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

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

Step by step solution

01

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): \(1.1\, W/m\cdot K\)
02

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: \(\Delta T = -\frac{Q\cdot d}{k}\)
03

Calculate the temperature drop (ΔT).

Now that we have the formula, we can plug in the given values to find the temperature drop: \(\Delta T = -\frac{275\, W/m^2\cdot 0.15\, m}{1.1\, W/m\cdot K}\) \(\Delta T = -\frac{41.25\, W/m^2}{1.1\, W/m\cdot K}\) \(\Delta T \approx 37.5\, ^\circ C\)
04

Choose the correct option.

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

An 8-m-internal-diameter spherical tank made of \(1.5\)-cm-thick stainless steel \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located in a room whose temperature is \(25^{\circ} \mathrm{C}\). The walls of the room are also at \(25^{\circ} \mathrm{C}\). The outer surface of the tank is black (emissivity \(\varepsilon=1\) ), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat transfer coefficients at the inner and the outer surfaces of the tank are \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -h period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\).

Ice slurry is being transported in a pipe \((k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}, D_{o}=3 \mathrm{~cm}\), and \(L=\) \(5 \mathrm{~m}\) ) with an inner surface temperature of \(0^{\circ} \mathrm{C}\). The ambient condition surrounding the pipe has a temperature of \(20^{\circ} \mathrm{C}\), a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and a dew point of \(10^{\circ} \mathrm{C}\). If the outer surface temperature of the pipe drops below the dew point, condensation can occur on the surface. Since this pipe is located in a vicinity of high voltage devices, water droplets from the condensation can cause electrical hazard. To prevent such incident, the pipe surface needs to be insulated. Determine the insulation thickness for the pipe using a material with \(k=0.95 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to prevent the outer surface temperature from dropping below the dew point.

A plane brick wall \((k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(10 \mathrm{~cm}\) thick. The thermal resistance of this wall per unit of wall area is (a) \(0.143 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (b) \(0.250 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (c) \(0.327 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (d) \(0.448 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (e) \(0.524 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\)

Consider two finned surfaces that are identical except that the fins on the first surface are formed by casting or extrusion, whereas they are attached to the second surface afterwards by welding or tight fitting. For which case do you think the fins will provide greater enhancement in heat transfer? Explain.

A room at \(20^{\circ} \mathrm{C}\) air temperature is loosing heat to the outdoor air at \(0^{\circ} \mathrm{C}\) at a rate of \(1000 \mathrm{~W}\) through a \(2.5\)-m-high and 4 -m-long wall. Now the wall is insulated with \(2-\mathrm{cm}\) thick insulation with a conductivity of \(0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Determine the rate of heat loss from the room through this wall after insulation. Assume the heat transfer coefficients on the inner and outer surface of the wall, the room air temperature, and the outdoor air temperature to remain unchanged. Also, disregard radiation. (a) \(20 \mathrm{~W}\) (b) \(561 \mathrm{~W}\) (c) \(388 \mathrm{~W}\) (d) \(167 \mathrm{~W}\) (e) \(200 \mathrm{~W}\)

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