Question 733: Thermodynamics & KTG

Question

A cylindrical rod is having temperatures $	heta_1$ and $	heta_2$ at its ends. The rate of heat flow is $\mathrm{Q} J / \mathrm{S}$. All the linear dimensions of the rod are doubled by keeping the temperature constant. The new rate of flow of heat is

Accepted Answer

The rate of heat flow through a cylindrical rod can be expressed using the formula:
$ \left(\frac{Q}{t}ight) = \frac{K \pi r^2\left(	heta_1 - 	heta_2ight)}{\Delta x} $
In this equation, $ \frac{Q}{t} $ represents the rate of heat flow, $ K $ is the thermal conductivity, $ r $ is the radius of the cylinder, and $ \Delta x $ is the length of the rod. The rate of heat flow is proportional to $ \frac{r^2}{\Delta x} $.
When all linear dimensions of the rod are doubled (i.e., both the radius and the length are twice as large), the new dimensions are:
New radius, $ r_2 = 2r_1 $
New length, $ \Delta x_2 = 2\Delta x_1 $
The new rate of heat flow, $\frac{Q'}{t}$, is proportional to $ \frac{r_2^2}{\Delta x_2} $. Thus, we can write the ratio of the original to the new rate of heat flow as:
$ \frac{Q}{Q'} = \left(\frac{r_1}{r_2}ight)^2 \left(\frac{\Delta x_2}{\Delta x_1}ight) = \left(\frac{1}{2}ight)^2 	imes \left(\frac{2}{1}ight) = \frac{1}{2} $
By rearranging, the new rate of heat flow $ Q' $ is:
$ Q' = 2Q $
Therefore, the new rate of flow of heat is twice the original rate, $ 2Q $.

Answer

$\mathrm{4Q}$

Answer

$\mathrm{2 Q}$

Answer

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

Answer

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

Thermodynamics & KTG

A cylindrical rod is having temperatures \theta_1 and \theta_2 at its ends. The rate of heat flow is \mathrm{Q} J / \mathrm{S}. All the linear dimensions of the rod are doubled by keeping the temperature constant. The new rate of flow of heat is

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