Energy Measuring Heat Calorimetry Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Heat Equation The heat equation. q = m x c x T Heat(q) = g x °C x cal = cal g°C The amount of heat lost or gained by a substance is calculated from the Mass of substance (g). Temperature change (T). Specific heat of the substance (cal/g°C).

Specific Heat 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/gC. In the metric system heat has units of cal/gC. Water has a specific heat of 1 cal/gC

Examples of Specific Heats Table 3.7 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Using Specific Heat A layer of copper on a pan has a mass of 135 g. How much heat in calories will raise the temperature of the copper from 26°C to 328°C if the specific heat of copper is 0.092 cal/g°C? The temperature change is 328°C - 26°C = 302°C. heat (cal) = m x T x c(Cu) 135 g x 302°C x 0.092 cal g°C = 3,751 cal or 3.75 kcal

Learning Check How many calories (kcal) are needed to raise the temperature of 325 g of water from 15.0°C to 77.0°C? 1) 20.15 kcal 2) 77.7 kcal 3) 84.3 kcal

Solution How many kilocalories are needed to raise the temperature of 325 g of water from 15.5°C to 77.5°C? 1) 20.15 kcal 77.0°C – 15.0°C = 62.0°C 325 g x 62.0°C x 1.00 cal x 1 kcal g °C 1000 cal = 20.15 kcal

Learning Check What is the specific heat of a metal if 24.8 g absorbs 66 cals of energy and the temperature rises from 20.2C to 24.5C?

Solution What is the specific heat of a metal if 24.8 g absorbs 66 cals of energy and the temperature rises from 20.2C to 24.5C? Equation: q = m c ΔT Given: 24.8 g, 66 cal, 20.2C to 24.5C Need: cal/gC Plan: c (specific heat) = Heat/gC ΔT = 24.5C – 20.2C = 4.3 C Rearrange Equation: c = heat (q) (mass)(T) Set Up: 66 cal = 0.62 cal/gC (24.8 g)(4.3C)

Calculating Mass Aluminum is used to make kitchen utensils. What is the mass of an aluminum spatula if 777 cal of heat raise its temperature from 20.0°C to 45.0°C. SHAl = 0. 22 cal/g°C? Given: 777 cal, 20.0°C to 45.0°C ΔT = 25.0°C Plan: Solve heat equation for mass m = heat ΔT x SH Set Up: 777 cal = 141 g Al 25.0°C x 0.22 cal/g°C

Transferring Heat Energy Flows from a warmer object to a colder object. Provides kinetic energy for the colder object. Heat lost by the warmer object is equal to the heat energy gained by the colder object.

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

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 1.00 cal/g°C = 210 cal Heat loss (-q) by tin = -210 cal SH tin = -210 cal = 0.053 cal/g°C (50.0 g)(-80.0°C)

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 cal 1 Cal = 1 kcal How Calories compare to Joules (another unit of hear): 1 Cal = 4184 J 1 Cal = 4.184 kJ

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

Energy Values for Some Foods Table 3.9 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Learning Check 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)

Solution 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