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: List what we know.
A metal plate is $3$ cm thick, so $d = 0.03$ m. The two faces differ by $\Delta T = 15$ degrees C. Heat flows through it at $900$ kcal per minute per square metre. We want the thermal conductivity $k$.

Step 2: The conduction law.
Heat conduction follows $\dfrac{Q}{t} = k A \dfrac{\Delta T}{d}$. Dividing by area gives the heat flux $\dfrac{Q}{tA} = k \dfrac{\Delta T}{d}$.

Step 3: Fix the time unit.
The flux is given per minute, but $k$ is usually per second. So $\dfrac{900}{60} = 15$ kcal per second per square metre.

Step 4: Rearrange for $k$.
From the law, $k = \dfrac{Q}{tA} \cdot \dfrac{d}{\Delta T}$.

Step 5: Put in the numbers.
$k = 15 \times \dfrac{0.03}{15} = 15 \times 0.002 = 0.03$ kcal per metre per second per degree C.

Step 6: Write in scientific form.
$0.03 = 3 \times 10^{-2}$, which is option (C).
\[ \boxed{k = 3 \times 10^{-2}\ \text{kcal m}^{-1}\text{s}^{-1}\,{}^{\circ}\text{C}^{-1}} \]