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Chapter 13: Problem 16

In a house the temperature at the surface of a window is \(25^{\circ} \mathrm{C} .\) The temperature outside at the window surface is \(5.0^{\circ} \mathrm{C} .\) Heat is lost through the window via conduction, and the heat lost per second has a certain value. The temperature outside begins to fall, while the conditions inside the house remain the same. As a result, the heat lost per second increases. What is the temperature at the outside window surface when the heat lost per second doubles?

Short Answer

Expert verified
The new outside temperature is -15掳C.

Step by step solution

01

Understanding Heat Conduction

Heat conduction through a material is given by Fourier's law of thermal conduction: \[ Q = \frac{kA(T_1 - T_2)}{d} \] where \( Q \) is the heat transferred per unit time, \( k \) is the thermal conductivity of the material, \( A \) is the area through which heat is transferred, \( T_1 \) and \( T_2 \) are the temperatures on either side of the material, and \( d \) is the thickness of the material.
02

Setting Up Initial Conditions

Initially, the heat loss per second \( Q_1 \) is given by:\[ Q_1 = \frac{kA(25^{\circ}\mathrm{C} - 5^{\circ}\mathrm{C})}{d} \] This simplifies to:\[ Q_1 = \frac{kA \cdot 20}{d} \]
03

Setting Up Conditions When Heat Loss Doubles

The heat loss doubles when \( Q_2 = 2Q_1 \). Therefore:\[ Q_2 = \frac{kA(25^{\circ}\mathrm{C} - T_{outside-new})}{d} \] And:\[ Q_2 = 2 \times \frac{kA \cdot 20}{d} = \frac{kA \cdot 40}{d} \] Thus:\[ \frac{kA(25^{\circ}\mathrm{C} - T_{outside-new})}{d} = \frac{kA \cdot 40}{d} \]
04

Solving for the New Outside Temperature

By equating the equations and solving for the new outside temperature:\[ 25^{\circ}\mathrm{C} - T_{outside-new} = 40 \] \[ T_{outside-new} = 25^{\circ}\mathrm{C} - 40 \] \[ T_{outside-new} = -15^{\circ}\mathrm{C} \]
05

Conclusion

The temperature at the outside window surface when the heat lost per second doubles is \(-15^{\circ}\mathrm{C}\).

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Key Concepts

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

Fourier's Law
Fourier's law of thermal conduction is an essential principle in understanding how heat is transferred through matter. It describes the rate at which heat energy moves through a material due to a temperature difference between two points. Mathematically, Fourier's law is expressed as: \[ Q = \frac{kA(T_1 - T_2)}{d} \]Here:
  • \[ Q \] represents the heat transferred per unit time, often measured in watts (W), indicating how much energy is being transferred.
  • \[ k \] is the thermal conductivity, which is a property of the material indicating its ability to conduct heat. High \( k \) values mean the material is a good conductor.
  • \[ A \] is the surface area through which heat is transferred, measured in square meters (m虏).
  • \[ T_1 \] and \[ T_2 \] are the temperatures at the two surfaces (or sides) of the material in degrees Celsius (掳C).
  • \[ d \] is the thickness of the material, measured in meters (m).
Fourier's law shows that heat transfer is directly proportional to the temperature difference across the material. This means greater temperature differences will result in more heat being transferred, as seen in our window example.
Temperature Gradient
The concept of a temperature gradient is crucial in the study of heat conduction as it represents the change in temperature with respect to distance. In simpler terms, it's the difference in temperature over a particular length.
The temperature gradient can be understood as:
  • It's the driver of heat flow, meaning that heat will naturally move from regions of higher temperature to regions of lower temperature.
  • In a uniformly thick material, a higher temperature gradient means faster heat flow.
  • In our window example, the inside surface temperature is \(25^{\circ} \mathrm{C} \) and the outside begins at \(5^{\circ} \mathrm{C} \). The difference, \(20^{\circ} \mathrm{C} \), is our initial temperature gradient.
When the temperature outside decreases, this gradient increases, and so does the rate of heat transfer according to Fourier's law. Thus, understanding the temperature gradient helps predict how the heat flow changes in different scenarios.
Thermal Conductivity
Thermal conductivity \( k \) is a measure of a material's ability to conduct heat. It's a vital property in heat conduction studies, as it determines how quickly heat can pass through a given material.
Some key points include:
  • Materials with high thermal conductivity, like metals, are excellent conductors, which means they transfer heat quickly.
  • In contrast, materials with low thermal conductivity, like wood or fiberglass, are good insulators.
  • In the exercise's context, knowing the window material's thermal conductivity helps determine how fast heat is lost from the room to the outside environment.
The unit of thermal conductivity is watt per meter-kelvin (W/m路K), and it plays a pivotal role in the computation of heat transfer using Fourier's law. By evaluating thermal conductivity, we can understand and predict heat loss in everyday applications, such as building energy efficiency, as illustrated by the window example.

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Most popular questions from this chapter

Light bulb 1 operates with a filament temperature of \(2700 \mathrm{K}\), whereas light bulb 2 has a filament temperature of \(2100 \mathrm{K}\). Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio \(A_{1} / A_{2}\) of the filament areas of the bulbs.

The Kelvin temperature of an object is \(T_{1},\) and the object radiates a certain amount of energy per second. The Kelvin temperature of the object is then increased to \(T_{2},\) and the object radiates twice the energy per second that it radiated at the lower temperature. What is the ratio \(T_{2} / T_{1} ?\)

A cubical piece of heat-shield tile from the space shuttle measures \(0.10 \mathrm{m}\) on a side and has a thermal conductivity of \(0.065 \mathrm{J} /\left(\mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}\right) .\) The outer surface of the tile is heated to a temperature of \(1150^{\circ} \mathrm{C},\) while the inner surface is maintained at a temperature of \(20.0^{\circ} \mathrm{C}\). (a) How much heat flows from the outer to the inner surface of the tile in five minutes? (b) If this amount of heat were transferred to two liters ( \(2.0 \mathrm{kg}\) ) of liquid water, by how many Celsius degrees would the temperature of the water rise?

A solid sphere has a temperature of \(773 \mathrm{K}\). The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature?

A closed box is filled with dry ice at a temperature of \(-78.5^{\circ} \mathrm{C}\), while the outside temperature is \(21.0^{\circ} \mathrm{C} .\) The box is cubical, measuring \(0.350 \mathrm{m}\) on a side, and the thickness of the walls is \(3.00 \times 10^{-2} \mathrm{m} .\) In one day, \(3.10 \times 10^{6} \mathrm{J}\) of heat is conducted through the six walls. Find the thermal conductivity of the material from which the box is made.
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