Question

Consider two walls of a house that are identical except that one is made of 10 -cm-thick wood, while the other is made of 25 -cm-thick brick. Through which wall will the house lose more heat in winter?

Short answer

Answer: The house will lose more heat in winter through the 25 cm-thick brick wall.

Step-by-step solution

Understand the concept of thermal conductivity

Thermal conductivity is a material property that represents the ability of a material to conduct heat. Materials with a higher thermal conductivity will allow heat to flow through them more easily than materials with a lower thermal conductivity. In this problem, we need to know the thermal conductivities of wood and brick to determine which wall will lose more heat in winter.

Know the thermal conductivities of wood and brick

The thermal conductivity of wood is typically around \(0.1 - 0.2 \; W/(m \cdot K)\), while the thermal conductivity of brick is around \(0.5 - 1.0 \; W/(m \cdot K)\). For this exercise, we will use the average values for both materials: \(k_{wood} = 0.15 \; W/(m \cdot K)\) and \(k_{brick} = 0.75 \; W/(m \cdot K)\).

Determine the formula for heat transfer rate

The formula for the rate of heat transfer through the walls is given by: \[Q = k\frac{A(T_{hot} - T_{cold})}{d}\] where Q is the heat transfer rate, k is the thermal conductivity of the material, A is the area of the wall, \(T_{hot}\) is the inside temperature, \(T_{cold}\) is the outside temperature, and d is the thickness of the wall.

Calculate the heat transfer rate for the wood wall

Let's assume the temperature inside the house is \(20^\circ C\), the temperature outside the house is \(0^\circ C\), and the area of both walls is the same. Plugging in the values into the heat transfer formula, we get: \[Q_{wood} = 0.15\frac{A(20 - 0)}{0.10}\] \[Q_{wood} = 30A\]

Calculate the heat transfer rate for the brick wall

Using the same temperature and area values, we can calculate the heat transfer rate for the brick wall: \[Q_{brick} = 0.75\frac{A(20 - 0)}{0.25}\] \[Q_{brick} = 60A\]

Compare the heat transfer rates for the wood and brick walls

Comparing the heat transfer rates for the wood and brick walls, we see that: \[Q_{brick} > Q_{wood}\] This means that the house will lose more heat in winter through the 25 cm-thick brick wall than through the 10 cm-thick wood wall.

Key concepts

Heat Transfer Rate

Understanding how quickly heat moves through materials is crucial when considering the efficiency of insulation in buildings. The heat transfer rate is a measure of the thermal energy that is transferred per unit time across a material. It is affected by several factors, such as temperature difference, material thickness, and thermal conductivity.

Using the formula
\[Q = k\frac{A(T_{hot} - T_{cold})}{d}\]
we can calculate how much heat is lost over time. The temperature difference (T_hot - T_cold) is a driving force for heat flow; greater differences will increase the heat transfer rate. The thickness (d) of the material acts as a barrier; thicker walls will usually slow down the heat transfer, assuming the material has the same thermal conductivity. Remember, this process is always seeking equilibrium, with heat moving from warmer to cooler spaces.

Material Property

When evaluating how materials behave under the influence of temperature, thermal conductivity is a key material property to consider. It is denoted as k in heat transfer equations and represents the ability of a material to conduct thermal energy. Materials with higher thermal conductivity, like metals, allow heat to pass through more easily, while those with lower thermal conductivity, like wood or foam, resist heat flow.

In the context of insulation, materials with low thermal conductivity are preferred, as they reduce the heat transfer rate. It's essential to understand this property because it allows us to determine the right materials for keeping a home warm in the winter or cool in the summer, thus improving energy efficiency and comfort.

Insulation Materials

Choosing the right insulation materials can determine the comfort and energy efficiency of a home. These materials are specifically designed to limit heat transfer between the inside and outside of a structure. Insulation is all about managing the thermal conductivity — typically, effective insulators have low thermal conductivity values.

Examples of good insulation materials include fiberglass, rock wool, and certain foams. Not only do these materials have low thermal conductivities, but they also often have a structure that traps air, further reducing the movement of heat. These attributes make such materials highly effective at keeping the heat inside during the winter and outside during the summer, leading to less energy usage and lower utility bills.