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: Start from Fourier's law.
The heat flow per area is $\tfrac{Q}{tA} = k\tfrac{\Delta T}{d}$. We solve for $k = \tfrac{Q}{tA}\cdot\tfrac{d}{\Delta T}$.

Step 2: Make units per second.
The flux $900$ kcal per minute per $\text{m}^2$ becomes $\tfrac{900}{60} = 15$ kcal per second per $\text{m}^2$.

Step 3: Put in the numbers.
With $d = 0.03\ \text{m}$ and $\Delta T = 15^{\circ}\text{C}$: \[ k = 15\times\frac{0.03}{15} = 15\times0.002 = 0.03. \]

Step 4: Write in scientific form.
\[ \boxed{3\times10^{-2}\ \text{kcal}\,\text{m}^{-1}\text{s}^{-1}\,^{\circ}\text{C}^{-1}} \]