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Chapter 17: Problem 100

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 W is required to keep the interior surface of the box 65.0 C\(^\circ\) (about 120 F\(^\circ\)) above the temperature of the outer surface. The total area of the box is 2.18 m\(^2\), and the wall thickness is 3.90 cm. Find the thermal conductivity of the material in SI units.

Short Answer

Expert verified
The thermal conductivity is approximately 0.0495 W/m路K.

Step by step solution

01

Understand the Heat Transfer Formula

The formula to find the thermal conductivity (\(k\)) is derived from the heat transfer equation: \[ Q = \frac{k \, A \, \Delta T}{d} \] Where:- \(Q\) is the heat transfer rate (in Watts).- \(A\) is the total surface area (in m\(^2\)).- \(\Delta T\) is the temperature difference (in degrees Celsius or Kelvin).- \(d\) is the thickness of the material (in meters).- \(k\) is the thermal conductivity (in W/m路K).We will solve this for \(k\).
02

Convert Units to SI Units

We need to ensure all measurements are in SI units: - Wall thickness: 3.90 cm = 0.039 m (since 1 cm = 0.01 m) - Maintain the power and area as they are already in proper units.
03

Plug Values into the Heat Transfer Formula

Given that:- \( Q = 180 \, \text{W} \)- \( A = 2.18 \, \text{m}^2 \)- \( \Delta T = 65.0 \, \text{C} \)- \( d = 0.039 \, \text{m} \)The formula rearranges to solve for \(k\):\[ k = \frac{Q \, d}{A \, \Delta T} \]
04

Calculate Thermal Conductivity

Substitute the known values into the rearranged formula:\[ k = \frac{180 \, \text{W} \, \times \, 0.039 \, \text{m}}{2.18 \, \text{m}^2 \, \times \, 65.0 \, \text{C}} \]Calculate:- Numerator: \( 180 \, \text{W} \, \times \, 0.039 \, \text{m} = 7.02 \, \text{W路m} \)- Denominator: \( 2.18 \, \text{m}^2 \, \times \, 65.0 = 141.7 \, \text{m}^2路\text{C} \)\[ k = \frac{7.02}{141.7} \approx 0.0495 \, \text{W/m路K} \]
05

Interpret the Result

The thermal conductivity of the material is approximately 0.0495 W/m路K. This value suggests that the material is a relatively poor conductor of heat, which is consistent with its use as an insulating material.

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Key Concepts

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

Understanding Heat Transfer
Heat transfer is a process where heat energy moves from a hotter object to a cooler one. This movement occurs until the temperatures of the two objects are the same. Heat can be transferred in three different ways:
  • Conduction: Transfer of heat through a material without any movement of the material itself. It鈥檚 like heat passing through a steel rod that鈥檚 been heated at one end.
  • Convection: Transfer of heat by the movement of a fluid (like air or water). An example is warmer water rising in a pot while cooler water sinks.
  • Radiation: Transfer of heat through electromagnetic waves, like the heat we feel from the sun.
In this exercise, we focus on conduction, especially as it relates to insulating materials. To analyze this, we use a simple formula involving thermal conductivity, surface area, temperature difference, and thickness.
Role of Insulating Materials
Insulating materials play a key role in reducing heat transfer. They are typically poor conductors of heat, meaning they don't easily allow heat to pass through. This property makes them useful in keeping heat where you want it, whether inside your home on a cold day or outside on a warm day.
Choosing the right insulating material involves considering its thermal conductivity, which quantifies how easily heat can flow through a material. Lower thermal conductivity values indicate better insulating properties. In the given exercise, we calculated the thermal conductivity to understand how effective an unknown material is as an insulator.
Converting to SI Units
To solve physics problems accurately, especially those involving thermal conductivity, it鈥檚 crucial to convert measurements to SI units. The International System of Units (SI) is a globally accepted system for consistency in science and engineering. Here are some common conversions needed in thermal calculations:
  • Length: Convert centimeters to meters by dividing by 100 (since 1 cm = 0.01 m).
  • Temperature: Often remains in Celsius or Kelvin in thermal problems, but adjustment might be needed if specified in another unit.
  • Area: Square meters (m2) is the standard, so ensure any given units are converted if necessary.
In the exercise, the wall thickness was initially in centimeters. We converted it to meters for the calculation to obtain the thermal conductivity in SI units properly.
Key Principles of Thermal Insulation
Thermal insulation is aimed at minimizing unwanted heat transfer. It works by combining low thermal conductivity materials with appropriate application techniques. A few guidelines help improve thermal insulation:
  • Use materials with low thermal conductivity to minimize heat transmission.
  • Ensure the insulating material is thick enough to be effective, as thickness directly affects insulation effectiveness.
  • Consider external factors such as weather and temperature fluctuations that might impact the insulation performance.
With the right choice and application, insulation materials significantly improve energy efficiency in buildings and processes, leading to energy cost savings and enhanced comfort.

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Most popular questions from this chapter

A vessel whose walls are thermally insulated contains 2.40 kg of water and 0.450 kg of ice, all at 0.0\(^\circ\)C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 28.0\(^\circ\)C? You can ignore the heat transferred to the container.

In a container of negligible mass, 0.200 kg of ice at an initial temperature of -40.0\(^\circ\)C is mixed with a mass m of water that has an initial temperature of 80.0\(^\circ\)C. No heat is lost to the surroundings. If the final temperature of the system is 28.0\(\circ\)C, what is the mass m of the water that was initially at 80.0\(^\circ\)C?

In very cold weather a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is -20\(^\circ\)C, what amount of heat is needed to warm to body temperature (37\(^\circ\)C) the 0.50 L of air exchanged with each breath? Assume that the specific heat of air is 1020 J / kg \(\cdot\) K and that 1.0 L of air has mass \(1.3 \times 10{^-}{^3} kg\). (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

A technician measures the specific heat of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat transferred to the liquid for 120 s at a constant rate of 65.0 W. The mass of the liquid is 0.780 kg, and its temperature increases from 18.55\(^\circ\)C to 22.54\(^\circ\)C. (a) Find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. (b) Suppose that in this experiment heat transfer from the liquid to the container or surroundings cannot be ignored. Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat? Explain.

A laboratory technician drops a 0.0850-kg sample of unknown solid material, at 100.0\(^\circ\)C, into a calorimeter. The calorimeter can, initially at 19.0\(^\circ\)C, is made of 0.150 kg of copper and contains 0.200 kg of water. The final temperature of the calorimeter can and contents is 26.1\(^\circ\)C. Compute the specific heat of the sample.
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