Question

In the steady state of temperature, the flow of heat across the body depends upon its:
A. thermal capacity
B. thermal conductivity
C. temperature difference across its opposite faces
D. thermal resistivity
Choose the CORRECT answer from the options given below:

• A. A,B and D only
• B. A,B and C only
• C. A, B, C and D
• D. B, C and D only

Hint:

Think of the analogy with electrical circuits (Ohm's Law, \(I = V/R\)): - Heat Flow \(H \leftrightarrow\) Current \(I\) - Temperature Difference \(\Delta T \leftrightarrow\) Voltage \(V\) - Thermal Resistance \(R_T = L/(kA) \leftrightarrow\) Electrical Resistance \(R\) Thermal capacity is analogous to electrical capacitance, which is irrelevant for steady DC current.

Answer 1

The Correct Option is D

Step 1: Recall Fourier's Law of Heat Conduction. In the steady state, the rate of heat flow (\(H\) or \(\frac{dQ}{dt}\)) through a body is given by: \[ H = -kA \frac{dT}{dx} \] For a slab of material with cross-sectional area A, thickness L, and a temperature difference \(\Delta T\) across its faces, the law is often written as: \[ H = \frac{kA\Delta T}{L} \] where \(k\) is the thermal conductivity.
Step 2: Analyze the factors in the equation. The flow of heat \(H\) depends on: - k (thermal conductivity): Yes. This is statement B. - \(\Delta T\) (temperature difference): Yes. This is statement C. - The geometry (A and L). Thermal resistivity (\(\rho_T\)) is the reciprocal of thermal conductivity (\(\rho_T = 1/k\)). Since the heat flow depends on \(k\), it also depends on \(\rho_T\). So, statement D is also correct. Thermal capacity (\(C = mc\), where c is specific heat) relates to the amount of heat a body can store to change its temperature. It does not determine the rate of heat flow in the steady state, where by definition, the temperature at any point in the body is constant. So, statement A is incorrect.
Step 3: Conclude the correct factors. The steady-state flow of heat depends on thermal conductivity (B), temperature difference (C), and thermal resistivity (D).