Question 907: Thermodynamics & KTG

Question

A cylindrical rod has temperatures '$$T_1$$' and '$$T_2$$' at its ends. The rate of flow of heat is '$$Q_1$$' cal $$\mathrm{s}^{-1}$$. If length and radius of the rod are doubled keeping temperature constant, then the rate of flow of heat '$$\mathrm{Q}_2$$' will be

Accepted Answer

To find the rate of flow of heat $$Q_2$$ when both the length and radius of the cylindrical rod are doubled while keeping the temperatures $$T_1$$ and $$T_2$$ at its ends constant, we need to understand how thermal conductivity works in a cylindrical shape.
The heat flow rate through a cylindrical rod is given by Fourier's law of heat conduction:
$$ Q = \frac{KA(T_1 - T_2)}{L} $$
where:
$$Q$$ is the rate of heat transfer
$$K$$ is the thermal conductivity of the material
$$A$$ is the cross-sectional area of the rod
$$T_1$$ and $$T_2$$ are the temperatures at the ends of the rod
$$L$$ is the length of the rod
The cross-sectional area $$A$$ of a cylindrical rod with radius $$r$$ is given by:
$$ A = \pi r^2 $$
Now, suppose the initial length and radius of the rod are $$L$$ and $$r$$, and this gives us the initial heat flow rate $$Q_1$$:
$$ Q_1 = \frac{K \pi r^2 (T_1 - T_2)}{L} $$
If both the length and radius of the rod are doubled, the new length becomes $$2L$$ and the new radius becomes $$2r$$. Hence, the new cross-sectional area $$A'$$ of the rod becomes:
$$ A' = \pi (2r)^2 = 4\pi r^2 $$
Substituting these into the heat flow equation, we get the new heat flow rate $$Q_2$$:
$$ Q_2 = \frac{K A' (T_1 - T_2)}{2L} $$
$$ Q_2 = \frac{K (4\pi r^2) (T_1 - T_2)}{2L} $$
$$ Q_2 = 2 \frac{K \pi r^2 (T_1 - T_2)}{L} $$
$$ Q_2 = 2 Q_1 $$
Thus, the rate of flow of heat $$Q_2$$ when the length and radius of the rod are doubled is:
$$\mathrm{Q}_2 = 2 \mathrm{Q}_1$$
Therefore, the correct answer is:
Option D: $$\mathrm{Q}_2 = 2 \mathrm{Q}_1$$

Answer

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

Answer

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

Answer

$$\mathrm{Q_2=4 Q_1}$$

Answer

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

Thermodynamics & KTG

A cylindrical rod has temperatures '$T_1' and 'T_2' at its ends. The rate of flow of heat is 'Q_1' cal \mathrm{s}^{-1}. If length and radius of the rod are doubled keeping temperature constant, then the rate of flow of heat '\mathrm{Q}_2$' will be

Choose the correct answer: