Question 926: Thermodynamics & KTG

Question

A cylindrical rod is having temperatures $$	heta_1$$ and $$	heta_2$$ at its ends. The rate of heat flow is '$$Q$$' $$\mathrm{J}{\mathrm{s}}^{-1}$$. All the linear dimensions of the rod are doubled by keeping the temperatures constant. What is the new rate of flow of heat?

Accepted Answer

To solve this problem, we need to understand how the rate of heat flow, $$Q$$, through a cylindrical rod changes when its linear dimensions are doubled. The rate of heat flow through a rod can be expressed using Fourier's law of heat conduction:
$$ Q = \frac{k \cdot A \cdot (	heta_1 - 	heta_2)}{L} $$
where:
$$Q$$ is the rate of heat flow.
$$k$$ is the thermal conductivity of the material.
$$A$$ is the cross-sectional area of the rod.
$$	heta_1$$ and $$	heta_2$$ are the temperatures at the ends of the rod.
$$L$$ is the length of the rod.
When the linear dimensions of the rod are doubled, both the length and the cross-sectional area will change. Let's look at each dimension:
If the original length is $$L$$, the new length will be $$2L$$.
If the original cross-sectional area is $$A$$, the new cross-sectional area will be $$4A$$ (since doubling the diameter of a circle quadruples its area).
Substituting the new dimensions into the formula, we get the new rate of heat flow, $$Q'$$:
$$ Q' = \frac{k \cdot 4A \cdot (	heta_1 - 	heta_2)}{2L} $$
Simplifying the equation:
$$ Q' = \frac{4 \cdot k \cdot A \cdot (	heta_1 - 	heta_2)}{2L} $$
We can further simplify this to:
$$ Q' = 2 \cdot \frac{k \cdot A \cdot (	heta_1 - 	heta_2)}{L} $$
Since $$\frac{k \cdot A \cdot (	heta_1 - 	heta_2)}{L}$$ is the original rate of heat flow $$Q$$, we have:
$$ Q' = 2Q $$
Therefore, the new rate of flow of heat is twice the original rate of flow. The correct answer is:
Option C $$2 \mathrm{Q}$$

Answer

$$\frac{Q}{2}$$

Answer

$$\frac{Q}{4}$$

Answer

$$2 \mathrm{Q}$$

Answer

$$\frac{3 Q}{2}$$

Thermodynamics & KTG

A cylindrical rod is having temperatures $\theta_1 and \theta_2 at its ends. The rate of heat flow is 'Q' \mathrm{J}{\mathrm{s}}^{-1}$. All the linear dimensions of the rod are doubled by keeping the temperatures constant. What is the new rate of flow of heat?

Choose the correct answer: