Question

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_1$ and $T_2 (T_1 > T_2).$ The rate of heat transfer, through the rod in a steady state is given by :

• A. $ \frac{ dQ}{dt}= \frac{ k(T_1 -T_2)}{LA}$
• B. $ \frac{ dQ}{dt}= k LA(T_1 -T_2)$
• C. $ \frac{ dQ}{dt}= \frac{ kA(T_1 -T_2)}{L}$
• D. $ \frac{ dQ}{dt}= \frac{ kL(T_1 -T_2)}{A}$

Answer 1

The Correct Option is C

 To determine the rate of heat transfer through a rod, we start by applying the formula for heat conduction, which is based on Fourier's law of heat conduction. This law states that the rate of heat transfer through a material is proportional to the negative gradient of temperatures and the area through which the heat flows. Mathematically, this is expressed as:

\(\frac{dQ}{dt} = -kA \frac{dT}{dx}\)

Where:

• \(\frac{dQ}{dt}\) is the rate of heat transfer (W, watts).
• \(k\) is the thermal conductivity of the material (W/m·K).
• \(A\) is the cross-sectional area through which the heat is being transferred (m²).
• \(\frac{dT}{dx}\) is the temperature gradient along the length of the rod (K/m).

Given that the temperature at one end of the rod is \(T_1\) and at the other end is \(T_2\) (where \(T_1 > T_2\)), and the length of the rod is \(L\), we can express the temperature gradient as:

\(\frac{dT}{dx} = \frac{T_1 - T_2}{L}\)

Replacing this into the heat conduction formula, we have:

\(\frac{dQ}{dt} = -kA \frac{T_1 - T_2}{L}\)

Since \(T_1 > T_2\), the negative sign indicates that heat flows from higher to lower temperature. Hence, we can write the rate of heat transfer as:

\(\frac{dQ}{dt} = \frac{kA(T_1 - T_2)}{L}\)

Thus, the correct option is:

1. \(\frac{dQ}{dt} = \frac{kA(T_1 - T_2)}{L}\)

This formula expresses the rate of heat transfer through a rod in the steady-state condition. The other options presented do not correctly account for the dimensions or setup of this heat conduction problem.