A concrete wall of thickness 15 cm has inside temperature 25°C and outside temperature 5°C. The rate of heat loss through per square metre of the wall (thermal conductivity 0.81 J/(s m K)) is:

Correct option is D

​$$\textbf{Given:}$$

$$\bullet \quad \text{Thickness (width) of the wall} = 15 \text{ cm} = 0.15 \text{ m}$$​​​

​$$\bullet \quad \text{Inside temperature} \quad T_1 = 25^\circ C$$

$$\bullet \quad \text{Outside temperature} \quad T_2 = 5^\circ C$$

$$\bullet \quad \text{Thermal conductivity} \quad h = 0.81 \ J/(s \cdot m \cdot K)$$

$$\bullet \quad \text{Area} \quad A = 1 \ m^2$$

$$\textbf{Formula Used:}$$ $$\text{The rate of heat transfer (dq/dt) is given by Fourier’s law of heat conduction:}$$

$$\frac{dq}{dt} = \frac{h A (T_1 - T_2)}{dL}$$

$$\text{where:}$$

$$\bullet \quad h = \text{ Thermal conductivity}$$

$$\bullet \quad A = \text{ Area of the wall}$$

$$\bullet \quad T_1 - T_2 = \text{ Temperature difference}$$

$$\bullet \quad dL = \text{ Thickness of the wall}$$

$$\frac{dq}{dt} = \frac{0.81 \times 1 \times (25 - 5)}{0.15}$$

$$= \frac{0.81 \times 20}{0.15}$$

$$= \frac{16.2}{0.15}$$  $$= 108 \ J/s$$​​​​​​​​​​​​​​​