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One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 \(\mathrm{W}\) is required to keep the interior surface of the box 65.0 \(\mathrm{C}^{\circ}\) (about 120 \(\mathrm{F}^{\circ}\) ) above the temperature of the outer surface. The total area of the box is 2.18 \(\mathrm{m}^{2}\) , and the wall thickness is 3.90 \(\mathrm{cm}\) . Find the thermal conductivity of the material in SI units.

Short Answer

Expert verified
The thermal conductivity of the material is approximately 0.049 W/mâ‹…K.

Step by step solution

01

Identify Known Values and Formula

To find the thermal conductivity (\( k \)), we'll use the formula for heat transfer rate: \[ Q = \frac{k \cdot A \cdot \Delta T}{d} \]where the heat transfer rate \( Q = 180 \, \text{W} \), \( A = 2.18 \, \text{m}^2 \), \( \Delta T = 65.0^{\circ} \text{C} \), and the wall thickness \( d = 3.90 \, \text{cm} = 0.039 \, \text{m} \).
02

Rearrange Formula to Solve for Thermal Conductivity

Rearrange the heat transfer formula to solve for thermal conductivity \( k \):\[ k = \frac{Q \cdot d}{A \cdot \Delta T} \]
03

Substitute Known Values into the Formula

Substitute the known values into the formula:\[ k = \frac{180 \, \text{W} \cdot 0.039 \, \text{m}}{2.18 \, \text{m}^2 \cdot 65.0 \,^{\circ}\text{C}} \]
04

Calculate the Thermal Conductivity

Perform the calculation:\[ k = \frac{180 \times 0.039}{2.18 \times 65.0} \approx 0.049 \frac{\text{W}}{\text{m} \cdot \text{K}} \]
05

Interpret the Result

The calculated thermal conductivity of the material is \( 0.049 \, \frac{\text{W}}{\text{m} \cdot \text{K}} \), indicating the insulating properties of the material.

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Key Concepts

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

Heat Transfer
In simple terms, heat transfer refers to the movement of thermal energy from one place to another due to a temperature difference. This concept is essential in understanding thermal conductivity. Heat can be transferred in three main ways: conduction, convection, and radiation.
Conduction is the process where heat moves through a solid material without the movement of the material itself. A good way to imagine this is heating one end of a metal rod; the temperature increase travels through the rod to the other side.
Convection involves the movement of heat by the physical movement of fluid (like air or water) from one place to another. This is why the air in a room can feel cooler near an open window.
Radiation occurs when heat is transferred through electromagnetic waves, such as the heat from the sun warming your skin.
  • Conduction - Heat moves through materials.
  • Convection - Heat carried by fluids.
  • Radiation - Heat transferred through waves.
Understanding heat transfer is vital for insulating materials, as they typically work by reducing the amount of heat transferred through conduction.
Insulating Materials
Insulating materials are designed to slow down the heat transfer process. They are commonly used to maintain temperature differences in various applications, like keeping homes warm in winter or cool in summer.
These materials have low thermal conductivity, meaning they don't allow heat to pass through them easily. This quality makes them effective at resisting heat flow. Examples include materials like fiberglass, foam, and even certain textiles.
  • Low thermal conductivity: Insulators resist heat flow.
  • Common types include fiberglass and foam.
  • Used to manage thermal environments in buildings.
In the given problem, the insulating material's effectiveness is demonstrated by how much power is needed to maintain a temperature difference inside the insulated box.
SI Units
SI Units, or the International System of Units, are the global standard for measuring physical quantities. They ensure consistency in scientific communication across the world. For thermal conductivity, the SI unit is watts per meter per Kelvin, written as \( ext{W/m} \cdot \text{K}\).
Using consistent units is crucial in calculations involving thermal conductivity, as seen in the problem where power (180 W) and area (2.18 my) are given in SI units.
  • Standard unit for thermal conductivity: \(\text{W/m} \cdot \text{K}\).
  • Ensures consistency in international scientific communication.
  • Allows clear comparisons of insulating effectiveness across materials.
Proper use of SI units helps accurately convey the results of thermal measurements and predictions.
Thermal Insulation Measurement
Thermal insulation measurement is crucial to understanding how well a material can prevent heat transfer. In the exercise, thermal conductivity was measured by calculating how much energy was required to maintain a desired temperature inside a box made of the insulating material.
The key formula used was \(Q = \frac{k \cdot A \cdot \Delta T}{d}\), where:
  • Q is the rate of heat transfer or power in watts.
  • A is the area through which heat is being transferred in square meters.
  • \(\Delta T\) is the temperature difference across the material in degrees Celsius.
  • d is the thickness of the material in meters.
This formula allows scientists and engineers to compare different materials' insulating capabilities by calculating their thermal conductivity. A lower thermal conductivity value indicates better insulating properties.

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Most popular questions from this chapter

"The Ship of the Desert" Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400 \(\mathrm{kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day \(\left(12 \text { hours) instead of letting it rise to } 40.0^{\circ} \mathrm{C} \text { . (Note: The }\right.\) specific heat of a camel or other mammal is about the same as that of a typical human, 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) )

(a) You feel sick and are told that you have a temperature of \(40.2^{\circ} \mathrm{C}\) . What is your temperature in "F? Should you be concerned? (b) The morning weather report in Sydney gives a current temperature of \(12^{\circ} \mathrm{C}\) . What is this temperature in \(^{\circ} \mathrm{F} ?\)

Two beakers of water, \(A\) and \(B\) , initially are at the same temperature. The temperature of the water in beaker \(A\) is increased \(10 F^{\circ},\) and the temperature of the water in beaker \(B\) is increased 10 \(\mathrm{K}\) . After these temperature changes, which beaker of water has the higher temperature? Explain.

A tube leads from a \(0.150-\mathrm{kg}\) calorimeter to a flask in which water is boiling under atmospheric pressure. The calorimeter has specific heat capacity 420 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , and it originally contains 0.340 \(\mathrm{kg}\) of water at \(15.0^{\circ} \mathrm{C}\) . Steam is allowed to condense in the calorimeter at atmospheric pressure until the temperature of the calorimeter and contents reaches \(71.0^{\circ} \mathrm{C}\) , at which point the total mass of the calorimeter and its contents is found to be 0.525 \(\mathrm{kg}\) . Compute the heat of vaporization of water from these data.

The celling of a room has an area of 125 \(\mathrm{ft}^{2}\) . The ceiling is insulated to an \(R\) value of 30 (in units of \(\mathrm{ft}^{2} \cdot \mathrm{F}^{\circ} \cdot \mathrm{h} / \mathrm{Btu} )\) . The surface in the room is maintained at \(69^{\circ} \mathrm{F}\) , and the surface in the attic has a temperature of \(35^{\circ} \mathrm{F}\) . What is the heat flow through the ceiling into the attic in 5.0 \(\mathrm{h} ?\) Express your answer in Btu and in joules.

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