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

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 joules of energy to raise the temperature of one gram of water 1 o C. –Kilojoule (kJ) = 1000 joules 1 cal = J 1 kcal = kJ

Section 10.2 The Flow of Energy 1. How many calories are there in J? 2. How many joules of energy are there in a candy bar that contains 350. Calories? J ÷ J/cal = 23.9 cal Remember, a Calorie is really 1000 calories Cal x 1000 cal/Cal= cal cal x J/cal = 1.46x10 6 J

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.

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.

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.

Section 10.2 The Flow of Energy B. Measuring Energy Changes To calculate the energy required for a reaction: Q = s  m   t

Section 10.2 The Flow of Energy 1. How much energy does it take to heat mL of water from 20.0 o C to 35.0 o C. Q = smΔT s = J/g o C m = mL = g ΔT = 15.0 o C Q = (4.184 J/g o C)(500.0 g)(15.0 o C) = J

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 = 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 o C = 23.5 o C This is the change in temperature, not the final temperature. x = 1.52 o C

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 g sample from 30.1 o to 75.0 o C. Q = smΔT s = x m = g ΔT = 75.0 – 30.1 = 44.9 o C 225 J = (x)(300.0 g)(44.9 o C) Q = 225 J x = J/g o C

Section 10.2 The Flow of Energy 4. If calories of heat are added to g of water, how much will the temperature increase? Q = smΔT s = 1.00 cal/g o C m = g ΔT = x cal = (1.00 cal/g o C)(100.0 g)(x) Q = cal x = 4.50 o C

Section 10.2 The Flow of Energy B. Measuring Energy Changes To calculate the energy required for a reaction:

Section 10.2 The Flow of Energy 1. How much energy does it take to heat mL of water from 20.0 o C to 35.0 o C.

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?

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 g sample from 30.1 o to 75.0 o C.

Section 10.2 The Flow of Energy 4. If calories of heat are added to g of water, how much will the temperature increase?