Entropía

An energy transformation occurs whenever a chemical change occurs (see Figure 4.6). If energy is absorbed during the change, the products will have more chemical potential energy than the reactants. Conversely, if energy is given off in a chemical change, the products will have less chemical potential energy than the reactants. Water can be decomposed in an elec-trolytic cell by absorbing electrical energy. The products, hydrogen and oxygen, have a greater chemical potential energy level than that of water (see Figure 4.6a). This potential energy is released in the form of heat and light when the hydrogen and oxygen are burned to form water again (see Figure 4.6b). Thus, energy can be changed from one form to another or from one substance to another and, therefore, is not lost.

The energy changes occurring in many systems have been thoroughly studied. No system has ever been found to acquire energy except at the expense of energy possessed by anoth-er system. This is the law of conservation of energy: Energy can be neither created nor destroyed, though it can be transformed from one form to another.

4.5 heAT: QuAnTiTATive MeAsureMenT

Calculate the amount of heat lost or gained in a given system.

The SI-derived unit for energy is the joule (pronounced jool, rhyming with tool, and abbrevi-ated J). Another unit for heat energy, which has been used for many years, is the calorie (ab-breviated cal). The relationship between joules and calories is

4.184 J = 1 cal (exactly)

To give you some idea of the magnitude of these heat units, 4.184 joules, or 1 calorie, is the quantity of heat energy required to change the temperature of 1 gram of water by 1°C, usually measured from 14.5°C to 15.5°C.

Since joules and calories are rather small units, kilojoules (kJ) and kiloca lories (kcal) are used to express heat energy in many chemical processes. The kilocalorie is also known as the nutritional Calorie (spelled with a capital C and abbreviated Cal). In this book, heat energy will be expressed in joules.

The difference in the meanings of the terms heat and temperature can be seen by this example: Visualize two beakers, A and B. Beaker A contains 100 g of water at 20°C, and bea-ker B contains 200 g of water also at 20°C. The beabea-kers are heated until the temperature of the water in each reaches 30°C. The temperature of the water in the beakers was raised by exactly the same amount, 10°C. But twice as much heat (8368 J) was required to raise the temperature of the water in beaker B as was required in beaker A (4184 J).

key terms joule calorie specific heat

learning ObjeCtive

1 kJ = 1000 J

1 kcal = 1000 cal = 1 cal

4.5 • Heat: Quantitative Measurement 71

The greek letter D is used to mean “change in.”

Lowercase t is used for Celsius and upper case T for Kelvin.

Table 4.3 Specific Heat of Selected Substances

Substance Specific heat (J/g°C) Specific heat (cal/g°C)

Water 4.184 1.000

Ethyl alcohol 2.138 0.511

Ice 2.059 0.492

Aluminum 0.900 0.215

Iron 0.473 0.113

Copper 0.385 0.0921

Gold 0.131 0.0312

Lead 0.128 0.0305

In the middle of the eighteenth century, Joseph Black (1728–1799), a Scottish chemist, was experimenting with the heating of elements. He heated and cooled equal masses of iron and lead through the same temperature range. Black noted that much more heat was needed for the iron than for the lead. He had discovered a fundamental property of matter—namely, that every substance has a characteristic heat capacity. Heat capacities may be compared in terms of specific heats. The specific heat of a substance is the quantity of heat (lost or gained) required to change the temperature of 1 g of that substance by 1°C. It follows then that the specific heat of liquid water is 4.184 J>g°C (or 1.000 cal>g°C). The specific heat of water is high compared with that of most substances. Aluminum and copper, for example, have specific heats of 0.900 J>g°C and 0.385 J>g°C, respectively (see ^{Table 4.3}). The relation of mass, specific heat, temperature change (t), and quantity of heat lost or gained by a system is expressed by this general equation:

a mass of

substance b aspecific heat

of substance b(Dt)  heat

Thus the amount of heat needed to raise the temperature of 200. g of water by 10.0°C can be calculated as follows:

(200. g )a4.184 J

g° C b(10.0 °C ) = 8.37 * 10³ J e x a m p l e 4 . 4

Calculate the specific heat of a solid in J>g°C and cal>g°C if 1638 J raises the tempera-ture of 125 g of the solid from 25.0°C to 52.6°C.

SOLUTION

Read • Knowns: 125 g of the solid

t = 52.6 - 25.0 = 27.6°C Heat = 1638 J

Solving for: specific heat of the solid plan • Use the equation

(mass)(specific heat)(t) = heat solving for specific heat

specific heat = heat g * t Calculate • specific heat = 1638 J

125 g * 27.6°C = 0.475 J>g°C Convert joules to calories using 1.00 cal>4.184 J

specific heat = a0.475 J g°C b a

1.000 cal

4.184 J b = 0.114 cal>g°C Check • Note that the units in the answer agree with the units for specific heat.

Mass is in grams, specific heat is in cal/g°C or J/g°C and Dt is in °C.

ENHANCED EXAMPLE

72 chapter 4 • Properties of Matter

4.6 energy in the real World

Define a hydrocarbon compound and explain its role in the world’s energy supply.

Coal, petroleum, natural gas, and woody plants provide us with a vast resource of energy, all of which is derived from the sun. Plants use the process of photosynthesis to store the sun’s energy, and we harvest that energy by burning the plants or using the decay products of the plants. These decay products have been converted over millions of years to fossil fuels. As the plants died and decayed, natural processes changed them into petroleum deposits that we now use in the forms of gasoline and natural gas.

Learning objective

P r a c t i c e 4 . 3

Calculate the quantity of energy needed to heat 8.0 g of water from 42.0°C to 45.0°C.

P r a c t i c e 4 . 4

A 110.0-g sample of metal at 55.5°C raises the temperature of 150.0 g of water from 23.0°C to 25.5°C. Determine the specific heat of the metal in J>g°C.

e x a m P l e 4 . 5

A sample of a metal with a mass of 212 g is heated to 125.0°C and then dropped into 375 g water at 24.0°C. If the final temperature of the water is 34.2°C, what is the specific heat of the metal? (Assume no heat losses to the surroundings.)

SoLUtion

read • Knowns: mass (metal) = 212 g mass (water) = 375 g

t (water) = 34.2 - 24.0 = 10.2°C

t (metal) = 125.0 - 34.2 = 90.8°C Solving for: specific heat of the metal

Plan • When the metal enters the water, it begins to cool, losing heat to the water.

At the same time, the temperature of the water rises. This continues until the temperature of the metal and the water is equal (34.2°C), after which no net flow of heat occurs.

Heat lost or gained by a system is given by

(mass)(specific heat)(t) = energy change (heat) heat gained by the water  heat lost by the metal

heat gained by the water = (mass water)(specific heat)(t)

= (375 g )a4.184 J

g °C b(10.2°C)= 1.60 * 10⁴ J calculate • heat gained by the water = heat lost by the metal = 1.60 * 10⁴ J

(mass metal)(specific heat metal)(t) = 1.60 * 10⁴ J (212 g)(specific heat metal)(90.8°C) = 1.60 * 10⁴ J

To determine the specific heat of the metal, we rearrange the equation (mass)(specific heat)(t) = heat

solving for the specific heat of the metal specific heat metal = 1.60 * 10⁴ J

(212 g)(90.8°C) = 0.831 J>g°C check • Note the units in the answer agree with units for specific heat.

4.6 • energy in the real World 73