Question

Heat is lost at a rate of \(275 \mathrm{~W}\) per \(\mathrm{m}^{2}\) area of a \(15-\mathrm{cm}-\) thick wall with a thermal conductivity of \(k=1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The temperature drop across the wall is (a) \(37.5^{\circ} \mathrm{C}\) (b) \(27.5^{\circ} \mathrm{C}\) (c) \(16.0^{\circ} \mathrm{C}\) (d) \(8.0^{\circ} \mathrm{C}\) (e) \(4.0^{\circ} \mathrm{C}\)

Short answer

Answer: The temperature drop across the wall is approximately \(37.5^{\circ} C\).

Step-by-step solution

Write down the given parameters.

We are given the following information: - Heat loss rate (Q): \(275\, W/m^2\) - Wall thickness (d): \(15\, cm = 0.15\, m\) (convert to meters) - Thermal conductivity (k): \(1.1\, W/m\cdot K\)

Write the formula for Fourier's law of heat conduction.

Fourier's law of heat conduction can be stated as: \(Q = -k\cdot A \cdot \frac{\Delta T}{d}\) Where Q is the rate of heat loss, k is the thermal conductivity, A is the area of the wall, ΔT is the temperature drop across the wall, and d is the wall thickness. We want to find the temperature drop (ΔT), so we'll rearrange the formula for ΔT: \(\Delta T = -\frac{Q\cdot d}{k\cdot A}\) However, since Q is given as \(275\, W/m^2\), we don't have to worry about the area, and the formula becomes: \(\Delta T = -\frac{Q\cdot d}{k}\)

Calculate the temperature drop (ΔT).

Now that we have the formula, we can plug in the given values to find the temperature drop: \(\Delta T = -\frac{275\, W/m^2\cdot 0.15\, m}{1.1\, W/m\cdot K}\) \(\Delta T = -\frac{41.25\, W/m^2}{1.1\, W/m\cdot K}\) \(\Delta T \approx 37.5\, ^\circ C\)

Choose the correct option.

The calculated temperature drop is \(\approx 37.5\, ^\circ C\). Thus, the correct option is: (a) \(37.5^{\circ} \mathrm{C}\)

Key concepts

Thermal Conductivity

Thermal conductivity is a material property that measures a substance's ability to conduct heat. It is typically denoted by the symbol \( k \) and is defined as the amount of heat (in watts) transferred through a unit area of the substance (in square meters) for a temperature difference (in Kelvin) per unit thickness (in meters). Materials with high thermal conductivity, like metals, are excellent heat conductors, whereas materials like wood or fiberglass with low thermal conductivity are considered insulators.

Understanding thermal conductivity is crucial when designing thermal insulation systems or analyzing heat transfer in materials. Higher values of \( k \) indicate that the material can transfer heat more efficiently, which is often desirable in heat sinks but undesirable in thermal insulation scenarios.

Temperature Gradient

A temperature gradient is the rate at which temperature changes from one point to another in space. Mathematically, it is expressed as \( \frac{\Delta T}{d} \), where \( \Delta T \) is the temperature difference between two points and \( d \) is the distance between those points. The temperature gradient is the driving force behind heat transfer by conduction, and the direction of heat flow is from the region of higher temperature to the region of lower temperature.

This concept is essential when calculating the rate of heat loss or gain in materials, such as in our exercise, where you need to find the temperature drop across a wall. A steeper temperature gradient indicates a more significant temperature change over a short distance, which results in a higher rate of heat transfer.

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics that involves the movement of thermal energy from one place to another. There are three main mechanisms of heat transfer: conduction, convection, and radiation. In the context of Fourier's law, we're concerned with conduction, which is the transfer of heat through a material without any actual movement of the material itself. It occurs at the molecular level as atoms and molecules vibrate and transfer energy to neighboring atoms and molecules.

The rate of heat transfer (\( Q \)) for conduction can be calculated using Fourier's law, which relates the heat transfer to the temperature gradient, the thermal conductivity, and the material's cross-sectional area. This allows us to deduce the amount of heat that flows through a specific material over time, given the temperature difference across it, as exemplified by the exercise provided.