Heat Transfer Short Notes for JEE Main: Revision Notes PDF

Conduction

Heat current (Fourier's law)

5× asked, last 2025

H=\frac{dQ}{dt}=\frac{KA\,\Delta T}{L},\qquad H=-KA\,\frac{dT}{dx}

Steady state, no heat lost from the sides.

K

in W m⁻¹ K⁻¹;

dT/dx

is the temperature gradient.

Not one face of a box. Heat enters through every wall. Use the total area of all six faces, with the wall thickness as $L$ .
Not $H$ as the melting rate. $H$ is in watts. Divide by the latent heat: mass melted per second $=H/L_f$ .

Thermal resistance

5× asked, last 2025

R=\frac{L}{KA},\qquad H=\frac{\Delta T}{R}

Like Ohm's law:

\Delta T

acts as

V

H

I

. Unit K W⁻¹. Series and parallel rules are the same as for resistors.

Area or $K$ changing along the length: add thin slices, $R=\int dx/(KA)$ , then heat current $=\Delta T/R$ .
Not the radius ratio as the area ratio. A rod has $A=\pi r^2$ , so square the radius ratio before putting it in $R=L/KA$ .
Quick method. Networks: replace each rod by $R=L/KA$ and solve like a resistor circuit, using series, parallel or a node equation at a joint. A balanced bridge carries no heat in its middle rod.

Slabs in series

7× asked, last 2025

R=R_1+R_2,\qquad K_{\text{eq}}=\frac{L_1+L_2}{L_1/K_1+L_2/K_2}

Same area; heat passes through one after the other, so

H

is the same in each. Equal lengths:

K_{\text{eq}}=2K_1K_2/(K_1+K_2)

x

T

joint10075good,

3K

poor,

K

steeper

Same heat current: the poorer conductor takes the bigger drop (junction at 75 °C)

Equal $H$ means the temperature drop across each slab is proportional to its $R$ .
Steady state: temperature falls linearly along a uniform rod. In a composite wall each slope is proportional to $1/K$ , so the poorer conductor has the steeper part of the graph.
Not $K_1K_2/(K_1+K_2)$ . The joined rod is twice as long, so equal lengths give $2K_1K_2/(K_1+K_2)$ . Also invert the sum of reciprocals before quoting $K$ .

Slabs in parallel

1× asked, last 2025

\frac1R=\frac1{R_1}+\frac1{R_2},\qquad K_{\text{eq}}=\frac{K_1A_1+K_2A_2}{A_1+A_2}

Same length and same end temperatures; heat currents add. Equal areas:

K_{\text{eq}}=(K_1+K_2)/2

Junction temperature

7× asked, last 2025

\theta=\frac{\theta_1R_2+\theta_2R_1}{R_1+R_2},\qquad \sum_i\frac{\theta_i-\theta}{R_i}=0

Two slabs in series, outer faces at

\theta_1

and

\theta_2

. Rods meeting at a joint: net heat into the joint is zero.

Not the average of the end temperatures (true only if $R_1=R_2$ ). The slab with smaller $R=L/KA$ takes the smaller drop, so $\theta$ is nearer that slab's outer-face temperature (the better conductor's if $L$ , $A$ are equal).
Quick method. T-joint: a rod joined at its middle splits into two halves of $R/2$ . Solve the node equation for $\theta$ , then the heat current in any rod is its $\Delta\theta/R$ .

Radial flow through a shell

1× asked, last 2021

R_{\text{sph}}=\frac{r_2-r_1}{4\pi K r_1r_2},\qquad R_{\text{cyl}}=\frac{\ln(r_2/r_1)}{2\pi KL}

r_1,r_2

= inner, outer radii. Steady state: integrate

H=-KA\,d\theta/dr

A=4\pi r^2

2\pi rL

(

L

= cylinder length).

Not $\Delta\theta\times R$ or a flat-slab formula. Divide $\Delta\theta$ by the shell's resistance. $K\Delta\theta/(r_2-r_1)$ forgets that the area grows as $4\pi r^2$ .

Growth of ice on a pond

\frac{dx}{dt}=\frac{K\theta}{\rho L_f\,x},\qquad t=\frac{\rho L_f}{2K\theta}\,(x_2^2-x_1^2)

Air at

-\theta

°C above water at 0 °C;

x

= ice thickness;

K

\rho

of ice;

L_f

= latent heat of fusion.

t

thickness

x

t_0

4t_0

9t_0

123

x\propto\sqrt t

: each extra layer of the same thickness takes longer (1 : 3 : 5)

$t\propto x^2$ : growing from 0 to $x$ , $x$ to $2x$ and $2x$ to $3x$ takes times in the ratio 1 : 3 : 5. Ice thickens more and more slowly.

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Convection

Natural and forced convection

Radiation and Stefan–Boltzmann law

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Kirchhoff's law
Stefan–Boltzmann law3×
Net exchange with surroundings1×
Comparing cooling of bodies
Radiation received at a distance1×

Wien's law and black-body spectrum

Wien's displacement law
Energy distribution curve

Newton's law of cooling

Newton's law of cooling11×
Exponential cooling6×
Average-temperature form4×

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Heat Transfer revision notes

Conduction

12 more concepts in 4 topics

Convection

Radiation and Stefan–Boltzmann law

Wien's law and black-body spectrum

Newton's law of cooling

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Heat Transfer revision notes

1Conduction7 concepts

12 more concepts in 4 topics

2Convection1 concept

3Radiation and Stefan–Boltzmann law6 concepts

4Wien's law and black-body spectrum2 concepts

5Newton's law of cooling3 concepts

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Conduction

Convection

Radiation and Stefan–Boltzmann law

Wien's law and black-body spectrum

Newton's law of cooling