1 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Chapter 3 Matter and Energy 3.5 Specific Heat.

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Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

Chapter 3 Matter and Energy

3.5

Specific Heat

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Specific heat

• Is different for different substances.

• Is the amount of heat that raises the temperature of 1 g of a substance by 1°C.

• In the SI system has units of J/gC.

• In the metric system has units of cal/gC.

Specific Heat

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Examples of Specific Heats

Table 3.7

Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Learning Check

What is the specific heat of a metal if 24.8 g

absorbs 275 J of energy and the temperature

rises from 20.2C to 24.5C?

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Solution

What is the specific heat of a metal if 24.8 g absorbs275 J of energy and the temperature rises from 20.2C to 24.5C?

Given: 24.8 g, 275 J, 20.2C to 24.5C Need: J/gC

Plan: SH = Heat/gC

ΔT = 24.5C – 20.2C = 4.3 C SH Equation: SH = heat (q) (mass)(T)

Set Up: 275 J = 2.6 J/gC

(24.8 g)(4.3C)

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Rearranging the specific heat expression givesthe heat equation.

Heat(q) = g x °C x J = J g°C

The amount of heat lost or gained by a substanceis calculated from the• Mass of substance (g).• Temperature change (T).• Specific heat of the substance (J/g°C).

Heat Equation

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A layer of copper on a pan has a mass of 135 g. How much heat in joules will raise the temperature of the copper from 26°C to 328°C if the specific heat of copper is 0.385 J/g°C?

The temperature change is 328°C - 26°C = 302°C.

heat (cal) = g x T x SH(Cu)

135 g x 302°C x 0.385 J g°C

= 15 700 J or 1.57 x 104 J

Using Specific Heat

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How many kilojoules are needed to raise the temperature of 325 g of water from 15.0°C to 77.0°C?

1) 20.4 kJ

2) 77.7 kJ

3) 84.3 kJ

Learning Check

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How many kilojoules are needed to raise the temperature of 325 g of water from 15.5°C to 77.5°C?

3) 84.3 kJ

77.0°C – 15.0°C = 62.0°C

325 g x 62.0°C x 4.184 J x 1 kJ

g °C 1000 J

= 84.3 kJ

Solution

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Calculating Mass

Aluminum is used to make kitchen utensils. What is the mass of an aluminum spatula if 3.25 kJ of heat raise its temperature from 20.0°C to 45.0°C. SHAl = 0. 897 J/g°C?

Given: 3.25 kJ (3250 J), 20.0°C to 45.0°C

ΔT = 25.0°C

Plan: Solve heat equation for mass

m = heat

ΔT x SH

Set Up: 3250 J/g°C = 145 g Al

25.0°C x 0.897 J

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Transferring Heat Energy

Heat energy • Flows from a warmer object to a colder object. • Provides kinetic energy for the colder object.• Lost by the warmer object is equal to the heat

energy gained by the colder object.

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Calorimeters and Heat Transfer

A calorimeter • Is used to measure heat transfer.• Can be made with a coffee cup,

water, and a thermometer.• Indicates the heat lost by a

sample and gained by water.

Heat lost (-q) = Heat (q) gainedCopyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

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Measuring Heat Changes

A 50.0-g sample of tin is heated to 99.8°C and dropped into 50.0 g water at 15.6°C. If the final temperature is 19.8°C, what is the specific heat of tin?

Heat gain (q) by water

= 50.0 g x 4.2°C x 4.184 J/g°C = 880 J

Heat loss (-q) by tin = -880 J

SH tin = -880 J = 0.22 J/g°C

(50.0 g)(-80.0°C)

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Energy and Nutrition

On food labels, energy is shown as the nutritional Calorie, written with a capital C. In countries other than the U.S., energy is shown in kilojoules (kJ).

1 Cal = 1000 cal1 Cal = 1 kcal1 Cal = 4184 J

1 Cal = 4.184 kJ

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Caloric Food Values

The caloric or energy

values for 1 g of a food

is given in• kJ or • kcal (Cal)

Table 3.8

Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

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Energy Values for Some Foods

Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Table 3.9

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Energy Requirements

The amount of energy needed eachday depends on• Age• Sex• Physical activity

Table 3.11

Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

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A cup of whole milk contains 12 g carbohydrate, 9.0 g fat, and 9.0 g protein. How many kcal (Cal) does a cup of milk contain?

1) 48 kcal (48 Cal)

2) 81 kcal (81 Cal)

3) 165 kcal (165 Cal)

Learning Check

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3) 165 kcal

12 g carbohydrate x 4 kcal/g = 48 kcal

9.0 g fat x 9 kcal/g = 81 kcal

9.0 g protein x 4 kcal/g = 36 kcal

Total kcal = 165 kcal

= 165 Cal

Solution