PHY 202 - Chapter 12
Revised on April 11, 2005


The illustrations and most of the content on this page are from "Physics", Cutnell and Johnson, 6th edition, Wiley.

Temperature and Heat

Common Temperature Scales

Celsius and Fahrenheit: The Fahrenheit scale, °F, was defined so that freezing water occurs at 32 °F and boiling water occurs at 212 °F. The Celsius scale, °C, was defined so that freezing water occurs at 0 °C and boiling water occurs at 100 °C. Compare the differences in number of degrees between boiling and freezing in the two temperature scales. This ratio is used in converting between the two scales.

    °F: 212 - 32 = 180 °F
    °C: 100 - 0  = 100 °C
    180°F/100°C = 9°F/5°C
The following steps can be taken to convert from one temperature scale to another:

  1. Determine the magnitude difference between the stated temperature and the ice point on the initial scale
  2. Convert this number of degrees from one scale to the other scale by using the conversion factor above (9/5 or 5/9) in the right direction so that the units cancel
  3. Add or subtract the number of degrees on the new scale to or from the ice point on the new scale

Kelvin: Scientists use the Kelvin temperature scale. Each Kelvin is the same size as each degree Celsius but the lowest temperature is zero on the Kelvin scale. That means that all Kelvin temperatures are positive. The freezing point of water is 273.15 K. The point of 0K is called absolute zero. Nothing can ever be colder than absolute zero.


Let's work through Problem 2 (p 366).        Solution


In groups, solve problem 3 (p 366).              Solution

Linear Thermal Expansion

Normal Solids: Most material expand when heated and contract when cooled. Focusing on one dimension, this is known as linear expansion. Experiments indicate that the amount of expansion is linearly dependent on the change in temperature. It is also dependent on the initial length of the object in question. This leads to the following linear expansion equation:

    ΔL = αLoΔT
where α is the coefficient of linear expansion and is dependent on the type of material in question. Table 12.1 in the text lists the coefficients for several materials. What are the units of α?

Bimetallic Strip: Thermostats are often built using a bimetallic strip, which is made by bonding two small strips of different metals together, where each metal has a different coefficient of linear expansion. Referring to the figure below, explain how this works?

The Expansoin of Holes: When there is a hole in a material that is heated, does the hole get bigger of smaller? The hole behaves just as if it were filled with the same material as the surrounding material. Therefore, the hole gets bigger.

Volume Thermal Expansion

The volume of most materials expands as the temperature increases. This is in direct analogy to linear expansion discussed previously. In fact, the formula for volume expansion is very similar to that for linear expansion, but the coefficient of linear expansion, α, is replaced with the coefficient of volume expansion, β

    ΔV = βVoΔT


In groups, solve Conceptual Question 5 (p 365).


Let's work through Problem 34 (p 366).            Solution

Heat and Internal Energy

When heat flows from hot to cold, it originates in the internal energy of the hot substance. The internal energy of a substance is the sum of:

Heat is measured in units of energy, or work. The SI unit of heat (energy) is the joule.

Heat and Temperature Change: Specific Heat Capacity

More heat (energy) is required to raise the temperature of a given substance by 10 degrees than be 5 degrees. Also, more heat is required to raise the temperature of 2 kg of a given material by a given number of degrees than 1 kg of the same material. Experiments indicate that this relationship can be expressed by the following equation:

    Q = cmΔT
where c is referred to as the specific heat capacity of a given material. What are the units of c?


In groups, discuss the answers to Conceptual Question 14 (p 365).

There are three other commonly used units of heat energy: kcal, C(c)alorie, and BTU (British Thermal Unit).

A device known as a calorimeter can be used to measure the specific heat of a given material. A calorimeter is like a glorified thermos, with great insulation and a thermometer to measure the temperature of the contained fluid. A sample of a given material is raised or lowered to a different temperature from the fluid and then dropped into it. Assuming a negligible amount of heat escapes the walls of the calorimeter, the change in measured temperature can be used to determine the specific heat of the material. The energy that left/entered the material must equal the energy that left/entered the fluid because energy is conserved.

Heat and Phase Change: Latent Heat

Generally, adding heat to a substance causes its temperature to rise. But this is not always the case. When the temperature reaches a certain point, the phase of the substance changes between one of: liquid, solid, and gas. At these points, energy is going into, or being released from the substance without a temperature change. Instead, the energy is going into or being released from the phase change. In these terms, put meaning to the following terms: melt (fuse), freeze, evaporate, condense, sublime (sublimation - CO2).

The amount of heat required to change the phase of a substance is determined by the following equation:

    Q = mL
where L is the latent heat of the substance. The latent heat of fusion, Lf, refers to the change between solid and liquid phases (either direction). The latent heat of vaporization, Lv, refers to the change between liquid and gas phases (either direction). The latent heat of sublimation, Ls, refers to the change between solid and gas phases (either direction).

The illustration below shows the temperature as heat is added to a given amount of water, that first starts out as ice at -30°C.

Based on the above figure, which is bigger for water, the latent heat of fusion or the latent heat of vaporization?


Let's work through Problem 53 (p 368).        Solution


In groups, solve problem 59 (p 368).              Solution

Skip Remaining Two Sections 12.9 & 12.10

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