Water changes its phase from liquid to vapour at a constant temperature of 11 degree C, consuming 4000 KJ of heat. The total entropy changed of water in KJ/K is.
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Phase changes
When a system receives an amount of energy q at a constant temperature, T, the entropy increase DS is defined by the following equation. Hence, the magnitude of ΔS for a reversible process such as a phase change is calculated
ΔS=qrevT(18.4.1)
(18.4.1) ΔS=
q
r
e
v
T
with the temperature in Kelvin. Since entropy changes are much smaller than enthalpy changes, they are usually reported in J K–1 mol–1.
Examples of reversible processes are
Boiling: As temperature is constant, ΔS = ΔHvap/T
Melting: As temperature is constant, ΔS = ΔHfus/T
For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, qsurr
q
s
u
r
r
is a good approximation of qrev
q
r
e
v
, and the second law may be stated as the following:
ΔSuniv=ΔSsys+ΔSsurr=ΔSsys+qsurrT(18.4.2)
(18.4.2)
Δ
S
univ
=
Δ
S
sys
+
Δ
S
surr
=
Δ
S
sys
+
q
surr
T
We may use this equation to predict the spontaneity of a process as illustrated in Example 18.4.1
18.4.
1
.
Please give me thanks or brainlist answer
When a system receives an amount of energy q at a constant temperature, T, the entropy increase DS is defined by the following equation. Hence, the magnitude of ΔS for a reversible process such as a phase change is calculated
ΔS=qrevT(18.4.1)
(18.4.1) ΔS=
q
r
e
v
T
with the temperature in Kelvin. Since entropy changes are much smaller than enthalpy changes, they are usually reported in J K–1 mol–1.
Examples of reversible processes are
Boiling: As temperature is constant, ΔS = ΔHvap/T
Melting: As temperature is constant, ΔS = ΔHfus/T
For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, qsurr
q
s
u
r
r
is a good approximation of qrev
q
r
e
v
, and the second law may be stated as the following:
ΔSuniv=ΔSsys+ΔSsurr=ΔSsys+qsurrT(18.4.2)
(18.4.2)
Δ
S
univ
=
Δ
S
sys
+
Δ
S
surr
=
Δ
S
sys
+
q
surr
T
We may use this equation to predict the spontaneity of a process as illustrated in Example 18.4.1
18.4.
1
.
Please give me thanks or brainlist answer
Answered by
1
Answer:
Explanation:
When a system receives an amount of energy q at a constant temperature, T, the entropy increase DS is defined by the following equation. Hence, the magnitude of ΔS for a reversible process such as a phase change is calculated
ΔS=qrevT(18.4.1)
with the temperature in Kelvin. Since entropy changes are much smaller than enthalpy changes, they are usually reported in J K–1 mol–1.
Examples of reversible processes are
Boiling: As temperature is constant, ΔS = ΔHvap/T
Melting: As temperature is constant, ΔS = ΔHfus/T
For many realistic applications, the surroundings are vast in
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