What is temperature gradient. Find the dimension of thermal conductivi

1.	What is temperature gradient. Find the dimension of thermal conductivity K.
Answer» The quantity \(\frac{\theta_1-\theta_2}{x}\) or \(\frac{d\theta}{dx}\) represents the rate of fall of temperature w.r.t. distance. The quantity \(\frac{d\theta}{dx}\) represents the rate of change of temperature w.r.t. distance and is called temperature gradient. Q = −KA[ \(\frac{d\theta}{dx}\)] t Q represent energy and its dimensions are : [Q] = [ML²T^-2 ] [dx] = [L] [A] = [L² ] [dθ] = [θ] [t] = [T] Dimension of K [K] = \(\frac{[ML^2T^{-2}][L]}{[L^2][\theta][T]}\) = [MLT^-3θ^-1]

What is temperature gradient. Find the dimension of thermal conductivity K.

Answer»

The quantity \(\frac{\theta_1-\theta_2}{x}\) or \(\frac{d\theta}{dx}\) represents the rate of fall of temperature w.r.t. distance.

The quantity \(\frac{d\theta}{dx}\) represents the rate of change of temperature w.r.t. distance and is called temperature gradient.

Q = −KA[ \(\frac{d\theta}{dx}\)] t

Q represent energy and its dimensions are :

[Q] = [ML²T^-2 ]

[dx] = [L]

[A] = [L² ]

[dθ] = [θ]

[t] = [T]

Dimension of K

[K] = \(\frac{[ML^2T^{-2}][L]}{[L^2][\theta][T]}\)

= [MLT^-3θ^-1]