SME
University Physics I: Mechanics · Problem type
The entropy gained heating water
Use this for “Find the entropy change of 1 kg of water heated from 20 °C to 80 °C.”
Chart
Shaded: ∫ from 293.15 to 353.15 of f(T) dT = 779.5
This is an example. Type your own values to start a new problem.
ΔS (
)
=
m (
)
×
c (
)
×
ln
T₂ (K)
T₁ (K)
t₁
Starting temperature
example · °C
t₂
Final temperature
example · °C
Q
Heat taken in
worked out (not typed) · J
Formulas
T₁
=
t₁
+
273.15
293.15 = 20 + 273.15
T₂
=
t₂
+
273.15
353.15 = 80 + 273.15
ΔS
=
m
×
c
×
ln(
T₂
T₁
)
+779.5
=
1
×
4,186
×
ln(
353.15
293.15
)
Q
=
m
×
c
×
(
T₂
−
T₁
)
+251,160 = 1 × 4,186 × (353.15 − 293.15)
Assumptions
•
The specific heat c stays the same over the range and nothing boils or freezes; water: c = 4186 J/(kg·K).
•
dS = dQ ÷ T with dQ = mc dT, added up from T₁ to T₂, so the log takes kelvins: T = t + 273.15.
•
Warming gives ΔS above 0 and cooling below 0; the heat itself is Q = mc(T₂ − T₁).
Step-by-step
WE KNOW
m = 1 kg, c = 4,186 J/(kg·K), t₁ = +20 °C, t₂ = +80 °C
FIND
starting absolute temperature (T₁), final absolute temperature (T₂), entropy change (ΔS), heat taken in (Q)
Step 1 · Find starting absolute temperature (T₁)
Use T₁ = t₁ + 273.15
Kelvins count from absolute zero, which is 273.15 degrees below 0 °C, so add 273.15.
T₁
=
20
+
273.15
T₁
=
293.15
K
Step 2 · Find final absolute temperature (T₂)
Use T₂ = t₂ + 273.15
Kelvins count from absolute zero, which is 273.15 degrees below 0 °C, so add 273.15.
T₂
=
80
+
273.15
T₂
=
353.15
K
Step 3 · Find entropy change (ΔS)
Use ΔS = m × c × ln(T₂ ÷ T₁)
Add
up
dQ
T
=
mc
dT
T
from
T₁
to
T₂
:
mc
comes
out
front,
and
∫
dT
T
is
ln(
T₂
T₁
).
ΔS
=
1
×
4,186
×
ln(
353.15
293.15
)
ΔS = 4,186 × ln(1.2047)
ΔS = 4,186 × 0.18623
ΔS = +779.5 J/K
Step 4 · Find heat taken in (Q)
Use Q = m × c × (T₂ − T₁)
The
heat
to
change
the
temperature
by
T₂
−
T₁
,
the
same
in
kelvins
as
in
degrees
Celsius.
Q
=
1
×
4,186
×
(353.15
−
293.15)
Q
=
4,186
×
60
Q
=
+251,160
J
Check
Put the numbers back in. Both sides should match.
293.15 = 20 + 273.15 ✓
353.15 = 80 + 273.15 ✓
+779.4687
=
1
×
4,186
×
ln(
353.15
293.15
)
✓
+251,160 = 1 × 4,186 × (353.15 − 293.15) ✓
Topic check
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