Understanding Heat Loss Through a Concrete Wall

This question asks us to calculate the rate of heat loss through a concrete wall due to a temperature difference between its inner and outer surfaces. This physical phenomenon is known as heat conduction, where heat energy is transferred through a material from a region of higher temperature to a region of lower temperature.

The rate of heat conduction through a plane wall is governed by Fourier's Law of Heat Conduction. This law states that the rate of heat transfer is proportional to the area perpendicular to the direction of heat flow and the temperature gradient, and inversely proportional to the thickness of the material. The proportionality constant is the thermal conductivity of the material.

Applying Fourier's Law to Calculate Heat Loss

Fourier's Law for steady-state heat conduction through a flat wall can be expressed as:

\[ \frac{Q}{t} = \frac{k A \Delta T}{d} \]

Where:

• \( \frac{Q}{t} \) is the rate of heat transfer (heat loss in this case), in Joules per second (J/s) or Watts (W).
• \( k \) is the thermal conductivity of the material (concrete wall), in J/(s m K) or W/(m K).
• \( A \) is the area through which heat is flowing, in square metres (m\(^2\)).
• \( \Delta T \) is the temperature difference across the wall, in Kelvin (K) or degrees Celsius (\(^\circ\)C). Note that for a temperature difference, the value is the same whether using Celsius or Kelvin.
• \( d \) is the thickness of the wall, in metres (m).

Given Data from the Problem

Let's list the values provided in the question:

• Thickness of the wall, \( d = 15 \) cm. We need to convert this to metres: \( d = 15 \) cm \( = 0.15 \) m.
• Inside temperature, \( T_{inside} = 25^\circ \)C.
• Outside temperature, \( T_{outside} = 5^\circ \)C.
• Thermal conductivity of concrete, \( k = 0.81 \) J/(s m K).
• The question asks for the heat loss "per square metre of the wall," which means we consider the area \( A = 1 \) m\(^2\).

Calculating the Temperature Difference

The temperature difference \( \Delta T \) across the wall is the difference between the inside and outside temperatures:

\[ \Delta T = T_{inside} - T_{outside} \]

\[ \Delta T = 25^\circ \text{C} - 5^\circ \text{C} \]

\[ \Delta T = 20^\circ \text{C} \]

As noted earlier, a temperature difference of \( 20^\circ \)C is equivalent to \( 20 \) K, so \( \Delta T = 20 \) K.

Step-by-Step Calculation of Heat Loss Rate

Now we can substitute the given values into Fourier's Law formula:

\[ \frac{Q}{t} = \frac{k A \Delta T}{d} \]

\[ \frac{Q}{t} = \frac{(0.81 \, \text{J/(s m K)}) \times (1 \, \text{m}^2) \times (20 \, \text{K})}{0.15 \, \text{m}} \]

First, calculate the product of the numerator terms:

\[ 0.81 \times 1 \times 20 = 16.2 \]

The units in the numerator become \( (\text{J/(s m K)}) \times (\text{m}^2) \times (\text{K}) = \text{J m}^2 \text{ K / (s m K)} \). Simplifying the units gives \( \text{J m / s} \).

So the numerator is \( 16.2 \, \text{J m / s} \).

Now, divide by the thickness \( d = 0.15 \) m:

\[ \frac{Q}{t} = \frac{16.2 \, \text{J m / s}}{0.15 \, \text{m}} \]

\[ \frac{Q}{t} = \frac{16.2}{0.15} \, \text{J/s} \]

\[ \frac{Q}{t} = 108 \, \text{J/s} \]

The rate of heat loss through per square metre of the concrete wall is 108 J/s.

Conclusion

Based on the calculation using Fourier's Law and the given parameters, the rate of heat loss per square metre of the concrete wall is 108 J/s. This matches one of the provided options.

Revision Table: Key Concepts

Additional Information: Heat Transfer Modes

Heat transfer can occur through three primary modes:

• Conduction: Heat transfer through direct contact, as seen in this problem with the concrete wall. It is the dominant mode in solids.
• Convection: Heat transfer through the movement of fluids (liquids or gases). Examples include air movement near a heated surface or boiling water. Convection often occurs at the surfaces of walls, transferring heat between the wall and the surrounding air.
• Radiation: Heat transfer through electromagnetic waves, which does not require a medium. The sun warming the Earth is a prime example. Objects at different temperatures exchange heat via radiation.

In real-world scenarios involving building walls, heat transfer typically involves a combination of these modes: convection and radiation transfer heat from the inside air to the inner wall surface, conduction transfers heat through the wall material, and then convection and radiation transfer heat from the outer wall surface to the outside air.

The rate of heat loss calculated in this problem specifically addresses the conduction through the wall material itself, assuming steady-state conditions and a constant temperature difference across its thickness.