Chapter 12 Notes
Heat, Temperature and Thermal Energy
There are several ways in which we can describe heat.
Temperature is harder to define.
Energy can flow via three means:
We can determine exactly how much heat will be gained or lost by an object if we know the SPECIFIC HEAT of the body. Specific heat is the amount of heat gained or lost by 1 kg of mass during a temperature change of 1 degree C or K. (note that a change of 1 degree Celsius is the same as a change of 1 degree Kelvin). Units of Specific Heat are
For example, the thermal energy Q transferred when 2.00 kg of water, Specific heat C = 4180 cools from 35 to 25 can be calculated as follows :
= 2.00 kg X 4180 X 10
= 8.36 x 104 Joules
You try... Do the practice problems 6-7 in 12.1
This formula does not work when we have temperature changes in matter that span phase-change points. For example, you can't use this simple relationship if you want to know how much heat is gained when ice at -20 is heated to +35 . This would involve the phase change from solid to liquid since ice under normal conditions cannot exist at temperatures above 0 C.
Changes in Phase
When matter is at one of its phase change points, such as melting/freezing or vaporization/condensation, addition or subtraction of thermal energy does NOT result in changes in temperature. Instead, the energy only effects the change in phase. For example, if we have ice at 0 , and continue to add heat, the ice will begin to melt, but it doesn't increase in temperature. This means that the KE of its particles does not increase. Instead, the extra heat is simply allowing the particles to break free of those bonds which held them together as a solid. Once all the ice has melted, it's temperature will still be 0, but now additional heat addition will begin to cause temperature change again. We can find the temperature needed to produce a change from solid to liquid (or vice-versa) for a particular mass if we know the mass, and a physical property of the matter known as Heat of Fusion (Hf) by using the following relationship:
the thermal energy needed to melt 1 kilogram of ice, heat of fusion = 3.35 x 105 J/kg is
Q = 1.00 kg X 3.35 x 105 J/kg
=3.35 x 105 Joules
Finding the energy involved in changing matter from liquid to gas, and vice versa, is handled by the formula
where Hv is the Heat of Vaporization for that kind of matter. Hv for water is 2.26 x 106 J/kg.
You try: do the practice problems 13-15 on page 255
Then do this honker:
Find the total energy required to raise the temperature of 2.00 kg of ice at -20to steam at +120.
You will need the Specific Heats for ice, water and steam (they are ALL DIFFERENT!) and the Heat of Fusion for ice and the Heat of Vaporization for water. These values are all in your text.
This will involve 5 steps:
1. Find the heat required to convert ice from -20 to 0. Remember.. ice can't exist beyond 0 under normal conditions. You can't use the Specific Heat formula to cross any of these phase change barriers, but you can use it here to make the temperature change indicated. Note that the specific heat formula from above is the ONLY formula we have been working with that deals with temperature change!
2. Find the heat required to convert the ice at 0 to water at 0. This is the phase change part. There is NO CHANGE IN TEMPERATURE while changing phase.. so don't be usin' any formula that has in it!
You want the Heat of Fusion formula here.
3. Find the heat required to convert the water from 0 to 100.00. Water can't exist beyond 100 degrees under normal conditions, so you can't go "all the way" to 120 deg. in this step. There IS a temperature change involved here.. so what formula uses ? The specific heat formula.
4. Now that the water is at 100 degrees (one of those nasty old phase-change points), additional heat won't change the temp,, just convert it to steam.. so, Find the heat required to convert the water at 100 degrees to steam at 100 degrees. There is no temp change. You need the Heat of Vaporization formula here.
5. After converting the water to steam, the temp is still 100 and we need to raise it's temp to 120. Use the specific heat formula.
Now, add all the values you got for Thermal Energy (Q) in each of the 5 steps together to get the total amount required!
Continued. CLICK HERE
homework: set 1.. ch. 12 questions 1-7 page 295-96
set 2: ch 12 probs 21-30
set 3: ch 12 probs 31-37