Knowledge of temperature and temperature scales is an important part of the study of thermodynamics. Specific heat values are given in degrees Celsius, so converting from Fahrenheit to the other temperature scales is an important skill.

1. A thermometer in a hot liquid reads 77° F. What temperature is this on the Celsius scale and on the Kelvin scale?

T_C = 5/9 (T_F - 32) = 5/9 (77° - 32) = 5/9 (45°) = 25°C

T_K = 25° + 273 = 25° + 273 = 298 K

Notice that we first convert to Celsius, and then to Kelvin. Also, there is no misprint in units. The degree marking is not used in the Kelvin designation of temperature, even though it is used in both the Fahrenheit and Celsius notations.

Heat is a form of energy, and this knowledge can help us to calculate heat loses and gains. If an object is dropped near Earth's surface, some of its energy is converted into heat when it strikes the ground, as shown in the following problem.

2. A 5.00-kg ball of clay falls from a height of 6.50 meters. If 92.0% of its energy is converted into distortional work that flattens the clay upon impact, how much heat is produced when the clay strikes the ground?

The total energy at the top of the fall is converted into kinetic energy just before impact with the ground (h = 0), and the kinetic energy is transformed into work and heat when the ball actually strikes the ground. We could, therefore, calculate the kinetic energy from the velocity of the clay just before impact. (This velocity is 11 m/s if you decide to do this calculation for practice.)

Because the total energy remains the same, however, it is easier to use conservation of energy and convert directly from potential energy into heat and work. The change in potential energy of the clay must be equal to the work done to distort the clay plus any heat energy that the clay retains after impact.

The change in potential energy of the clay is given by:

E_p = m g h = 5.00 kg (9.80 m/s²)(6.50 m) = 319 J

This means a total of 319 J of energy is available from the fall. The work done to distort the clay on impact is given as 92.0%, but we must use the decimal equivalent 0.920 in our calculation. The work causing distortion is then negative because it is work done on the clay.

W = -0.920E = - 0.920 (319 J) = -293 J

The heat retained by the clay ball can then be found using the second law of thermodynamics:

H = E + W = 319 J + (-293 J) = 26 J

Thus we see that most of the energy goes to distort the molecular structure of the clay and only a small fraction of the energy causes an increase in the total internal KE of the molecules. The joule is a perfectly acceptable unit for heat, but we can also express the heat in kilocalories.

26 J (1 kcal / 4186 J) = 0.0062 kcal

3. How much increase in temperature does the clay in question 2 experience upon impact?

Here we must use the concept of specific heat in our calculation. Since the specific heat is given in Table 5.1 (see textbook) in both kcal/kg-C° and J/kg-C°, we can use either 26 J or 0.0062 kcal in our determination, as long as we choose the proper specific heat value. Since "clay" is not given specifically, we will use the value for "soil," which should be close enough for our example problem. There are tables similar to 5.1 in other books that contain specific heat values for almost any material that you might need to work with, so we could find an exact value for the specific heat of clay if we really needed to. Such tables might be found in your local library.

H = m c T, but we must first solve for T

T = H / m c = 26 J / [5.00 kg (1050 J/kg-C°)] = 0.005° C

Remember that this is the change in temperature, not the final temperature of the clay. If the clay has begun its fall with a temperature of 21.000°, the temperature of the clay would now be only 21.005°, which is much too small a change to notice. We had to use 5 significant figures in our energy calculation to find such a small change, but heat really is generated in this process and it is not always negligible in real-life situations. As an example, consider the work done by the brakes to slow down your car if you are going 55 mph and must make a sudden stop. Almost all of the work done to reduce the kinetic energy of the moving car is converted by friction into heat, and the temperature of your braking system will increase dramatically.

Now let's look at a problem that requires phase transitions as well as specific heat calculations.

4. How many calories are required to raise the temperature of 10 kg of ice from -20° C to 50° C? (The specific heat of ice is 0.50 kcal/kg-C°)

The calculation must be done in three steps:

(1) raising the temperature of the ice from -20° C to 0° C

(2) converting the solid ice at 0° C to a liquid water at 0° C

(3) raising the temperature of the liquid water from 0° C to 50° C

The total number of calories required is the sum of the three steps.

Step 1, raising the temperature of the ice from -20° C to 0° C

H₁ = m cT

where the ice temperature change is: T = T_F - T_I = 0° - (-20° C) = 20° C

H₁ = 10 kg (0.50 kcal/kg-C°)(20° C) = 100 kcal

Step 2, converting the solid ice at 0° C to a liquid water at 0° C

The latent heat of fusion (L_F) of water is 80 kcal/kg. This is found using Table 5.2 in the textbook.

H₂ = m L_F = 10 kg (80 kcal/kg) = 800 kcal

Notice that there is no temperature-change term in this calculation. The transition from ice to water takes place at a constant temperature, 0° C. As long as there is any ice left, the temperature of the ice-water mixture will remain at the melting point temperature, 0° C. As soon as all of the ice has melted, the heat being added begins to increase the temperature of the water again.

Step 3, raising the temperature of the liquid water from 0° C to 50° C

H₃ = m cT

where the water temperature change is: T= 50° C - 0° C = 50° C

and c for water is found using Table 5.1 in the textbook.

H₃ = 10 kg (1.00 kcal/kg-°C)(50° C) = 500 kcal

The total number of calories (H_Total) required to heat the ice at -20° C to water at 50° C is the sum:

H_Total = H₁ + H₂ + H₃ = 100 kcal + 800 kcal + 500 kcal = 1400 kcal

5. A 350-g glass beaker holding 100 g of water is at a temperature of 25.0° C. How much heat must be added to bring the beaker and water to 30.0° C?

Since no phase changes are involved for water or glass in this temperature range we can simply use the specific heat equation.

H = m cT

Calculations must be made separately for the glass and the water. All calculations must be done in the same units.

We were given:mass of glass = m_g = 0.35 kg mass of water = m_w = 0.10 kg

From Table 5.1:

spec. heat (glass) = c_g = 0.16 kcal/kg-C°

spec. heat (water) = c_w = 1.00 kcal/kg-C°

The temperature difference is: T = 30.0°C - 25.0°C = 5.0°C for both materials

Now we can calculate the heat that must be added.

H = m_g c_gT + m_w c_wT

H = 0.35 kg (0.16 kcal/kg-C°)(5.0 C°) + 0.10 kg (1 kcal/kg-C°)(5.0° C)

H = 0.28 kcal + 0.50 kcal = 0.78 kcal

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