SME
Thermal & Statistical Physics
Entropy and ensembles
Use this for “2 kg of water at 90 °C is poured into 3 kg of water at 10 °C in an insulated jug. What is the final temperature, and how much entropy does the mixing make?”
Chart
ΔS = ΔS₂ + ΔS₁
+407.9 + (−371.6) = +36.23
The flows in add and the flows out take away: ΔS = +36.23.
This is an example. Type your own values to start a new problem.
×
T₁ (
)
+
×
T₂ (
)
m₁ (kg)
+
m₂ (kg)
=
Tf (
)
ΔS
=
ΔS₁ (J/K)
+
ΔS₂ (J/K)
=
ΔS (J/K)
c
Specific heat
example
Formulas
Tf
=
m₁
×
T₁
+
m₂
×
T₂
m₁
+
m₂
323.15
=
1 × 353.15 + 1 × 293.15
1 + 1
ΔS₁
=
m₁
×
c
×
ln(
Tf
T₁
)
−371.6
=
1
×
4,186
×
ln(
323.15
353.15
)
ΔS₂
=
m₂
×
c
×
ln(
Tf
T₂
)
+407.9
=
1
×
4,186
×
ln(
323.15
293.15
)
ΔS = ΔS₁ + ΔS₂
+36.23 = −371.6 + 407.9
Assumptions
•
No heat leaks to the room or the jug, and c stays the same over the range (water: 4186 J/(kg·°C)).
•
Water 1 is the hotter. Each part’s ΔS is ∫ dQ ÷ T with dQ = mc dT, so the logs take kelvins.
•
The total ΔS is above 0 whenever T₁ ≠ T₂: mixing is irreversible, the second law.
Step-by-step
WE KNOW
m₁ = 1 kg, T₁ = 353.15 K, m₂ = 1 kg, T₂ = 293.15 K, c = 4,186 J/(kg·°C)
FIND
final temperature (Tf), hot water’s entropy change (ΔS₁), cold water’s entropy change (ΔS₂), total entropy change (ΔS)
Step 1 · Find final temperature (Tf)
Use Tf = (m₁ × T₁ + m₂ × T₂) ÷ (m₁ + m₂)
Heat
out
of
the
hot
water,
m₁
c
(
T₁
−
Tf),
is
heat
into
the
cold,
m₂
c
(Tf
−
T₂
);
c
cancels,
so
Tf
is
the
mass-weighted
average.
Tf
=
1 × 353.15 + 1 × 293.15
1 + 1
Tf
=
646.3
2
Tf = 323.15 K
Step 2 · Find hot water’s entropy change (ΔS₁)
Use ΔS₁ = m₁ × c × ln(Tf ÷ T₁)
Add
up
dQ
T
as
the
hot
water
cools:
∫
m₁
c
dT
T
from
T₁
to
Tf
is
m₁
c
ln(
Tf
T₁
),
below
0.
ΔS₁
=
1
×
4,186
×
ln(
323.15
353.15
)
ΔS₁ = 4,186 × ln(0.91505)
ΔS₁ = 4,186 × (−0.088777)
ΔS₁ = −371.62 J/K
Step 3 · Find cold water’s entropy change (ΔS₂)
Use ΔS₂ = m₂ × c × ln(Tf ÷ T₂)
The
same
sum
for
the
cold
water
warming
from
T₂
to
Tf:
above
0.
ΔS₂
=
1
×
4,186
×
ln(
323.15
293.15
)
ΔS₂ = 4,186 × ln(1.10234)
ΔS₂ = 4,186 × 0.0974352
ΔS₂ = +407.9 J/K
Step 4 · Find total entropy change (ΔS)
Use ΔS = ΔS₁ + ΔS₂
Entropy adds. The cold water gains more than the hot loses: the same heat enters at a lower T.
ΔS = −371.62 + 407.85
ΔS = +36.23 J/K
Check
Put the numbers back in. Both sides should match.
323.15
=
1 × 353.15 + 1 × 293.15
1 + 1
✓
−371.62
=
1
×
4,186
×
ln(
323.15
353.15
)
✓
+407.9
=
1
×
4,186
×
ln(
323.15
293.15
)
✓
+36.23 = −371.62 + 407.9 ✓
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Heat capacities of an ideal gas
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An Einstein solid: multiplicity and entropy
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