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Heat Problems
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There are a few problems that deal with
heat:
– Heat as enthalpy
– Specific heat
– Heat of fusion
– Heat of vaporization
Enthalpy
Enthalpy is the total energy of a system
 It is represented by a H.
 If the pressure remains constant, the
enthalpy increase of a sample of matter
equals the energy as heat that is received.
 This remains true even during a chemical
reaction or a change of state occurs.
 It also indicates the total kinetic energy of
the particles in a sample
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So to calculate Enthalpy, we have
to visit molar heat capacity
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Molar heat capacity of a pure substance is the energy
as heat needed to increase the temperature of 1 mol
of a substance by 1K
It is symbolized by C
It has a unit of J/(mol.K)
Formula: q=nCT
Heat = (amount in moles) (molar heat capacity)
(change in temperature)
Note: moles are used so you may have to do some
gram to mole problems.
Table 1 on Page 343 has many Molar Heat Capacities
of Elements and Compounds.
Examples
The molar heat capacity of tungsten is 24.2
J/(mol.K). Calculate the energy as heat
needed to increase the temperature of 0.40
mol of tungsten by 10.0 K.
 Suppose a sample of water increased in
temperature by 3.5 K when the sample
absorbed 856 J of energy as heat. Calculate
the number of moles of water if the molar heat
capacity is 50.5 J /(mol.K). What is the mass
of that water?
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Molar Enthalpy Change
Since enthalpy “equals the energy as heat that is
received”, we can rewrite the heat equation to:
 H = nCT
 Enthalpy change = (moles)(molar heat capacity)
(change in temperature)
 The enthalpy change for one mole of a pure
substance is known as the molar enthalpy change.
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Examples
The molar heat capacity of Al(s) is 24.2
J/K.mol. Calculate the molar enthalpy
change when Al(s) is cooled from 128.5 C
to 22.6 C.
 Lead has a molar heat capacity of 26.4
J/K.mol. What molar enthalpy change
occurs when lead is cooled from 302 C to
275 C?
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Molar Heat Capacity is related to
Specific Heat
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Specific heat is the energy as heat needed to raise the
temperature of one gram of substance by one Kelvin.
Symbolized by cp
 This is why an iron gets hot faster than a pot of water
 There are two equations that involve specific heat
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1) Molar Mass (g/mol) x cp (J/K.g) = C (J/K.mol)
2) q = m x cp x T
Example Problem
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If the molar heat capacity of nitrogen (N2) is
29.1 (J/K.mol), what is nitrogen’s specific heat?
Calculate the specific heat of a substance
if a 35 g sample absorbs 48 J as the
temperature is raised from 293 K to 313 K
 cp= 48 J/[35gx(313K-293K)]
= 0.069 J/gK
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Two more heat equations
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Heat of fusion (Lf)
– Heat involved in melting or freezing a substance
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Heat of vaporization (Lv)
– Heat involved in evaporation or condensing a substance
As a phase change occurs, all energy is being used
for the change so the temperature will not change.
 So, during a phase change, heat equals the mass
(m) times a value (L) for that material (q=m x L)
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Example Problem
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If the heat of fusion of water is 80 cal/g,
what is the amount of heat energy
required to change 15.0 grams of ice at
0C to 15.0 grams of water at 0C?
How do they all fit together?
This is Figure 17 from
page 394 from the book
As heat is added,
temperature increases
linearly (q=nCT)
Except, when a phase
change occurs, because
all the heat is going into
changing the phase
(q=m L, note no T)
Try Practice Problems
2-4 on page 61
1.
2.
3.
0.385 J/gK
329 K
3.6 kJ