: Shells and Cones

When the area changes along the flow, split the conductor into thin layers of thickness $d x$ , each with $d R = \frac{d x}{k A ( x )}$ , and add them in series by integration:

Spherical shell (radii $r_{1} < r_{2}$ ): $A = 4 π x^{2}$ , $R = \int_{r_{1}}^{r_{2}} \frac{d x}{4 π k x ^{2}} = \frac{r _{2} - r _{1}}{4 π k r _{1} r _{2}}$ .
Cylindrical shell (length $L$ ): $A = 2 π xL$ , $R = \frac{ln ( r _{2} / r _{1} )}{2 π k L}$ . Radial heat current $H = \frac{2 π k L ( T _{1} - T _{2} )}{ln ( r _{2} / r _{1} )}$ .
Truncated cone (length $ℓ$ , end radii $r_{1}$ , $r_{2}$ ): $R = \frac{ℓ}{π k r _{1} r _{2}}$ .

Figure 8: Radial flow. The area

4 π x^{2}

(sphere) or

2 π xL

(cylinder) changes with radius, so add thin-shell resistances:

d R = \frac{d x}{k ( 4 π x ^{2} )}

\frac{d x}{k ( 2 π xL )}

and integrate from

r_{1}

r_{2}

Figure 9: Solved Example 6. Each thin disc has resistance

d R = \frac{d x}{k π r ^{2}}

with

r

growing linearly; integrating gives

R = \frac{ℓ}{π k r _{1} r _{2}}

, i.e. a uniform rod of the geometric-mean area

π r_{1} r_{2}

JEE Advanced

Variable conductivity: if $k = k_{0} (1 + a T)$ or $k$ depends on $x$ , keep $H$ constant (steady state) and integrate $H d x = - k (T) A d T$ or $\frac{H d x}{k ( x ) A} = - d T$ . Ingen-Hausz experiment: identical wax-coated rods of different metals dipped in hot water; in the steady state the lengths $l$ over which wax melts satisfy $\frac{k _{1}}{k _{2}} = \frac{l _{1}^{2}}{l _{2}^{2}}$ . Rod losing heat from its sides (not lagged): the temperature falls exponentially along it rather than linearly.

7. Time-dependent Conduction Problems

When temperatures change, write a heat balance over a short time $d t$ and integrate:

Heating a body through a rod from a large reservoir at $θ_{0}$ : $\frac{k A}{L} (θ_{0} - θ) d t = m s d θ$ , which gives an exponential approach to $θ_{0}$ .
Ice forming on a pond with air at $- θ$ : heat released by freezing a layer $d x$ ( $ρ A d x L$ ) conducts through the ice of thickness $x$ : $\frac{k A θ}{x} d t = ρ A L d x$ , so $t = \frac{ρ L}{2 k θ} x^{2}$ .

Figure 10: Solved Example 12. Heat released by freezing the new layer

d x

must conduct through the ice already formed, so growth slows:

x^{2} = \frac{2 k θ t}{ρ L}

, and doubling the thickness takes four times the time.

Figure 11: Solved Examples 11 and 13. The heat current

\frac{K A ( θ _{0} - θ )}{L}

shrinks as the small vessel warms, so

θ

approaches

θ_{0}

exponentially:

t = \frac{L m s}{K A} ln \frac{θ _{0} - θ _{1}}{θ _{0} - θ _{2}}

8. Convection

Convection is heat transfer by the actual movement of the heated fluid. Fluid near a hot surface expands, becomes less dense and rises; cooler, denser fluid sinks to take its place, setting up a convection current. It cannot happen in solids and needs gravity for natural convection.

Natural (free) convection: driven by density differences (boiling water, room heaters placed low and ACs placed high, chimneys).
Forced convection: a pump or fan moves the fluid (car radiator, blood circulation, fan heater, cooling of electronics).
Sea breeze and land breeze: by day land heats faster (lower specific heat), air rises over land and the breeze blows from sea to land; at night the reverse.
Trade winds: air heated at the equator rises and moves towards the poles at high altitude; cooler surface air blows towards the equator, deflected westward by the Earth's rotation.
Convection losses are roughly proportional to the temperature difference (the basis of Newton's law of cooling for forced convection).

Figure 12: Natural convection on a large scale. Water's high specific heat keeps the sea cooler by day and warmer by night; air rises over the warmer surface and the surface wind (blue) blows towards it.

Figure 13: The electrical-analogy method. Temperature is like potential, heat current like electric current,

R = \frac{L}{k A}

\frac{ℓ}{σ A}

Figure 14: Mind map of this concept for quick revision (radiation is on the next page).

9. Solved Examples

Solved Example 1

A rod of length

10 m

, area

0.5 m^{2}

and

k = 2 W m^{- 1}^{\circ} C^{- 1}

joins reservoirs at

10 0^{\circ} C

and

0^{\circ} C

. Find the heat current and the temperature at a distance

x

from the hot end (steady state).

Solution:

$R = \frac{L}{k A} = \frac{10}{2 \times 0.5} = 10^{\circ} C W^{- 1}$ ; $H = \frac{100 - 0}{10} = 10 W$ .

The same $H$ flows through the first $x$ metres: $\frac{k A ( 100 - T )}{x} = \frac{k A ( 100 - 0 )}{L} \Rightarrow T = \frac{100 ( L - x )}{L}$ .

Answer: $H = 10 W$ ; $T = 100 - 10 x$ (in $^{\circ} C$ , $x$ in m).

Solved Example 2

A uniform rod AC of length

9 m

has its ends at

T_{H} = 8 0^{\circ} C

and

T_{L} = 2 0^{\circ} C

. Find the temperature

5 m

from the hot end in the steady state.

Solution:

Same heat current throughout: $\frac{T _{H} - T _{L}}{9} = \frac{T _{H} - T}{5} \Rightarrow \frac{60}{9} = \frac{80 - T}{5}$ .

Answer: $T = \frac{140}{3} \approx 46. 7^{\circ} C$ .

Solved Example 3

The outer wall of a hill-resort house is teak wood (thickness

L_{1}

, conductivity

k_{1}

), two layers of an unknown material (each thickness

L

, conductivity

k

) and brick (thickness

L_{2} = 5 L_{1}

, conductivity

k_{2} = 5 k_{1}

). In the steady state

T_{1} = 2 5^{\circ} C

T_{2} = 2 0^{\circ} C

and

T_{5} = - 2 0^{\circ} C

. Find the interface temperatures

T_{3}

and

T_{4}

Solution:

Wood: $R_{1} = \frac{L _{1}}{k _{1} A}$ ; brick: $R_{2} = \frac{5 L _{1}}{5 k _{1} A} = R_{1}$ . Let each middle layer have resistance $R$ . The same heat current flows through all four:

\frac{25 - 20}{R _{1}} = \frac{20 - T _{3}}{R} = \frac{T _{3} - T _{4}}{R} = \frac{T _{4} - ( - 20 )}{R _{1}}

First and last: $T_{4} + 20 = 5 \Rightarrow T_{4} = - 1 5^{\circ} C$ . Middle two: $20 - T_{3} = T_{3} - T_{4} \Rightarrow T_{3} = \frac{20 + T _{4}}{2}$ .

Answer: $T_{4} = - 1 5^{\circ} C$ , $T_{3} = 2. 5^{\circ} C$ (and $R = 3.5 R_{1}$ ).

Figure 15: Solved Example 3 to scale. The same heat current flows through every layer, so each temperature drop is proportional to that layer's thermal resistance:

5^{\circ} C

across wood and brick (

R_{1}

each),

17. 5^{\circ} C

across each insulating layer (

R = 3.5 R_{1}

Solved Example 4

A single-room house has

100 m^{2}

of such wall with

k_{1} = 0.125

k_{2} = 0.625

k = 0.25 W m^{- 1}^{\circ} C^{- 1}

L_{1} = 4 cm

L_{2} = 20 cm

L = 10 cm

. Inside is kept at

2 5^{\circ} C

and outside is

- 2 0^{\circ} C

. Find the power of the heater needed.

Solution:

$R_{1} = R_{2} = \frac{0.04}{0.125 \times 100} = 32 \times 1 0^{- 4}^{\circ} C W^{- 1}$ ; $R = \frac{0.10}{0.25 \times 100} = 40 \times 1 0^{- 4}^{\circ} C W^{- 1}$ .

$R_{eq} = R_{1} + R_{2} + 2 R = 144 \times 1 0^{- 4}^{\circ} C W^{- 1}$ . $H = \frac{25 - ( - 20 )}{144 \times 1 0 ^{- 4}}$ .

Answer: $H \approx 3.1 kW$ ; the heater must supply this to make up the loss.

Solved Example 5

Two thin concentric copper shells of radii

r_{1}

and

r_{2}

(

r_{2} > r_{1}

) have a material of conductivity

k

between them. The inner shell is kept at

T_{H}

by a heater of power

P

at the centre and the outer at

T_{C}

. Find

P

Solution:

A shell of radius $x$ and thickness $d x$ has $d R = \frac{d x}{k \cdot 4 π x ^{2}}$ , so $R = \frac{1}{4 π k} (\frac{1}{r _{1}} - \frac{1}{r _{2}}) = \frac{r _{2} - r _{1}}{4 π k r _{1} r _{2}}$ .

Answer:

P = \frac{T _{H} - T _{C}}{R} = \frac{4 π k ( T _{H} - T _{C} ) r _{1} r _{2}}{r _{2} - r _{1}}

Solved Example 6

A rod of conductivity

k

is a truncated cone of length

ℓ

with end radii

r_{1}

and

r_{2}

. Find its thermal resistance between the two ends.

Solution:

At distance $x$ from the narrow end the radius is $r = r_{1} + \frac{( r _{2} - r _{1} ) x}{ℓ}$ , so $d r = \frac{r _{2} - r _{1}}{ℓ} d x$ .

$R = \int_{0}^{ℓ} \frac{d x}{k π r ^{2}} = \frac{ℓ}{k π ( r _{2} - r _{1} )} \int_{r_{1}}^{r_{2}} \frac{d r}{r ^{2}} = \frac{ℓ}{k π ( r _{2} - r _{1} )} (\frac{1}{r _{1}} - \frac{1}{r _{2}})$ .

Answer: $R = \frac{ℓ}{π k r _{1} r _{2}}$ .

Solved Example 7

Three rods of thermal resistances

R

2 R

and

2 R

meet at a point

x

. Their other ends are at

0^{\circ} C

10 0^{\circ} C

and

5 0^{\circ} C

respectively. Find the temperature of

x

in the steady state.

Solution:

Take all currents as leaving $x$ and set their sum to zero: $\frac{x - 0}{R} + \frac{x - 100}{2 R} + \frac{x - 50}{2 R} = 0$ .

$2 x + x - 100 + x - 50 = 0 \Rightarrow 4 x = 150$ .

Answer: $x = 37. 5^{\circ} C$ .

Solved Example 8

Four rods of resistances

R_{1}

R_{2}

(upper path) and

R_{3}

R_{4}

(lower path) join ends at

T_{1}

and

T_{2}

; a fifth rod

R^{'}

joins the two middle points. Find the condition for no heat to flow through

R^{'}

Solution:

With no heat in $R^{'}$ , both middle points are at the same $T$ , and each path carries its own current $H_{1}$ , $H_{2}$ : $T_{1} - T = H_{1} R_{1} = H_{2} R_{3}$ and $T - T_{2} = H_{1} R_{2} = H_{2} R_{4}$ .

Dividing: $\frac{R _{1}}{R _{2}} = \frac{R _{3}}{R _{4}}$ .

Answer: $R_{1} R_{4} = R_{2} R_{3}$ (balanced thermal Wheatstone bridge).

Solved Example 9

In the bridge of Figure 7 (upper path

6 R

3 R

; lower path

4 R

2 R

; middle rod

2 R

; ends at

10 0^{\circ} C

and

0^{\circ} C

), find the temperature of the upper middle junction.

Solution:

$\frac{6 R}{3 R} = \frac{4 R}{2 R} = 2$ : the bridge is balanced, so the middle rod carries no heat.

Upper path: $\frac{100 - T}{6 R} = \frac{T - 0}{3 R} \Rightarrow 100 - T = 2 T$ .

Answer: $T = \frac{100}{3} \approx 33. 3^{\circ} C$ .

Solved Example 10

A square frame is made of four identical rods, each of thermal resistance

R

. Two adjacent corners are held at

10 0^{\circ} C

and

0^{\circ} C

. Find the ratio of the heat current

H_{2}

through the three-rod path to

H_{1}

through the direct rod.

Solution:

Both paths have the same temperature difference: $H_{1} = \frac{100}{R}$ , $H_{2} = \frac{100}{3 R}$ .

Answer: $\frac{H _{2}}{H _{1}} = \frac{1}{3}$ . The total current splits $3 : 1$ , i.e. $\frac{3}{4}$ of it through the direct rod.

Solved Example 11

A container of negligible heat capacity holds

1 kg

of water at

0^{\circ} C

. A steel rod (

L = 10 m

A = 10 cm^{2}

k = 46 W m^{- 1}^{\circ} C^{- 1}

) connects it to a large steam chamber at

10 0^{\circ} C

. How long does the water take to reach

5 0^{\circ} C

? (

s = 4180 J kg^{- 1}^{\circ} C^{- 1}

; no losses, rod's heat capacity negligible)

Solution:

At temperature $T$ : $\frac{100 - T}{R} = m s \frac{d T}{d t}$ , with $R = \frac{L}{k A} = \frac{10}{46 \times 1 0 ^{- 3}} = 217^{\circ} C W^{- 1}$ .

$\int_{0}^{50} \frac{d T}{100 - T} = \int_{0}^{t} \frac{d t}{R m s} \Rightarrow ln 2 = \frac{t}{R m s}$ .

$t = R m s ln 2 = 217 \times 1 \times 4180 \times 0.693$ .

Answer: $t \approx 6.3 \times 1 0^{5} s \approx 175 h$ .

Solved Example 12

On a winter day the air is at

- θ^{\circ} C

. A cylindrical drum of height

h

, made of a bad conductor and open at the top, is filled with water at

0^{\circ} C

. Find the time for all the water to freeze. (Ice: conductivity

k

, density

ρ

, latent heat

L

; neglect expansion on freezing.)

Solution:

When the ice is $x$ thick, heat conducted up in time $d t$ is $\frac{k A θ}{x} d t$ . It comes from freezing a layer $d x$ : $ρ A d x L$ .

$\frac{k A θ}{x} d t = ρ A L d x \Rightarrow d t = \frac{ρ L}{k θ} x d x$ . Integrate from $0$ to $h$ .

Answer: $t = \frac{ρ L h ^{2}}{2 k θ}$ . Time grows as the square of the thickness.

Solved Example 13

A large tank of water at constant temperature

θ_{0}

is joined to a small vessel holding water of mass

m

θ_{1} < θ_{0}

by a metal rod of length

L

, area

A

and conductivity

k

. Find the time for the small vessel to reach

θ_{2}

(

θ_{1} < θ_{2} < θ_{0}

). Specific heat of water

s

; other heat capacities negligible.

Solution:

In time $d t$ : $\frac{k A}{L} (θ_{0} - θ) d t = m s d θ \Rightarrow d t = \frac{L m s}{k A} \frac{d θ}{θ _{0} - θ}$ .

Answer:

t = \frac{L m s}{k A} ln \frac{θ _{0} - θ _{1}}{θ _{0} - θ _{2}}

Solved Example 14

A steel rod (

15 cm

k = 50.2 W m^{- 1} K^{- 1}

) and a copper rod (

10 cm

k = 385 W m^{- 1} K^{- 1}

) are joined end to end. The free end of the steel is at

30 0^{\circ} C

and of the copper at

0^{\circ} C

. The steel rod's area is twice the copper rod's. Find the junction temperature (rods lagged, steady state).

Solution:

Same heat current: $\frac{50.2 ( 2 A ) ( 300 - T )}{0.15} = \frac{385 A ( T - 0 )}{0.10}$ .

$669.3 (300 - T) = 3850 T \Rightarrow 200800 = 4519 T$ .

Answer: $T \approx 44. 4^{\circ} C$ .

Solved Example 15

A brass boiler has base area

0.15 m^{2}

and thickness

1.0 cm

. It boils away water at

6.0 kg min^{- 1}

on a gas stove. Estimate the temperature of the part of the flame touching the boiler. (

k_{brass} = 109 W m^{- 1} K^{- 1}

L_{v} = 2256 kJ kg^{- 1}

)

Solution:

Heat current $H = \frac{6.0}{60} \times 2.256 \times 1 0^{6} = 2.256 \times 1 0^{5} W$ .

$H = \frac{k A ( T - 100 )}{d} \Rightarrow T - 100 = \frac{( 2.256 \times 1 0 ^{5} ) ( 0.01 )}{109 \times 0.15} = 138$ .

Answer: $T \approx 23 8^{\circ} C$ .

Solved Example 16

Two rods of the same length and area but conductivities

k_{1}

and

k_{2}

are joined in series. The equivalent conductivity is
(A)

k_{1} + k_{2}

(B)

\frac{k _{1} + k _{2}}{2}

(C)

\frac{2 k _{1} k _{2}}{k _{1} + k _{2}}

(D)

k_{1} k_{2}

Solution:

Answer: (C). $\frac{2 L}{k _{eq} A} = \frac{L}{k _{1} A} + \frac{L}{k _{2} A}$ , giving the harmonic mean. (In parallel it would be (B).)

Practice Questions

A glass window is $1 m^{2}$ and $4 mm$ thick ( $k = 0.8 W m^{- 1} K^{- 1}$ ). Inside $2 0^{\circ} C$ , outside $4^{\circ} C$ . Find the heat lost per second (surface effects ignored).Answer: $3.2 kW$
Two rods A and B of the same length and area ( $k_{A} = 2 k_{B}$ ) are in series between $10 0^{\circ} C$ (A's end) and $0^{\circ} C$ . Find the junction temperature.Answer: $66. 7^{\circ} C$
Three identical rods form an equilateral triangle; one corner is at $10 0^{\circ} C$ , another at $0^{\circ} C$ . Find the temperature of the third corner.Answer: $5 0^{\circ} C$
An ice layer $1 cm$ thick forms on a pond in $1 h$ . How long for it to grow from $1 cm$ to $2 cm$ ?Answer: $3 h$ (total time $\propto x^{2}$ )
Why are cooking pans given copper bottoms and wooden handles?Answer: Copper conducts heat quickly to the food; wood is a poor conductor and protects the hand
Why is a room heater kept near the floor but an AC near the ceiling?Answer: Warm air rises and cool air sinks, so each sets up a convection current through the room
Find the thermal resistance of a hollow sphere with $r_{1} = 10 cm$ , $r_{2} = 20 cm$ , $k = 0.04 W m^{- 1} K^{- 1}$ .Answer: about $9.9 K W^{- 1}$

Common Mistakes to Avoid

Watch out

Using $H = k A Δ T / L$ before the steady state is reached, when the current still varies along the rod.
Adding conductivities for rods in series. Add resistances in series; add conductances ( $\frac{1}{R}$ ) in parallel.
Forgetting to convert cm and $cm^{2}$ to m and $m^{2}$ in $R = \frac{L}{k A}$ .
Taking the larger temperature drop across the better conductor. In series it falls across the poorer conductor.
Calling heat current a vector (or a tensor). It is a scalar; only its sense along a conductor matters.
Using $R = \frac{L}{k A}$ for a sphere, cylinder or cone. The area changes: integrate $\frac{d x}{k A ( x )}$ .
Assuming ice thickness grows linearly with time. It grows as $t$ .
Saying convection can occur in solids or without gravity. Natural convection needs a fluid and gravity.

Frequently Asked Questions

What are the three modes of heat transfer?

Heat is transferred by conduction, convection and radiation. Conduction passes energy between neighbouring particles without bulk motion, convection carries heat by the movement of a fluid, and radiation sends energy as electromagnetic waves, which can cross a vacuum.

What is Fourier's law of heat conduction?

Fourier's law says the rate of heat flow through a layer is proportional to its area and to the temperature gradient across it: $H = - k A \frac{d T}{d x}$ . The constant $k$ is the thermal conductivity, measured in $W m^{- 1} K^{- 1}$ .

What is thermal resistance?

Thermal resistance of a slab is $R = \frac{L}{k A}$ , in $K W^{- 1}$ . The heat current is $H = \frac{Δ T}{R}$ , just as electric current is $I = \frac{V}{R}$ .

What is meant by steady state in heat conduction?

In the steady state the temperature at each point of a conductor no longer changes with time. No part stores heat, so the same heat current passes through every cross-section, and in a uniform lagged rod the temperature falls linearly along its length.

Why do metals feel colder than wood at the same temperature?

Metals have a much larger thermal conductivity, mostly because of their free electrons. They draw heat out of your hand much faster than wood does, so the skin cools quickly and the metal feels colder, although both are at the same temperature.

Is heat current a vector quantity?

No. Heat current is a scalar, like electric current. It has a sense of flow along a conductor, from hot to cold, but heat currents meeting at a junction add algebraically, not by the vector law.

How is heat transfer asked in NEET?

NEET asks about the three modes of heat transfer, conductivity values and good or bad conductors, rods in series and parallel, junction temperature, the equivalent conductivity of two rods and everyday examples of convection such as sea breeze.

What kind of conduction problems come in JEE Main?

JEE Main uses thermal resistance networks with series, parallel, junction and bridge rods, spherical and cylindrical shells, tapered rods, ice forming on a pond and heating through a rod, where the heat balance must be integrated over time.

Previous year questions on Heat Transfer

7 questions from past papers, each with a step-by-step solution.

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