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 ( $\Theta$ )?

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

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Dimensional Analysis of Physical Constants

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru