Two rectangular blocks, having identical dimensions, can be arranged in either configuration-I or configuration-II as shown in the figure. One of the blocks has thermal conductivity \kappa and the other 2 \kappa . The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in configuration-I. The time to transport the same amount of heat in configuration-II is

Heat and Thermodynamics PYQ — JEE Advanced 2013 · Shift 1 · Q54 with Solution

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Correct answer: A

Use Fourier’s law of heat conduction

For steady conduction through a slab,

\frac{Q}{t} = \frac{k A \Delta T}{L}

where:

$k$ = thermal conductivity,
$A$ = cross-sectional area,
$L$ = length along heat flow,
$\Delta T$ = temperature difference.

Since the same amount of heat $Q$ is transported in both cases and $\Delta T$ is also the same,

t \propto \frac{1}{\text{equivalent conductance}}

where conductance is

G = \frac{kA}{L}.

Interpret the two configurations

Because the two identical rectangular blocks are arranged differently along the $x$ -axis:

Configuration-I: the blocks are side by side along the direction of heat flow, so heat passes through one block and then the other. This is series combination.
Configuration-II: the blocks are stacked so that both connect the same hot and cold faces. This is parallel combination.

Let each block have dimensions such that along heat flow its length is $L$ and cross-sectional area is $A$ .

The two conductivities are:

k_1 = \kappa, \qquad k_2 = 2\kappa

Equivalent conductance in configuration-I (series)

For series conduction, thermal resistances add:

R = \frac{L}{kA}

So,

R_1 = \frac{L}{\kappa A}, \qquad R_2 = \frac{L}{2\kappa A}

Hence,

R_{\text{I}} = R_1 + R_2 = \frac{L}{\kappa A} + \frac{L}{2\kappa A} = \frac{3L}{2\kappa A}

Thus conductance is

G_{\text{I}} = \frac{1}{R_{\text{I}}} = \frac{2\kappa A}{3L}

Equivalent conductance in configuration-II (parallel)

For parallel conduction, conductances add:

G_{\text{II}} = \frac{\kappa A}{L} + \frac{2\kappa A}{L} = \frac{3\kappa A}{L}

Relate time with conductance

Since

Q = G\,\Delta T\, t

for fixed $Q$ and fixed $\Delta T$ ,

t \propto \frac{1}{G}

Therefore,

\frac{t_{\text{II}}}{t_{\text{I}}} = \frac{G_{\text{I}}}{G_{\text{II}}} = \frac{\frac{2\kappa A}{3L}}{\frac{3\kappa A}{L}} = \frac{2}{9}

Given

t_{\text{I}} = 9\text{ s}

so,

t_{\text{II}} = 9 \times \frac{2}{9} = 2\text{ s}

Check options

The correct option is:

\boxed{\text{A: } 2.0\text{ s}}

More from Heat and Thermodynamics

View written solutionFree

Correct answer: A

Use Fourier’s law of heat conduction

For steady conduction through a slab,

\frac{Q}{t} = \frac{k A \Delta T}{L}

where:

$k$ = thermal conductivity,
$A$ = cross-sectional area,
$L$ = length along heat flow,
$\Delta T$ = temperature difference.

Since the same amount of heat $Q$ is transported in both cases and $\Delta T$ is also the same,

t \propto \frac{1}{\text{equivalent conductance}}

where conductance is

G = \frac{kA}{L}.

Interpret the two configurations

Because the two identical rectangular blocks are arranged differently along the $x$ -axis:

Configuration-I: the blocks are side by side along the direction of heat flow, so heat passes through one block and then the other. This is series combination.
Configuration-II: the blocks are stacked so that both connect the same hot and cold faces. This is parallel combination.

Let each block have dimensions such that along heat flow its length is $L$ and cross-sectional area is $A$ .

The two conductivities are:

k_1 = \kappa, \qquad k_2 = 2\kappa

Equivalent conductance in configuration-I (series)

For series conduction, thermal resistances add:

R = \frac{L}{kA}

So,

R_1 = \frac{L}{\kappa A}, \qquad R_2 = \frac{L}{2\kappa A}

Hence,

R_{\text{I}} = R_1 + R_2 = \frac{L}{\kappa A} + \frac{L}{2\kappa A} = \frac{3L}{2\kappa A}

Thus conductance is

G_{\text{I}} = \frac{1}{R_{\text{I}}} = \frac{2\kappa A}{3L}

Equivalent conductance in configuration-II (parallel)

For parallel conduction, conductances add:

G_{\text{II}} = \frac{\kappa A}{L} + \frac{2\kappa A}{L} = \frac{3\kappa A}{L}

Relate time with conductance

Since

Q = G\,\Delta T\, t

for fixed $Q$ and fixed $\Delta T$ ,

t \propto \frac{1}{G}

Therefore,

\frac{t_{\text{II}}}{t_{\text{I}}} = \frac{G_{\text{I}}}{G_{\text{II}}} = \frac{\frac{2\kappa A}{3L}}{\frac{3\kappa A}{L}} = \frac{2}{9}

Given

t_{\text{I}} = 9\text{ s}

so,

t_{\text{II}} = 9 \times \frac{2}{9} = 2\text{ s}

Check options

The correct option is:

\boxed{\text{A: } 2.0\text{ s}}

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