Question 802: Thermodynamics & KTG

Question

Rate of flow of heat through a cylindrical rod is ' $\mathrm{H}_1$ '. The temperature at the ends of the rod are ' $T_1$ ' and ' $T_2$ '. If all the dimensions of the rod become double and the temperature difference remains the same, the rate of flow of heat becomes ' $\mathrm{H}_2$ '. Then

Accepted Answer

Let $ l_1 $ be the initial length of the rod and $ r_1 $ be the radius of the rod. The initial rate of heat flow through the rod is given by:
$ \mathrm{H}_1 = \frac{\mathrm{kA}_1 \left(T_2 - T_1ight)}{l_1} $
When all dimensions of the rod are doubled, the rate of heat flow becomes:
$ \mathrm{H}_2 = \frac{\mathrm{kA}_2 \left(T_2 - T_1ight)}{l_2} $
The relationship between the rates can be expressed as:
$ \frac{\mathrm{H}_2}{\mathrm{H}_1} = \frac{\mathrm{A}_2}{\mathrm{A}_1} 	imes \frac{l_1}{l_2} $
If the radius is doubled, $ r_2 = 2r_1 $, then the new cross-sectional area is $ \mathrm{A}_2 = 4\mathrm{A}_1 $. Additionally, the new length is $ l_2 = 2l_1 $.
Thus,
$ \frac{\mathrm{H}_2}{\mathrm{H}_1} = 4 	imes \frac{1}{2} = 2 $
Therefore, the new rate of heat flow is:
$ \mathrm{H}_2 = 2 \mathrm{H}_1 $

Answer

$\mathrm{H}_2=4 \mathrm{H}_1$

Answer

$\mathrm{H}_2=2 \mathrm{H}_1$

Answer

$\mathrm{H}_2=\frac{\mathrm{H}_1}{2}$

Answer

$\mathrm{H}_2=\frac{\mathrm{H}_1}{4}$

Thermodynamics & KTG

Rate of flow of heat through a cylindrical rod is ' \mathrm{H}_1 '. The temperature at the ends of the rod are ' T_1 ' and ' T_2 '. If all the dimensions of the rod become double and the temperature difference remains the same, the rate of flow of heat becomes ' \mathrm{H}_2 '. Then

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