Question

The temperature difference between two sides of metal plate, 3 cm thick is 15°C. Heat is transmitted through plate at the rate of 900 kcal per minute per m² at steady state. The thermal conductivity of metal is

• A. \( 1.8 \times 10^{-2} \, \text{kcal} \, \text{ms}^{-1} \, \text{°C}^{-1} \)
• B. \( 4.5 \times 10^{-2} \, \text{kcal} \, \text{ms}^{-1} \, \text{°C}^{-1} \)
• C. \( 3 \times 10^{-2} \, \text{kcal} \, \text{ms}^{-1} \, \text{°C}^{-1} \)
• D. \( 6 \times 10^{-2} \, \text{kcal} \, \text{ms}^{-1} \, \text{°C}^{-1} \)

Hint:

Always ensure consistent units. Convert minutes to seconds and centimetres to metres before substituting into Fourier’s law. The formula \(k = \frac{Q}{tA} \cdot \frac{d}{\Delta T}\) directly gives the thermal conductivity.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a metal plate of thickness \(d = 3\ \text{cm} = 0.03\ \text{m}\) with a temperature difference \(\Delta T = 15\ \text{°C}\) across it. The rate of heat transfer per unit area is \(\frac{Q}{t \cdot A} = 900\ \text{kcal}/(\text{min} \cdot \text{m}^2)\). We need to find the thermal conductivity \(k\) of the metal under steady‑state conditions.

Step 2: Key Formula or Approach:
Fourier’s law of heat conduction for a plane slab is: \[ \frac{Q}{t} = k\,A\,\frac{\Delta T}{d} \] Rearranging for \(k\): \[ k = \frac{Q}{t \cdot A} \cdot \frac{d}{\Delta T} \]

Step 3: Detailed Explanation:
First convert the heat flux to per second: \[ \frac{Q}{t \cdot A} = 900\ \frac{\text{kcal}}{\text{min}\cdot\text{m}^2} = \frac{900}{60}\ \frac{\text{kcal}}{\text{s}\cdot\text{m}^2} = 15\ \frac{\text{kcal}}{\text{s}\cdot\text{m}^2}. \] Now substitute into the formula: \[ k = 15 \times \frac{0.03}{15} = 15 \times 0.002 = 0.03\ \frac{\text{kcal}}{\text{s}\cdot\text{m}\cdot\text{°C}}. \] Expressed in scientific notation: \(k = 3 \times 10^{-2}\ \text{kcal}\,\text{ms}^{-1}\,\text{°C}^{-1}\).

Step 4: Final Answer:
This matches option (C).