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An electric kitchen range has a total wall area of 1.40 \(\mathrm{m}^{2}\) and is insulated with a layer of fiberglass 4.00 \(\mathrm{cm}\) thick. The inside surface of the fiberglass has a temperature of \(175^{\circ} \mathrm{C}\) , and its outside surface is at \(35.0^{\circ} \mathrm{C}\) . The fiberglass has a thermal conductivity of 0.040 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) (a) What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of 1.40 \(\mathrm{m}^{2} ?(\mathrm{b})\) What electric- power input to the heating element is required to maintain this temperature?

Short Answer

Expert verified
(a) The heat current is 196 W. (b) The electric power input required is 196 W.

Step by step solution

01

Understanding the Problem

To find the heat current through the insulation, we can use the formula for heat transfer through a material: \[ Q = \frac{k imes A imes (T_1 - T_2)}{d} \] where \( Q \) is the heat current, \( k \) is the thermal conductivity, \( A \) is the area, \( (T_1 - T_2) \) is the temperature difference, and \( d \) is the thickness of the insulation.
02

Converting Units

The thickness of the insulation is given in centimeters, so first, convert this to meters: \[ d = 4.00 \text{ cm} = 0.040 \text{ m} \]
03

Calculating Temperature Difference

Calculate the temperature difference across the fiberglass: \[ (T_1 - T_2) = 175^{\circ} \text{C} - 35.0^{\circ} \text{C} = 140^{\circ} \text{C} \] Write temperature difference in Kelvin, but the difference itself stays the same.
04

Calculating Heat Current (Part a)

Substitute the known values into the heat transfer formula: \[ Q = \frac{0.040 \, \mathrm{W/m} \cdot \mathrm{K} \times 1.40 \, \mathrm{m}^2 \times 140 \, \mathrm{K}}{0.040 \, \mathrm{m}} \]\[ Q = \frac{7.84}{0.040} \]\[ Q = 196 \, \mathrm{W} \] The heat current through the insulation is 196 W.
05

Understanding Electric Power Input (Part b)

The electric power input required must equal the heat current through the insulation to maintain the temperature difference.

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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 how well heat flows through a material. It's denoted by the symbol \( k \), and it's expressed in units of watts per meter-kelvin (\( \mathrm{W/m} \cdot \mathrm{K} \)). This property is crucial in understanding how substances transfer heat. In our exercise, the fiberglass insulation has a thermal conductivity of 0.040 \( \mathrm{W/m} \cdot \mathrm{K} \).

A low thermal conductivity implies that the material is a good insulator since it resists heat flow effectively. When selecting an insulating material, one typically seeks materials with low thermal conductivity to minimize energy losses. In practical applications, knowing the thermal conductivity helps in calculating heat transfer and determining energy efficiency in systems.
  • Higher \( k \) value: good conductor, poor insulator (e.g., metals)
  • Lower \( k \) value: poor conductor, good insulator (e.g., fiberglass)
Insulation
Insulation is used to minimize heat transfer between different areas, conserving energy and maintaining desired temperatures. In the exercise, fiberglass serves as the insulation material, which is 4.00 cm thick, equivalent to 0.040 meters when converted.

The effectiveness of insulation is largely determined by its thickness and thermal conductivity. A thicker layer or a material with lower thermal conductivity leads to better insulation. Insulation keeps heat energy from escaping, maintaining the temperature of the electric kitchen range's interior, thus requiring less energy to operate.
  • Purpose: Reduce heat loss/gain
  • Materials: Often low \( k \)
  • Examples: Fiberglass, foam, and wool
Electric Power
Electric power is the rate at which energy is consumed or generated and is measured in watts (W). In this context, it refers to the power required to maintain the temperature of the electric kitchen range. To keep the temperature constant, the electric power input must match the heat current that flows through the insulation.

This ensures that any heat loss through the insulation is compensated. The calculation shows that to maintain a stable interior temperature, 196 W of electric power is needed to balance the heat current working away from the insulated surface.
  • Measured in: Watts (W)
  • Needed to balance heat loss
  • Crucial for energy efficiency
Temperature Difference
Temperature difference is the driving force behind heat transfer. The greater the difference in temperature across a material, the greater the heat flow. It's represented as \( (T_1 - T_2) \), where \( T_1 \) and \( T_2 \) are the temperatures on either side of the material.

In the problem, the difference is \( 140^{\circ} \text{C} \) between the inside and outside of the fiberglass insulation. This value stays the same when converted to Kelvin because we're measuring the difference, not the absolute temperature. A large temperature difference speeds up heat transfer, which means more energy will be required to maintain the desired temperature inside.
  • Measured in: degrees Celsius or Kelvin
  • Drives the rate of heat transfer
  • Critical in designing insulation

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Most popular questions from this chapter

Food Intake of a Hamster. The energy output of an animal engaged in an activity is called the basal metabolic rate (BMR) and is a measure of the conversion of food energy into other forms of energy. A simple calorimeter to measure the BMR consists of an insulated box with a thermometer to measure the temperature of the air. The air has density 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and specific heat 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . A 50.0 -g hamster is placed in a calorimeter that contains 0.0500 \(\mathrm{m}^{3}\) of air at room temperature. (a) When the hamster is running in a wheel, the temperature of the air in the calorimeter rises 1.60 \(\mathrm{C}^{\circ}\) per hour. How much heat does the running hamster generate in an hour? Assume that all this heat goes into the air in the calorimeter. You can ignore the heat that goes into the walls of the box and into the thermometer, and assume that no heat is lost to the surroundings. (b) Assuming that the hamster converts seed into heat with an efficiency of 10\(\%\) and that hamster seed has a food energy value of 24 \(\mathrm{J} / \mathrm{g}\) , how many grams of seed must the hamster eat per hour to supply this energy?

A copper calorimeter can with mass 0.446 kg contains 0.0950 \(\mathrm{kg}\) of ice. The system is initially at \(0.0^{\circ} \mathrm{C}\) (a) If 0.0350 \(\mathrm{kg}\) of steam at \(100.0^{\circ} \mathrm{C}\) and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

A copper calorimeter can with mass 0.100 kg contains 0.160 \(\mathrm{kg}\) of water and 0.0180 \(\mathrm{kg}\) of ice in thermal equilibrium at atmospheric pressure. If 0.750 \(\mathrm{kg}\) of lead at a temperature of \(255^{\circ} \mathrm{C}\) is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

Why. Liquid nitrogen is a relatively inexpensive material that is often used to perform entertaining low-temperature physics demonstrations. Nitrogen gas liquefies at a temperature of \(-346^{\circ} \mathrm{F}\) . Convert this temperature to (a) \(^{\circ} \mathrm{C}\) and \((\mathrm{b}) \mathrm{K}\) .

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.

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