How exactly is "heat" measured experimentally?
Calorimetry
Calorimetry = the science of measuring heat.
Calorimetry is based on observing the temperature change (ΔT) when a substance absorbs or gives off heat (q).
Some substances require a lot of heat energy (q) to raise their temperature by 1°C or 1K.
➞ these substances make good coolants (ex: H2O).
Other substances don't require very much heat energy (q) to raise their temperature by 1°C or 1K.
Metals are a good example (ex: Cu, Fe).
➞ a metal pan on a stove gets hotter much faster than the H2O in the pan!
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Heat Capacity (C)
Heat Capacity = the amount of energy (as heat) required to raise the temperature of a substance by 1°C or 1K.
The higher the "heat capacity," the smaller the change in temperature for a given amount of absorbed heat.
➞ H2O has a high heat capacity (good coolant).
➞ Metals have low heat capacities.
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** The term "Heat Capacity" does not mention the amount of the substance being considered.
** If the heat capacity is given per gram of substance, it is called the "specific heat capacity."
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Specific Heat Capacity (s)
Specific Heat Capacity = the energy (as heat) required to raise the temperature of 1g of a substance by 1°C or 1K.
Here are some common specific heat (s) values:
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The Calorimetry Equation
Actual measurements of heat gain and/or heat loss are performed using a device called a constant-pressure calorimeter.
Quite often, in your chemistry class or course, you'll hear the term "coffee-cup calorimeter." This is just a type of constant-pressure calorimeter.
Two Useful Calorimetry Equations:
1. Heat Gained: q = m . s . ΔT
2. Heat Lost: q = m . s . ΔT
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ex: The specific heat capacity of silver is 0.24 J/g.°C
a. Calculate the energy required to raise the temperature of 150.0 g of Ag from 273K to 298K.
q = m . s . ΔT , where ΔT = Tfinal - Tinitial
q = (150.0g) . (0.24 J/g.°C) . (298K - 273K)
q = 900. J
b. Calculate the energy required to raise the temperature of 1.0 mol of Ag by 1°C.
... Well, we know the specific heat capacity (s) is 0.24 J/g.°C.
So we just need to convert the grams (g) part of the units to moles (mol):
c. It takes 1.25 kJ of energy to heat a sample of pure silver from 12.0°C to 15.2°C. Calculate the mass of the silver sample.
(1.25 kJ / 1) . (1000 J / 1 kJ) = 1250 J
q = m . s . ΔT
1250 J = (m) . (0.24 J/g.°C) . (15.2°C - 12.0°C)
1250 J = (m) . (0.24 J/g.°C) . (+ 3.2°C)
m = 1.6 x 103 g Ag
Here's the same math in an easy-to-follow, handwritten format:
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Coffee-Cup Calorimetry Problems
Unlike the above example, where one substance is gaining or losing heat, often 2 substances at different initial temperatures are combined.
In situations like this, the heat lost by the hotter substance equals the heat gained by the cooler substance...
We'll do some coffee-cup calorimeter practice problems in the next video...
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In my next video blog entry from SECTION 6 - Thermochemistry,
We'll start with some coffee-cup calorimetry examples, and then discuss the change in enthalpy, ΔH...