Thermochemistry. Energy and Heat Energy = the ability to do work, measured in Joules (J) 1 joule = 1 Newton of force applied to a 1 kg object over the.

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Presentation transcript:

Thermochemistry

Energy and Heat Energy = the ability to do work, measured in Joules (J) 1 joule = 1 Newton of force applied to a 1 kg object over the course of 1 meter. Heat = form of energy (sometimes called thermal energy) that is transferred from hot substances to cold ones. Heat conduction occurs when higher temp. (faster) particles bump into lower temp. (slower) particles, transferring their kinetic energy. play the animation 

Measuring heat exchanged between objects: Ex) How much heat is needed to raise the temperature of 1200 g of water in a pot from 22 o C to 100 o C in order to begin boiling it. q = c x m x ∆T q = heat gained/lost, measured in Joules (J) c = specific heat = resistance to change in temp. J/(g*K) This value is a constant for a given substance. For example, water is always 4.18 J/(g* o C). See chart  m = mass (grams) ∆T = temp. change (K or °C) (question to think about: Why can we use either K or o C for ∆T) substance specific heat (j/g  C) aluminum copper0.385 ethanol (liquid)2.44 ethanol (gas)1.42 gold0.129 graphite0.709 iron0.449 lead0.129 silver0.233 water (gas)1.87 water (liquid)4.18 water (solid)2.06 zinc0.389

Now let’s answer the question: How much heat is needed to raise the temperature of 1200 g of water in a pot from 22 o C to 100. o C in order to begin boiling it. (click play to see explanation)

New Question: A 28.5 g sample of an unknown substance increases in temperature from 42.3  C to 87.7  C upon adding joules of heat. Find the specific heat of the material Steps: (Try each step on your own, in your notes before clicking ahead.) 1.Rearrange the equation: 2.Plug in knowns: 3.solve:

New problem: An unknown metal with a mass of 185 g is taken from an oven, where it had been heated to 350. o C, and is dropped into a 1.0 L container of water. The water rises in temperature from 23 o C to 31 o C. Find the specific heat of the metal. Unlike the previous two problems, here we have two substances (metal and water) that are experiencing temperature changes. We must assume that the heat lost by the metal (its q) would have been gained by the water (the water’s q). Therefore if you solve for q water you also know the q metal Try it. Do you understand… why we found the heat gained by the water, and then plugged it in as the heat lost by the metal? Why the final temperature was 31 o C for both the water and the metal? 1.0 L = 1000 mL = 1000 g (for water)