A brick wall having a thickness of 24 cm has an inner surface temperature of 25 degree C and an outer surface temperature of 5 degree C. The rate of heat loss through per square metre of the wall (thermal conductivity 0.15 J/ (s.m.K) is :

Correct option is A

​$$\textbf{Given Data:}\\\bullet \text{ Thickness of the wall } = d = 24 \text{ cm} = 0.24 \text{ m}\\\bullet \text{ Inner surface temperature } = T_1 = 25^\circ C\\\bullet \text{ Outer surface temperature } = T_2 = 5^\circ C\\\bullet \text{ Thermal conductivity } = k = 0.15 \text{ J/(s} \cdot \text{m} \cdot K)\\\bullet \text{ Area } = A = 1 \text{ m}^2 \text{ (per square meter of the wall)}\\\textbf{Step 1: Use Fourier's Law for heat conduction}\\\text{The rate of heat transfer } Q \text{ through a wall is given by:}\\Q = \frac{k A (T_1 - T_2)}{d}\\\text{Substituting the given values:}\\Q = \frac{0.15 \times 1 \times (25 - 5)}{0.24}\\Q = \frac{0.15 \times 20}{0.24}\\Q = \frac{3}{0.24}\\Q = 12.5 \text{ J/s}$$​​