Question

Rate of flow of heat through cylindrical rod is h1 temperature end of rod are T1 and T2 if all the dimensions of rod become double and temperature difference remain same and rate of heat flow become h2 then

Answer 1

\(H_1 = kA \frac{(T_2 - T_1)}l\)

⇒ \(H_1 = \frac{k\pi r^2(T_2 - T_1)}l\)   ......(1)

Now, the linear dimensions are doubled, that is, R = 2r and L = 2 while the temperature is same, then new rate of heat flow:

\(H_2 = \frac{kA'(T_2 - T_1)}L\)    ......(2)

⇒ \(H_2 = \frac{k\pi R^2(T_2 - T_1)}L\)

⇒ \(H_2 = \frac{k\pi (2r)^2(T_2 - T_1)}{2l}\)

⇒ \(H_2 = \frac{2k\pi r^2(T_2 - T_1)}{l}\)

After dividing the equation (2) by (1), we get;

H₂ = 2H₁