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Chapter 14: Problem 50

\(\cdot\) A slab of a thermal insulator with a cross-sectional area of 100 \(\mathrm{cm}^{2}\) is 3.00 cm thick. Its thermal conductivity is 0.075 \(\mathrm{W} /(\mathrm{m} \cdot \mathrm{K}) .\) If the temperature difference between opposite faces is \(80 \mathrm{C}^{\circ},\) how much heat flows the slab in 1 day?

Short Answer

Expert verified
The heat flow through the slab in 1 day is 259,200 J.

Step by step solution

01

Identify Given Parameters

We are given the following values:- Cross-sectional area, \( A = 100 \text{ cm}^2 = 0.01 \text{ m}^2 \) (converted to m虏)- Thickness \( d = 3.00 \text{ cm} = 0.03 \text{ m} \)- Thermal conductivity \( k = 0.075 \text{ W/mK} \)- Temperature difference \( \Delta T = 80 \text{ 掳C} \)- Duration \( t = 1 \text{ day} = 86400 \text{ seconds} \).
02

Recall the Heat Transfer Formula

The formula to calculate heat transfer due to conduction is:\[Q = \frac{kA\Delta T t}{d}\]where \( Q \) is the amount of heat transferred, \( k \) is the thermal conductivity, \( A \) is the cross-sectional area, \( \Delta T \) is the temperature difference, \( t \) is the time, and \( d \) is the thickness of the slab.
03

Substitute Given Values into the Formula

Substitute the numbers into the heat transfer formula:\[Q = \frac{0.075 \times 0.01 \times 80 \times 86400}{0.03}\]
04

Calculate the Heat Transfer

Perform the calculation:\[Q = \frac{0.075 \times 0.01 \times 80 \times 86400}{0.03} = \frac{5184}{0.03} = 2.592 \times 10^5 \text{ J}\]Thus, the heat flow through the slab in one day is \( 259,200 \text{ J} \).

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Key Concepts

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

Heat Transfer
Heat transfer refers to the movement of thermal energy from one object to another. This process occurs when there is a temperature difference between bodies or areas. Heat always moves from warmer regions to cooler ones, striving to reach thermal equilibrium. There are three main methods of heat transfer:
  • Conduction, which is the direct transfer through materials, like when heat moves through a pan to cook food.
  • Convection, where heat is carried by fluid movement, such as currents in water or air.
  • Radiation, the transfer through electromagnetic waves, like sunlight warming the Earth.
In the context of our exercise, heat transfer occurs through conduction. This is because the slab of thermal insulator transfers heat between its surfaces due to the temperature difference.
Thermal Insulators
Thermal insulators are materials that resist the flow of heat. These materials are designed to slow down heat transfer, making them vital for maintaining temperature in homes or cooking equipment. Insulators have low thermal conductivity, meaning they are less efficient at conducting thermal energy. Common thermal insulators include:
  • Wood
  • Styrofoam
  • Fiberglass
In the exercise problem, the slab has a thermal conductivity of 0.075 W/mK, indicating it restricts heat transfer to a moderate degree. This property is crucial for applications wanting to control heat movement, such as in building insulation.
Temperature Difference
Temperature difference is the driving force behind heat transfer. It is the variation in temperature between two points or surfaces. A greater temperature difference results in a more substantial heat flow. This is because the thermal energy naturally flows from areas of higher temperatures to those of lower temperatures as the system tries to reach equilibrium. In our exercise, a temperature difference of 80 掳C exists between the slab's faces. This significant difference forces heat to flow through the slab, making it a critical factor in calculating how much heat is transferred in a given time period. Mathematically, this is incorporated into the heat transfer equation that we used to find the heat flow, emphasizing its importance in such calculations.

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Most popular questions from this chapter

\(\cdot\) Evaporative cooling. The evaporation of sweat is an important mechanism for temperature control in some warmblooded animals. (a) What mass of water must evaporate from the skin of a 70.0 \(\mathrm{kg}\) man to cool his body 1.00 \(\mathrm{C}^{\circ}\) . The heat of vaporization of water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) The specific heat capacity of a typical human body is 3480 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}) .\) (b) What volume of water must the man drink to replenish the evaporated water? Compare this result with the volume of a soft-drink can, which is 355 \(\mathrm{cm}^{3} .\)

\(\cdot\) The emissivity of tungsten is \(0.35 .\) A tungsten sphere with a radius of 1.50 \(\mathrm{cm}\) is suspended within a large evacuated enclo- sure whose walls are at 290 \(\mathrm{K}\) . What power input is required to maintain the sphere at a temperature of 3000 \(\mathrm{K}\) if heat conduction along the supports is negligible?

Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales: (a) the midday temperature at the surface of the moon \((400 \mathrm{K}) ;\) (b) the temperature at the tops of the clouds in the atmosphere of Saturn \((95 \mathrm{K}) ;(\mathrm{c})\) the temperature at the center of the sun \(\left(1.55 \times 10^{7} \mathrm{K}\right)\) .

\(\bullet\) In a physics lab experiment, a student immersed 200 one-cent coins (each having a mass of 3.00 g) in boiling water. After they reached thermal equilibrium, she quickly fished them out and dropped them into 0.240 \(\mathrm{kg}\) of water at \(20.0^{\circ} \mathrm{C}\) in an insulated container of negligible mass. What was the final temperature of the coins? [One-cent coins are made of a metal alloy-mostly zinc-with a specific heat capacity of 390 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}) . ]\)

A pot with a steel bottom 8.50 \(\mathrm{mm}\) thick rests on a hot stove. The area of the bottom of the pot is 0.150 \(\mathrm{m}^{2} .\) The water inside the pot is at \(100.0^{\circ} \mathrm{C},\) and 0.390 \(\mathrm{kg}\) are evaporated every 3.00 min. Find the temperature of the lower surface of the pot, which is in contact with the stove.
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