1 copyright © 2008 by pearson education, inc. publishing as benjamin cummings chapter 3 matter and...
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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
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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
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Energy Values for Some Foods
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Table 3.9
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Energy Requirements
The amount of energy needed eachday depends on• Age• Sex• Physical activity
Table 3.11
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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