Calculating Brick Wall Heat Loss Rate Per Square Meter
This problem involves calculating the steady-state rate of heat transfer through a plane wall (a brick wall) under a given temperature difference.
Given Data
- Wall Thickness, \(L = 24\) cm \(= 0.24\) m
- Inner Surface Temperature, \(T_{inner} = 25^{\circ}C\)
- Outer Surface Temperature, \(T_{outer} = 5^{\circ}C\)
- Thermal Conductivity, \(k = 0.15\ J/(s\cdot m \cdot K)\)
- Area considered, \(A = 1\ m^2\) (per square metre)
Heat Transfer Formula
The rate of heat loss (\(Q/t\)) through a plane wall is calculated using Fourier's Law of Heat Conduction:
\( \frac{Q}{t} = \frac{k \cdot A \cdot (T_{inner} - T_{outer})}{L} \)Calculation Steps
- Calculate Temperature Difference (\(\Delta T\)): The temperature difference across the wall is: \( \Delta T = T_{inner} - T_{outer} = 25^{\circ}C - 5^{\circ}C = 20^{\circ}C \) Note: A temperature difference in Celsius is equal to the temperature difference in Kelvin (\(20^{\circ}C = 20\ K\)).
- Substitute Values into the Formula: Plug the given values and the calculated temperature difference into Fourier's Law. \( \frac{Q}{t} = \frac{(0.15\ J/(s\cdot m \cdot K)) \cdot (1\ m^2) \cdot (20\ K)}{0.24\ m} \)
- Compute the Heat Loss Rate: \( \frac{Q}{t} = \frac{0.15 \times 1 \times 20}{0.24}\ J/s \) \( \frac{Q}{t} = \frac{3}{0.24}\ J/s \) \( \frac{Q}{t} = \frac{300}{24}\ J/s \) \( \frac{Q}{t} = 12.5\ J/s \)
The rate of heat loss through the brick wall per square meter is \(12.5\ J/s\). This value matches Option D.
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