Question

Calculate the rate at which heat would be lost on a very cold winter day through a \(6.2 \mathrm{~m} \times 3.8 \mathrm{~m}\) brick wall \(32 \mathrm{~cm}\) thick. The inside temperature is \(26^{\circ} \mathrm{C}\) and the outside temperature is \(-18^{\circ} \mathrm{C} ;\) assume that the thermal conductivity of the brick is \(0.74 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

Short answer

The rate at which heat would be lost on a very cold winter day through the brick wall is around 24.76 kW.

Step-by-step solution

Identify the given quantities

The given quantities in this exercise are: Area of the brick wall \(A = 6.2 \mathrm{m} \times 3.8 \mathrm{m}\), the thickness of the wall \(d = 0.32 \mathrm{m}\), the inside and outside temperatures \(T_1 = 26^{\circ} \mathrm{C} \) and \(T_2 = -18^{\circ} \mathrm{C} \), and the thermal conductivity of the brick \(k = 0.74 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Use the formula for heat transfer

The formula for heat transfer via conduction is: \(Q = \frac{kA(T_1 - T_2)}{d}\). Substitute the given values into the formula: \(Q = \frac{0.74 \mathrm{~W}/\mathrm{m\cdot K} \times 6.2 \mathrm{m} \times 3.8 \mathrm{m} \times (26^{\circ} \mathrm{C} - (-18^{\circ} \mathrm{C}))}{0.32 \mathrm{m}}\)

Calculate the solution

Now, perform the arithmetic operations to compute the heat lost per unit time. Ensure that units in the temperature difference are converted to Kelvin before performing the multiplication. Note that a difference of 1 degree Celsius is equal to a difference of 1 Kelvin, so the absolute value of the difference is \(44 \mathrm{~K}\), which results into \(Q \approx 24760 \mathrm{~W}\) or \(24.76\ \mathrm{kW}\).

Key concepts

Thermal Conductivity

Understanding thermal conductivity is critical when evaluating how materials interact with heat. It's a measure of a material's ability to conduct heat and is denoted by the symbol \( k \). In the context of our exercise, thermal conductivity is essential to determine the rate of heat transfer through the brick wall.

For different materials, \( k \) varies – metals generally have high thermal conductivities, while non-metals and gases have lower values. The thermal conductivity of the brick is given as \( 0.74 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \), which suggests that brick is a comparatively poor conductor of heat. This quality helps in insulating buildings, maintaining a stable indoor temperature by slowing the transfer of heat to or from the environment.

When calculating heat loss or gain, one must multiply the thermal conductivity by the material's area and the temperature difference, and divide by the thickness. A higher thermal conductivity means that heat will transfer more quickly through the material, which isn't ideal for insulation purposes.

Conduction in Physics

Conduction is one of the three modes of heat transfer, the other two being convection and radiation. In the exercise, we explore conduction, which occurs when heat transfers through a solid material from one molecule to another without the molecules themselves moving significantly. It is the main form of heat transfer in solids and depends on the interaction between particles.

Heat naturally flows from regions of higher temperature to those of lower temperature. In our scenario, the inside of the house is warmer, so the heat moves through the brick wall towards the colder outside. The rate at which this heat transfer occurs is dependent on factors such as the thermal conductivity, the temperature difference across the material, and the material’s dimensions.

The formula \( Q = \frac{kA(T_1 - T_2)}{d} \) brought into play captures these factors: \( k \) represents the thermal conductivity, \( A \) the area of the wall, \( T_1 - T_2 \) the temperature difference, and \( d \) the thickness of the wall. Recognizing the role of each variable can greatly aid in understanding and solving conduction problems.

Physics Problem Solving

Physics problems can often seem daunting, but breaking them down into smaller steps, as seen in our heat transfer exercise, makes them more manageable. Problem-solving in physics usually involves a structured approach: identifying given quantities, applying relevant physical principles or formulas, and carrying out calculations meticulously while keeping track of units.

It is critical to understand the physical meaning of the quantities and their interrelationships. In our exercise, quantifying the heat transfer requires recognition of the underlying physical process – conduction, the knowledge of the material properties like thermal conductivity, and the capability to manipulate mathematical expressions.

When tackling physics problems, it's also important to convert units where necessary and ensure that the units of the quantities involved are consistent. As an improvement advice, remember to convert temperatures to Kelvin when working with thermal physics formulas, even though the Celsius to Kelvin conversion doesn't change the value of temperature differences. Also, checks on dimensional consistency throughout the calculation steps can help ferret out potential errors. Equipped with these strategies, students can confidently approach and solve a wide array of physics problems.