288KJ is conducted across two opposite faces of a 3m cube of temperature gradient 90ºCm$^{-1}$ in 7200s. Calculate the thermal conductivity.

1. Analyze the Question

We are asked to calculate the thermal conductivity ($k$) of a material given the following parameters:

• Heat energy ($Q$) = $288\text{ kJ} = 288,000\text{ J}$
• Side of the cube ($L$) = $3\text{ m}$
• Temperature gradient ($\frac{\Delta\theta}{\Delta x}$) = $90^\circ\text{Cm}^{-1}$
• Time ($t$) = $7,200\text{ s}$

2. Explain the Concept

The rate of heat transfer ($H$) through a material is given by the formula:

Where:

• $Q$ is the heat energy.
• $t$ is the time.
• $k$ is the thermal conductivity.
• $A$ is the cross-sectional area.
• $\frac{\Delta\theta}{\Delta x}$ is the temperature gradient.

Since the object is a cube with side length $L = 3\text{ m}$, the area of one face is $A = L^2 = 3^2 = 9\text{ m}^2$.

3. Show the Solution

Step 1: Calculate the rate of heat flow ($H$)

Step 2: Rearrange the formula to solve for $k$

Step 3: Substitute the values

Step 4: Convert to standard form

Rounding to two significant figures as presented in the options, we get $4.9 \times 10^{-2}\text{ Wm}^{-1}\text{K}^{-1}$.

4. Identify the Answer

Comparing our result with the given options, the correct value is A.

5. Tip

When dealing with thermal conductivity problems, always ensure your units are consistent. Convert kilojoules (kJ) to Joules (J) immediately, and remember that for a cube, the area $A$ is simply the square of the side length ($L^2$). If the temperature gradient is already given as a single value ($\frac{\Delta\theta}{\Delta x}$), do not divide by the length again!