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Published byGeorgina Sutton Modified over 8 years ago
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Section 10.2 The Flow of Energy 1.To convert between different energy units. 2.To understand the concept of heat capacity. 3.To solve problems using heat transfer equations. Objectives
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Section 10.2 The Flow of Energy B. Measuring Energy Changes The common energy units for heat are the calorie and the joule. –calorie (cal) – the amount of energy (heat) required to raise the temperature of one gram of water 1 o C. –Kilocalorie (kcal or Cal) = 1000 calories (or 1 Calorie) –These are the ‘Calories’ listed on food –joule (J) – it takes 4.184 joules of energy to raise the temperature of one gram of water 1 o C. –Kilojoule (kJ) = 1000 joules 1 cal = 4.184 J 1 kcal = 4.184 kJ
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Section 10.2 The Flow of Energy 1. How many calories are there in 100.0 J? 2. How many joules of energy are there in a candy bar that contains 350. Calories? 100.0 J ÷ 4.184 J/cal = 23.9 cal Remember, a Calorie is really 1000 calories. 350. Cal x 1000 cal/Cal= 350000 cal 350000 cal x 4.184 J/cal = 1.46x10 6 J
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Section 10.2 The Flow of Energy B. Measuring Energy Changes Factors that affect energy transfers. 1. Mass of sample (m) = proportional to energy. A bigger sample will require more energy to heat up. 2. Size of temperature change (ΔT) = proportional to energy. A bigger temperature change will require more energy.
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Section 10.2 The Flow of Energy B. Measuring Energy Changes Factors that affect energy transfers. 3. Type of material = varies depending on the material. We use a quantity called the specific heat capacity of the material. If you have 10 g samples of each material below, how much will the temperature increase when it absorbs 100 J of energy? 10 g water 10 g aluminum 10 g gold 2.39 o C 11.2 o C 76.9 o C You cannot predict this without testing the material.
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Section 10.2 The Flow of Energy B. Measuring Energy Changes Specific heat capacity is the energy required to change the temperature of one gram of a substance by one Celsius degree. The higher the specific heat capacity, the more energy it takes to increase the temperature of the material.
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Section 10.2 The Flow of Energy B. Measuring Energy Changes To calculate the energy required for a reaction: Q = s m t
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Section 10.2 The Flow of Energy 1. How much energy does it take to heat 500.0 mL of water from 20.0 o C to 35.0 o C. Q = smΔT s = 4.184 J/g o C m = 500.0 mL = 500.0 g ΔT = 15.0 o C Q = (4.184 J/g o C)(500.0 g)(15.0 o C) = 31400 J
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Section 10.2 The Flow of Energy 2. If you transfer 255 J of energy to 40.0 g of water at 22.0 o C, what is the final temperature? Q = smΔT s = 4.184 J/g o C m = 40.0 g ΔT = x 255 J = (4.184 J/g o C)(40.0 g)(x) Q = 255 J 22.0 o C + 1.52 o C = 23.5 o C This is the change in temperature, not the final temperature. x = 1.52 o C
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Section 10.2 The Flow of Energy 3. What is the specific heat capacity of a substance if 225 J of energy will heat a 300.0 g sample from 30.1 o to 75.0 o C. Q = smΔT s = x m = 300.0 g ΔT = 75.0 – 30.1 = 44.9 o C 225 J = (x)(300.0 g)(44.9 o C) Q = 225 J x = 0.0167 J/g o C
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Section 10.2 The Flow of Energy 4. If 450.0 calories of heat are added to 100.0 g of water, how much will the temperature increase? Q = smΔT s = 1.00 cal/g o C m = 100.0 g ΔT = x 450.0 cal = (1.00 cal/g o C)(100.0 g)(x) Q = 450.0 cal x = 4.50 o C
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Section 10.2 The Flow of Energy B. Measuring Energy Changes To calculate the energy required for a reaction:
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Section 10.2 The Flow of Energy 1. How much energy does it take to heat 500.0 mL of water from 20.0 o C to 35.0 o C.
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Section 10.2 The Flow of Energy 2. If you transfer 255 J of energy to 40.0 g of water at 22.0 o C, what is the final temperature?
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Section 10.2 The Flow of Energy 3. What is the specific heat capacity of a substance if 225 J of energy will heat a 300.0 g sample from 30.1 o to 75.0 o C.
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Section 10.2 The Flow of Energy 4. If 450.0 calories of heat are added to 100.0 g of water, how much will the temperature increase?
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