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Chapter 12 _ Temperature and Heat

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Chapter 12 _ Temperature and Heat Powered By Docstoc
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12.1 Common Temperature Scales
Example 1 At what temperature will the reading on the Fahrenheit scale be numerically equal to that on the Kelvin scale? The relationship between the two scales is given by TF = (9/5) TC + 32.0. Setting TF = TC and rearranging gives TF - (9/5) TF = 32.0 TF = - (5/4)(32.0) = - 40.0

Example 2 The surface temperature of the sun is approximately 5700 °C. What is this temperature on the Fahrenheit scale?

TF = (9/5) TC + 32.0 TF = (9/5)(5700) + 32.0 = 1.0 104 °F

Example 3 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 TC = (5/9)(TF - 32.0) TC = (5/9)(25 - 32.0) = - 4 °C

12.2 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.

614cd78f-c95b-4746-a61e-335f1d6b60ae.doc Page 2 of 7 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. Example 4 Many people keep their rooms at a temperature of 65 °F. What is this temperature on the Kelvin scale? First convert the temperature to Celsius. TC = (5/9)(TF - 32.0) = (5/9)(65 - 32.0) = 18 °C Now add 273.15 to get the temperature on the Kelvin scale. T = 18 + 273.15 = 291 K

12.4 Linear Thermal Expansion
Example 5 How much will the length of a 1.0 km section of concrete highway change if the temperature varies from - 15 °C in the winter to 41 °C in the summer? Using equation 12.2 and the value of the coefficient of linear expansion for concrete from Table 12.1, we have L = LT = (12 10- 6 /C °)(1.0 103 m)[(41 °C - (-15 °C)]

L = 0.67 m

Example 6 Two 12 ft sections of aluminum siding are placed end to end on the outside wall of a house. How large a gap should be left between the pieces to prevent buckling if the temperature can change by 55 C°? Each piece can expand an amount L = LT = (23 10-6 /°C)(12 ft)(55 °C) = 0.015 ft = 0.18 in.

If each piece can expand in both directions, then each will expand 0.09 in. toward the other. The gap should be at least 2(0.09 in.) = 0.18 in. to keep the pieces from contacting each other and buckling.

Example 7

614cd78f-c95b-4746-a61e-335f1d6b60ae.doc Page 3 of 7 A glass window pane (1.100 m 0.920 m) is fitted inside an aluminum frame with a clearance of 0.05 cm all around on a day when the temperature is 39 °C. How much must the temperature drop for the frame to begin to exert a stress on the window pane? The changes of the largest dimensions of the pane and frame are Lg = gLgT and La =aLaT. The pieces will contact when La -Lg = 0.10 cm. The necessary change in temperature is then

Similarly, for the smallest dimensions,

The window pane and frame will contact along the large dimension first when the temperature drops by 63 C°.

12.5 Volume Thermal Expansion
Example 8 A diesel powered automobile has its steel tank filled to the top with 22.0 gal of diesel fuel from an underground storage tank. The fuel in the storage tank is initially at 55 °F (13 °C) and warms up to 95 °F (35 °C) as it sits in the car's tank. How much fuel spills out of the car's tank? The coefficient of volume expansion of diesel fuel is 9.5 10-4 /C°. The interior volume of the car's tank expands as if it were made of steel. It will increase by Vt = tVT. The fuel will increase its volume by an amount Vf = fVT. The amount of fuel which will spill out of the tank, V, is then V = Vf -Vt = (f -t)VT V = (9.5 10-4 /C° - 36 10-6 /C°)(22 gal)(22 C°) V = 0.44 gal

12.7 Heat and Temperature Change: Specific Heat Capacity
In order to raise the temperature of a substance by an amount T, 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 T. The relationship between the heat required, the mass, and the change in temperature is

614cd78f-c95b-4746-a61e-335f1d6b60ae.doc of 7 Q = cm T

Page 4 (12.4)

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 9 How much heat is required to raise 5.0 kg of water from 25 °C to 55 ° C? Using equation (12.4) we have: Q = cm T = [4186 J/(kg C ° )](5.0 kg)(55 °C - 25 °C) = 6.3 105 J.

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 10 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 T. 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 11 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 °Cis dropped into the water. What is the final equilibrium temperature of the system assuming that no heat is lost to the outside environment?

614cd78f-c95b-4746-a61e-335f1d6b60ae.doc Page 5 of 7 Using equation (12.4) and the fact that the heat lost by the iron equals the heat gained by the aluminum + water, Heat Gained = Heat Lost (cm T)aluminum + (cm T)water = (cm T)iron [9.00 102J/(kg·C°)](0.250 kg)(Tf - 25.0 °C) + [4186 J/(kg·C°)](0.0500 kg)(Tf - 25.0 °C) = [452 J/(kg C ° )](0.0750 kg)(350.0 °C - Tf) Tf = 48.5 °C

12.8 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, Lf, refers to a phase change between the liquid and solid phases, and the latent heat of vaporization, Lv, refers to a change between liquid and vapor phases. Example 12 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 Qtotal = Qfusion + Qtemp. change + Qvaporization Qtotal = mLf + cm T + mLv Qtotal = (0.25 kg)(33.5 104 J/kg) + [4186 J/(kg · C°)](0.25 kg)(100 C°) + (0.25 kg)(22.6 Qtotal = 7.5 105J/kg) 105 J.

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12.9 Equilibrium Between Phases of Matter
Under certain conditions of temperature and pressure, a substance can exist in equilibrium in more than one phase at the same time. The pressure at which vapor coexists in equilibrium with liquid is called the equilibrium vapor pressure. For example, consider the following vaporization curve for water. We can see that, for water, at a pressure of 1.01 105 Pa, the corresponding vaporization temperature is 100 °C. Thus, liquid water and vapor can exist in equilibrium at this pressure and temperature. Water, therefore, boils at 100 °C at atmospheric pressure. However, if we were to change the pressure, the corresponding temperature needed for equilibrium would change. For example, as evident from the curve, if the pressure were about 2 105 Pa, the corresponding temperature required for equilibrium would be about 125 °C. So water would boil at 125 °C if it were subjected to two atmospheres of pressure.

12.10 Humidity
The partial pressure of water vapor is the part of the total atmospheric pressure that is due to water vapor in the air. For example, out of the total 1.01 105 Pa of pressure in our atmosphere, 4.0 103 Pa may be due to water vapor. The relative humidity is the partial pressure of water vapor at the existing temperature divided by the equilibrium vapor pressure of water at the existing temperature. That is (12.6) Example 13 On a certain day the partial pressure of water vapor is 1.0 103 Pa. Using the vaporization curve below, determine the relative humidity if the temperature is (a) 16 °C, and (b) 26 °C.

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(a)

At 16 °C we see that the equilibrium vapor pressure is 1.3 (12.6), we have

103 Pa. Therefore, using equation

(b)

At 26 °C the equilibrium vapor pressure is 3.0

103 Pa. Therefore,

When the partial pressure of water vapor equals the equilibrium vapor pressure of water at a given temperature, equation (12.6) tells us that the relative humidity is 100%. In this case, the air is said to be saturated. The temperature corresponding to this condition is known as the dew point. The dew point is therefore the temperature at which the water vapor in the air would condense in the form of liquid drops (dew or fog). In example 13 it can be seen that the dew point (the temperature corresponding to a vapor pressure of 1.0 103 Pa) is about 13 °C.


				
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