In this chapter you will become acquainted with the concept of temperature and the three most common temperature measurement scales: Celsius, Fahrenheit and Kelvin. You will learn to convert a temperature reading in one scale to any other. Additionally, you will study heat including the concepts of specific heat, latent heat of phase change, thermal expansion and the transfer of heat.
The common temperature scale which assigns 32° to the freezing point of water and 212° to the boiling point of water.
A scientific temperature scale which assigns 0° to the freezing point of water and 100° to the boiling point of water.
Kelvin (absolute) scale
A scientific temperature scale which has the same size degree as the Celsius scale, but assigns the lowest possible temperature as 0°. A degree is called a Kelvin (K) and is the SI unit for temperature.
The zero point on the Kelvin temperature scale. It is the lowest possible temperature that can be attained by cooling an object.
The change in physical size (length, area or volume) of a substance when its temperature changes. For most substances, the physical size increases with an increase in temperature and decreases with a decrease in temperature.
The energy associated with individual molecules in a gas, liquid, or solid. This energy may take the form of translational or rotational kinetic energy, vibrational energy, or potential energy.
The energy that flows from a higher temperature object to a lower temperature object because of the difference in temperatures. The SI unit of heat is the joule (J). Another unit for heat is the calorie (cal) or kilocalorie (kcal).
Specific heat capacity
The heat Q per unit mass per degree change in temperature that must be supplied or removed to change the temperature of a substance. The SI unit for specific heat is J/(kg C°).
The amount of heat energy per kilogram that must be added or removed when a substance changes from one phase to another (i.e., from solid to liquid - heat of fusion, or from liquid to gas - heat of vaporization). The SI unit of latent heat is J/kg.
Temperature Scale Conversion
Common Temperature Scales
Example 1 At what temperature will the reading on the Fahrenheit scale be numerically equal to that on the Celsius scale?
The relationship between the two scales is given by TF = (9/5)TC + 32.0. We want the temperature when TF = TC, so we make that substitution. Rearranging gives
Example 2 Most bank thermometers alternately display the temperature in Fahrenheit and Celsius. If a bank thermometer displays 25 °F, what will it display for the Celsius temperature?
The relationship between Fahrenheit and Celsius temperatures is
The Kelvin Temperature Scale
The only difference between the Kelvin and Celsius scales is in the choice of the zero point. The zero point for the Celsius scale was chosen to be the freezing point of water since water is abundant and its freezing point is easily measurable. The zero of the Kelvin scale is chosen to be the lowest possible temperature, that is, absolute zero. Absolute zero corresponds to −273.15 °C. To convert between the two scales it is only necessary to adjust for this difference in zero points by adding or subtracting 273.15.
If you should need to convert from Fahrenheit to Kelvin, simply convert the Fahrenheit temperature to Celsius and then add 273.15 to get the Kelvin temperature.
Heat and Temperature Change: Specific Heat Capacity
In order to raise the temperature of a substance by an amount ΔT, a certain amount of heat, Q, is required. The greater the mass m of the substance, the greater the amount of heat required to raise it by ΔT. The relationship between the heat required, the mass, and the change in temperature is
Q = mc T where the constant of proportionality, c, is referred to as the specific heat capacity or simply the specific heat. The specific heat has units of J/(kg C°).
Table 12.2 shows the specific heat capacities for various substances. For example, water has a specific heat of 4186 J/(kg C°). In other words, it would take 4186 J of heat to raise the temperature of 1 kilogram of water by 1 °C.
Example 4 How much heat is required to raise 5.0 kg of water from 25 °C to 55° C?
Using equation (12.4) we have:
A common unit for measuring heat is the kilocalorie (kcal), defined as the amount of heat energy required to raise the temperature of 1 kilogram of water by 1 °C. From the above discussion it is obvious that 4186 J = 1 kcal. This conversion factor is known as the mechanical equivalent of heat.
Example 5 If 15 kcal of heat are added to 5.0 kg of silver, by how much will its temperature rise?
Using equation (12.4) and the specific heat of silver found in Table 12.2, we can solve for ΔT. We have
Specific heat capacity can be measured using the technique of calorimetry. A calorimeter is an insulated container much like a thermos for hot coffee or iced tea. An ideal calorimeter would prevent any heat from leaking in or out. However, heat can flow between the materials inside the calorimeter which have different temperatures. Cooler materials gain heat while hotter materials lose heat until a common temperature is reached. That is, until thermal equilibrium is achieved. In the process of reaching thermal equilibrium, the total amount of heat gained equals the total amount of heat lost.
Example 6 An aluminum cup having a mass of 250.0 g is filled with 50.0 g of water. The initial temperature of the cup and water is 25.0 °C. A 75.0-g piece of iron initially at 350.0 °C is dropped into the water. What is the final equilibrium temperature of the system assuming that no heat is lost to the outside environment?
Using equation (12.4) and the fact that the heat lost by the iron equals the heat gained by the aluminum + water,
Heat and Phase Change: Latent Heat
There are three phases of matter that exist in nature; solid, liquid, and vapor (gas). Nearly all substances can exist in all three phases, depending on the conditions of temperature and pressure that prevail. For example, consider water at atmospheric pressure. For low temperatures (below 0 °C) water exists as a solid, in the form of ice. At intermediate temperatures (between 0 °C and 100 °C) water is a liquid, and at high temperatures (above 100 °C) it is a gas (water vapor).
We know that water can change phases, either from solid to liquid (or vice versa) or from liquid to gas. The simple act of ice melting represents a phase change.
It is important to note that phase changes always occur at a single temperature, with the addition or subtraction of heat. For example, in order to melt 1 kg of ice, the ice must be at 0 °C and we need to add 33.5 × 104 J of heat to get 1 kg of liquid water which will still be at 0 °C.
The amount of heat per kilogram required to produce a change of phase for a substance is known as the latent heat of that substance. The latent heat of fusion, , refers to a phase change between the liquid and solid phases, and the latent heat of vaporization, , refers to a change between liquid and vapor phases.
Example 7 How much heat is required to convert 250 g of ice at 0 °C to steam at 100 °C?
The problem includes three distinct parts. (1) The heat required to melt the ice, (2) the heat required to raise the temperature of the water from 0 °C to 100 °C, and (3) the heat required to convert the hot water to steam at 100 °C. Referring to Table 12.3 for the values of the latent heats of fusion and vaporization we can write
Practice Problems on Temperature and Heat
1. At what common temperature are the Fahrenheit and Kelvin temperatures the same?
2. What is the temperature on the Fahrenheit temperature scale that corresponds to absolute zero?
5. A 150.0-g glass beaker contains 750 g of water initially at 25.0 °C. A 250.0-g piece of copper
copper, glass, and water, the final temperature is 32.0 °C. What was the temperature of the oven? The specific heats for water, copper, and glass are 4186 J/kg-C, 387 J/kg-C, and 840 J/kg-C, respectively.
6. Calculate the specific heat of a metal from the following data. A container made of metal has a
mass of 1.50 kg and contains 6.00 kg of water. A 0.900 kg piece of the same metal initially at a temperature of 275 °C is dropped into the water. The container and water are initially at a
temperature of 17.0 °C and the final equilibrium temperature of the system is 26.0 °C.
The Transfer of Heat
In this chapter you will be introduced to the mechanisms of heat transfer - convection, conduction, and radiation. After completion of the chapter you will be able to explain how heat is transferred by convection and calculate the rate at which heat is transferred by conduction through various materials. In addition, you will be able to use the Stefan - Boltzmann law to calculate the rate at which energy is radiated from or absorbed by an object.
A process in which heat energy is transferred by the flow of a fluid.
The flow of a fluid when heat is transferred by convection.
A convection process in which the fluid flows due to buoyant forces produced because the heated fluid is less dense than the surrounding cooler fluid.
A process by which heat is transferred through a material without a bulk movement of the material.
Materials which conduct heat well.
Materials which conduct heat poorly.
A process by which energy is transferred by electromagnetic waves.
An idealized perfect absorber and perfect emitter of radiation.
The heat conducted through a bar is
The Stefan - Boltzmann law gives the radiant energy emitted or absorbed by an object:
The ability of a substance to transfer heat by conduction is measured by its thermal conductivity, k, which appears in equation (13.1). Metals have the highest thermal conductivities and are good conductors of heat, while gases generally have the lowest thermal conductivities and are poor conductors of heat. This is why most commercial insulating materials depend on trapped gases, such as air, for their insulating properties.
Example 1 Fiberglas insulation is composed of small glass fibers tangled together so that air is trapped in the spaces between them. Show that the insulating value of fiber glass insulation is due to the trapped air rather than the glass by comparing the heat which will flow through a 1.0 m2 sheet of 3 1/2 inch thick glass in 1.0 minute and a similar sheet of air. Assume that the temperature difference maintained across the sheets is 25 C°.
The heat conducted through the glass sheet is
and the heat conducted through the equivalent sheet of air is
The glass conducts about thirty times as much heat as the air.
Example 2 Two identically shaped bars of different materials are placed end to end. A temperature difference, ΔT, is maintained across the bars. Find an expression for the temperature difference across the first bar. Assume that no heat is lost through the sides of the bars.
If no heat is lost through the sides of the bars, then the heat transferred through each bar in a time, t, is Q. Equation 13.1 gives for the first bar
and for the second bar
Equating and rearranging yields
In the above example we can define an equivalent thermal conductivity of the two bars if we use the value that we found for in the first equation.
The two bars conduct heat energy as if they were a single bar of length, 2L, and thermal conductivity
Example 3 In example 2 the first bar is brass and the second bar is aluminum. The temperature of the left end is 95 °C and the temperature of the right end is 55 °C. What is the temperature at the junction between the bars? What is the equivalent conductivity of the bars?
Table 13.1 gives the thermal conductivity of brass, , and the conductivity of aluminum, . The temperature difference across the brass bar is then
The units of the conductivities have been suppressed for clarity.
The temperature at the junction between the bars is T = 95 °C − 27 °C = 68 °C.
The equivalent thermal conductivity of the bars is
Both emission and absorption of radiation by an object are governed by the Stefan - Boltzmann law, equation (13.2). The temperature, T, in the equation is the Kelvin temperature of the object in the case of emission and it is the Kelvin temperature of the object's surroundings in the case of absorption.
In order to use equation (13.2), we need to know the emissivity, e, of the object in question. It ranges from e = 1 for a perfect blackbody to e = 1 for a perfect reflector of radiation. The emissivity of real objects varies with the type of radiation (visible, infrared, etc.) and is usually determined by experiment.
Example 5 The sun shines directly on one side of a flat black panel on a spacecraft which has an area of 5.0 m2 and an emissivity ofe = 0.95. The spacecraft is located 1.5 1011 m from the sun. How much power does the panel absorb from the sun if the sun is considered to be an ideal blackbody with a temperature of 5700 K? Assuming that the panel can only lose energy through radiation, what is the equilibrium temperature of the panel?
The total power emitted by the sun is given by equation 13.2 with , the surface temperature of the sun. The radius of the sun is .
This energy spreads out over the surface of a sphere of radius, r, so that the amount reaching the panel is
The panel is in equilibrium if its temperature, Tp, remains constant. This means that the panel is losing as much power as it is gaining. It gains energy only through the one sunlit side, but it can radiate energy from both sides.
1. How much heat per square meter is conducted in one hour through a 0.64 cm thick window pane when the inside temperature is 22 °C and the outside temperature is 0 °C? The thermal conductivity of the window glass is 0.80 J/s-m-C.
2. How much heat per square meter is conducted in one hour through two 0.64 cm thick window panes separated by a 0.64 cm stagnant air gap when the inside temperature is 22 °C and the outside temperature is 0 °C? The thermal conductivity of the window glass is 0.80 J/s-m-C and that of air is 0.0256 J/s-m-C.
3. Two cooking pots are identical in every respect except that one has a 0.64 cm thick copper bottom and the other has an aluminum bottom. How thick should the aluminum bottom be in order for it to conduct the same amount of heat from a stove burner in the same time as the copper bottom? The thermal conductivity of copper is 390 J/s-m-C and that of aluminum is 240 J/s-m-C.
4. A star radiates as if it were a blackbody with a temperature of 3000 K. If it radiates the same power as the sun, what is the ratio of its radius to the radius of the sun? Assume the sun has a blackbody temperature of 6000 K. The surface area of a sphere is given by 4πR2.
5. An object of emissivity, 0.5, radiates a certain amount of power when it is at a temperature of 650 °C. If the object were a perfect blackbody, what temperature would be required for it to radiate the same power?
6. How much heat is lost in one hour through a 0.15 m 3.7 m 6.1 m concrete floor if the
inside temperature is 22.0 °C and the ground temperature is 13.0 °C?