Heat Exchange During Physical Changes - Parkway C-2

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Heat Exchange During Physical
Changes
Heating Curve: graph of temperature vs. time (or heat added)
As a substance is heated, (imagine a beaker of water in this
case) its temperature only rises when no phase change is
occurring. Why is this?
Answer: the phase change requires (or releases) energy.
Example: boiling water is endothermic, so any heat added
during the boiling process is used for boiling, not raising the
temperature.
The same can be said for melting.
The opposite can be said for condensing and freezing, because
they are exothermic.
Therefore, we can say that every sample of water that contains
at least a little liquid should be between ____ and ______oC.
Question: Look at the parts of the curve in which the
temperature is changing (3 separate places). What causes the
differences in the slopes?
Calculating heat exchanged during a phase change:
Heat of fusion (Hfus.):
Amount of heat released as liquid freezes to solid or heat absorbed as solid melts
to liquid
q = m x Hfus.
try:
How much heat is absorbed by 25.6 g of
ice as it melts?
q = m x Hfus. = 25.6 g x 334 J/g = 8550 J
Heat of vaporization (Hvap.)
heat released as gas condenses to liquid
or
heat absorbed as liquid vaporizes to gas
q = m Hvap.
Try these problems:
How much heat is released as 15.6 g of ethyl alcohol (ethanol)
condenses?
13700 J
The cooling tower in a nuclear power plant utilizes the process of water
vaporization to cool hot pipes from inside the plant. If the heat that
must be released is 4.56 megajoules per hour, how many liters of water
(per hour) must be vaporized to keep the pipes cool?
2.02 L
How many mL of sweat (assume sweat to be the same as water) would
cool your body down by 25000 joules?
11 mL
Combining specific heat with phase change:
How much heat would be required to raise a 200. mL
sample of water from 50.0 oC to 125oC?
This problem actually has 3 steps:
1. Heat liquid water from 50-100oC
No phase change is occurring. We’re just going up in
temp by 50oC, so we use q = c x m x ∆T
q = (4.18 J/g*oC)200. g(50oC) = 41800 J
2. Boil the water.
q = m x Hvap. = 200. g (2260 J/g) = 452000 J
3. Heat the water vapor from 100 – 125oC
Once again, no phase change, so…
q = c x m x ∆T = (1.87 J/g*oC)200 g(25oC) = 9350 J
Finally, add up all 3 answers:
41800 J + 452000 J + 9350 J = 503150 J
Step 3
Step 1
Step 2
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