All chemical reactions are accompanied by a change in energy, which is defined in scientific terms as the capacity to do work or supply heat (Figure 1.9). Detailed discussion of the various kinds of energy will be included in Chapter 7, but for now we will look at the various units used to describe energy and heat, and how heat energy can be gained or lost by matter.
Temperature, the measure of the amount of heat energy in an object, is commonly reported either in Fahrenheit (°F) or Celsius (°C) units. The SI unit for reporting tem- perature, however, is the kelvin (K). (Note that we say only “kelvin,” not “degrees kelvin”.) The kelvin and the celsius degree are the same size—both are 1/100 of the interval be- tween the freezing point of water and the boiling point of water at atmospheric pressure.
Energy The capacity to do work or supply heat.
ANALYSIS Knowing the patient’s body weight (in kg) and the recommended dosage (in mg/kg), we can calculate the appropriate amount of digitalis.
BALLPARK ESTIMATE Since a kilogram is roughly equal to 2 lb, a 160 lb patient has a mass of about 80 kg. At a dosage of 20 mg/kg, an 80 kg patient should receive 80 * 20 mg, or about 1600 mg of digitalis, or 1.6 mg.
SOLUTION
STEP 1: Identify known information.
STEP 2: Identify answer and units.
STEP 3: Identify conversion factors. Two conversions are needed. First, convert the patient’s weight in pounds to weight in kg. The correct dose can then be determined based on mg digitalis/kg of body weight. Finally, the dosage in mg is con- verted to mg.
STEP 4: Solve. Use the known information and the conversion factors so that units cancel, obtaining the answer in mg.
Temperature The measure of the amount of heat energy in an object.
Thus, a change in temperature of 1 °C is equal to a change of 1 K. The only difference between the Kelvin and Celsius temperature scales is that they have different zero points.
The Celsius scale assigns a value of 0 °C to the freezing point of water, but the Kelvin scale assigns a value of 0 K to the coldest possible temperature, sometimes called absolute zero, which is equal to -273.15 ⬚C. Thus, 0 K = -273.15 ⬚C, and +273.15 K = 0 ⬚C.
For example, a warm spring day with a temperature of 25 °C has a Kelvin temperature of 298 K (for most purposes, rounding off to 273 is sufficient):
Temperature in K = Temperature in ⬚C + 273.15 Temperature in ⬚C = Temperature in K - 273.15
For practical applications in medicine and clinical chemistry, the Fahrenheit and Celsius scales are used almost exclusively. The Fahrenheit scale defines the freez- ing point of water as 32 °F and the boiling point of water as 212 °F, whereas 0 °C and 100 °C are the freezing and boiling points of water on the Celsius scale. Thus, it takes 180 Fahrenheit degrees to cover the same range encompassed by only 100 celsius degrees, and a Celsius degree is therefore exactly 180>100 = 9>5 = 1.8 times as large as a Fahrenheit degree. In other words, a change in temperature of 1.0 °C is equal to a change of 1.8 °F. Figure 1.10 gives a comparison of all three scales.
Converting between the Fahrenheit and Celsius scales is similar to converting between different units of length or volume, but is a bit more complex because two corrections need to be made—one to adjust for the difference in degree size and one to adjust for the different zero points. The degree-size correction is made by using the relationship 1 °C = 19>52 °F and 1 °F = 15>92 °C. The zero-point correction is made by remembering that the freezing point is higher by 32 on the Fahrenheit scale than on the Celsius scale. These corrections are incorporated into the following formulas, which show the conversion methods:
Celsius to Fahrenheit: ⬚F = a9 ⬚F
5 ⬚C * ⬚Cb + 32 ⬚F Fahrenheit to Celsius: ⬚C = 5 ⬚C
9 ⬚F * 1⬚F - 32 ⬚F2
▲ Figure 1.9
The reaction of aluminum with bromine releases energy in the form of heat.
When the reaction is complete, the products undergo no further change.
Boiling water
Body temperature Room temperature Freezing water A cold day
“Crossover point”
Fahrenheit (°F)
“Absolute zero”
212
98.6 68 32
−4
−40
−459.67
Celsius (°C) 100
37 20 0
−20
−40
−273.15
Kelvin (K) 373.15
310 293 273.15 253 233
0 98.6
68
37
20
▲ Figure 1.10
A comparison of the Fahrenheit, Celsius, and Kelvin temperature scales.
One Fahrenheit degree is 5>9 the size of a kelvin or a celsius degree.
S E C T I O N 1 . 1 3 Temperature, Heat, and Energy 31
Temperature–Sensitive Materials
Wouldn’t it be nice to be able to tell if the baby’s formula bot- tle is too hot without touching it? Or to easily determine if the package of chicken you are buying for dinner has been stored appropriately? Temperature-sensitive materials are already being used in these and other applications. Although these materials have been used previously in many popular “fads,” like mood rings or clothes that changed color at different temperatures, more practical applications are emerging.
Most current applications use substances known as ther- mochromic materials that change color as their temperature increases, and they change from the liquid phase to a semi- crystalline ordered state. These “liquid crystals” can be incor- porated into plastics or paints and can be used to monitor the temperature of the products or packages in which they are in- corporated. For example, some meat packaging now includes a temperature strip that darkens when the meat is stored above a certain temperature, which makes the meat unsafe to eat. Some beverage containers turn color to indicate when the beverage has reached its optimal temperature for consumption. Hospitals and other medical facilities now routinely use strips that, when
placed under the tongue or applied to the forehead, change color to indicate the patient’s body temperature. In the future, we may even see road signs that change color to warn us of dangerous icy road conditions.
See Chemistry in Action Problems 1.98 and 1.99 at the end of the chapter.
CHEMISTRY IN ACTION
Energy is represented in SI units by the unit joule (J; pronounced “jool”), but the metric unit calorie (cal) is still widely used in medicine. In most of this text we will present energy values in both units of calories and joules. One calorie is the amount of heat necessary to raise the temperature of 1 g of water by 1 °C. A kilocalorie (kcal), often called a large calorie (Cal) or food calorie by nutritionists, equals 1000 cal:
Specific heat The amount of heat that will raise the temperature of 1 g of a substance by 1 °C.
1000 cal = 1 kcal 1000 J = 1 kJ 1 cal = 4.184 J 1 kcal = 4.184 kJ
Not all substances have their temperatures raised to the same extent when equal amounts of heat energy are added. One calorie raises the temperature of 1 g of water by 1 °C but raises the temperature of 1 g of iron by 10 °C. The amount of heat needed to raise the temperature of 1 g of a substance by 1 °C is called the specific heat of the substance. It is measured in units of cal/1g#⬚C2.
Specific heat = calories grams * ⬚C
Specific heats vary greatly from one substance to another, as shown in Table 1.10.
The specific heat of water, 1.00 cal/(g# ⬚C) (or 4.184 J/g °C) is higher than that of most other substances, which means that a large transfer of heat is required to change the temperature of a given amount of water by a given number of degrees. One conse- quence is that the human body, which is about 60% water, is able to withstand chang- ing outside conditions.
Knowing the mass and specific heat of a substance makes it possible to calculate how much heat must be added or removed to accomplish a given temperature change, as shown in Worked Example 1.15.
Heat 1cal2 = Mass 1 g 2 * Temperature change 1⬚C2 * Specific heata cal g # ⬚Cb
TABLE1.10 Specific Heats of Some Common Substances Substance Specific Heat [cal/g °C]; [J/g °C]
Ethanol 0.59; 2.5
Gold 0.031; 0.13
Iron 0.106; 0.444
Mercury 0.033; 0.14
Sodium 0.293; 1.23
Water 1.00; 4.18
STEP 1: Identify known information.
STEP 2: Identify answer and units.
STEP 3: Identify conversion factors. The amount of energy (in cal) can be calculated using the specific heat of water 1cal/g # ⬚C2, and will depend on both the mass of water (in g) to be heated and the total temperature change (in °C).
In order for the units in specific heat to cancel correctly, the mass of water must first be converted from kg to g.
STEP 4: Solve. Starting with the known information, use the conversion factors to cancel unwanted units.
Worked Example 1.14 Temperature Conversions: Fahrenheit to Celsius
A body temperature above 107 °F can be fatal. What does 107 °F correspond to on the Celsius scale?
ANALYSIS Using the temperature (in °F) and the appropriate temperature conversion equation we can convert from the Fahrenheit scale to the Celsius scale.
BALLPARK ESTIMATE Note in Figure 1.10 that normal body temperature is 98.6 °F, or 37 °C. A temperature of 107 °F is approximately 8 °F above normal; since 1 °C is nearly 2 °F, then 8 °F is about 4 °C. Thus, the 107 °F body tempera- ture is 41 °C.
SOLUTION
STEP 1: Identify known information.
STEP 2: Identify answer and units.
STEP 3: Identify conversion factors. We can convert from °F to °C using this equation.
STEP 4: Solve. Substitute the known temperature (in °F) into the equation.
Temperature = 107 ⬚F Temperature = ?? ⬚C
⬚C = 5 ⬚C
9 ⬚F * 1⬚F - 32 ⬚F2
⬚C = 5 ⬚C
9 ⬚F * 1107 ⬚F - 32 ⬚F2 = 42 ⬚C*
(Rounded off from 41.666 667 °C) BALLPARK CHECK Close to our estimate of 41 °C.
*It is worth noting that the 5>9 conversion factor in the equation is an exact conversion, and so does not impact the number of significant figures in the final answer.
Worked Example 1.15 Specific Heat: Mass, Temperature, and Energy
Taking a bath might use about 95 kg of water. How much energy (in calories and Joules) is needed to heat the water from a cold 15 °C to a warm 40 °C?
ANALYSIS From the amount of water being heated (95 kg) and the amount of the temperature change (40 ⬚C - 15 ⬚C = 25 ⬚C), the total amount of energy needed can be calculated by using specific heat
31.00 cal/1g#⬚C24 as a conversion factor.
BALLPARK ESTIMATE The water is being heated 25 °C (from 15 °C to 40 °C), and it therefore takes 25 cal to heat each gram. The tub contains nearly 100,000 g (95 kg is 95,000 g), and so it takes about 25 * 100,000 cal, or 2,500,000 cal, to heat all the water in the tub.
SOLUTION
Mass of water = 95 kg
Temperature change = 40 ⬚C - 15 ⬚C = 25 ⬚C Heat = ?? cal
Specific heat = 1.0 cal g# ⬚C 1 kg = 1000 gS1000 g
1 kg
95 kg * 1000 g
kg * 1.00 cal
g# ⬚C * 25 ⬚C = 2,400,000 cal
= 2.4 * 106 cal 1or 1.0 * 107 J2 BALLPARK CHECK Close to our estimate of 2.5 * 106 cal.
S E C T I O N 1 . 1 4 Density and Specific Gravity 33
PROBLEM 1.22
The highest land temperature ever recorded was 136 °F in Al Aziziyah, Libya, on September 13, 1922. What is this temperature on the kelvin scale?
PROBLEM 1.23
The patient in the photo in the Chemistry in Action on page 31 has a temperature of 39 °C. What is the body temperature of the patient in °F?
PROBLEM 1.24
Assuming that Coca-Cola has the same specific heat as water, how much energy in calories is removed when 350 g of Coca-Cola (about the contents of one 12 oz can) is cooled from room temperature (25 °C) to refrigerator temperature (3 °C)?
PROBLEM 1.25
What is the specific heat of aluminum if it takes 161 cal (674 J) to raise the tempera- ture of a 75 g aluminum bar by 10.0 °C?