The rate of heat transfer through a uniform metallic rod of cross-sectional area A and len… — JAMB UTME

The rate of heat transfer through a uniform metallic rod of cross-sectional area $A$ and length $d$ is given by $\frac{Q}{t} = \frac{k A \Delta T}{d}$ , where $Q$ is heat energy, $t$ is time, $\Delta T$ is temperature difference, and $k$ is thermal conductivity. What is the dimensional formula for $k$ in terms of mass ( $M$ ), length ( $L$ ), time ( $T$ ), and temperature ( $\Theta$ )?

$M L T^{-3} \Theta^{-1}$ Answer
B
$M L^2 T^{-3} \Theta^{-1}$
C
$M L T^{-2} \Theta^{-1}$
D
$M L^{-1} T^{-3} \Theta^{-1}$

Answer

The dimensional formula for thermal conductivity is

M L T^{-3} \Theta^{-1}

Expressing thermal conductivity explicitly gives

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

. Substituting fundamental dimensions yields

[k] = \frac{(M L^2 T^{-2})(L)}{(L^2)(\Theta)(T)} = M L T^{-3} \Theta^{-1}

. Thus, the dimensional formula

M L T^{-3} \Theta^{-1}

is correct.

Step-by-Step Solution

Make thermal conductivity

k

the subject of the formula.

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

Isolating

k

enables direct substitution of base dimensional quantities.

Substitute fundamental dimensions for each physical quantity.

[k] = \frac{[M L^2 T^{-2}][L]}{[L^2][\Theta][T]}

Heat energy

Q

has dimensions

[M L^2 T^{-2}]

, distance

d

[L]

, area

A

[L^2]

, temperature change

\Delta T

[\Theta]

, and time

t

[T]

Combine exponents for each fundamental dimension.

[k] = M^{1} L^{2+1-2} T^{-2-1} \Theta^{-1} = M L T^{-3} \Theta^{-1}

Applying exponent laws simplifies the expression to base units.

Key Concept

Dimensional Analysis of Physical Constants

Question

Answer

Step-by-Step Solution

Key Concept

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