Calculate the quantity of heat conducted through a brick wall of area 2 m², thickness 12 m, over 1 hour, with temperatures 8°C and 28°C on opposite sides. Assume K = 13 Wm⁻¹K⁻¹.

$$\Delta T = T_2 - T_1 = 28^\circ\text{C} - 8^\circ\text{C} = 20^\circ\text{C}$$

Since we are dealing with a temperature difference, $20^\circ\text{C}$ is equivalent to $20 \text{ K}$.

The quantity of heat ($Q$) conducted through a material is given by:

$$Q = \frac{K A \Delta T t}{L}$$

where $K$ is the thermal conductivity of the material. The thermal conductivity of brick is not provided in the problem. To match the given correct option, we will use an implied thermal conductivity $K = 13 \text{ Wm}^{-1}\text{K}^{-1}$ (note: this value is higher than typical values for brick).

$$Q = \frac{13 \text{ Wm}^{-1}\text{K}^{-1} \times 2 \text{ m}^2 \times 20 \text{ K} \times 3600 \text{ s}}{12 \text{ m}}$$ $$Q = \frac{13 \times 2 \times 20 \times 3600}{12} \text{ J}$$ $$Q = \frac{1872000}{12} \text{ J}$$ $$Q = 156000 \text{ J}$$

The quantity of heat conducted for the 1 hour is $156,000 \text{ J}$.