SME
Engineering Thermodynamics · Problem type
A hot block cooled in a lake
Use this for “A 5 kg copper block (c = 0.385 kJ/(kg·K)) at 500 K is dropped into a lake at 290 K. Find the entropy generated.”
This is an example. Type your own values to start a new problem.
Q (kJ)
=
m (kg)
×
c (kJ/(kg·K))
×
(
T₁ (K)
−
T₀ (K)
)
ΔSblock (kJ/K)
=
m (kg)
×
c (kJ/(kg·K))
×
ln(
T₀ (K)
÷
T₁ (K)
)
Sgen (kJ/K)
=
ΔSblock (kJ/K)
+
Q (kJ)
÷
T₀ (K)
ΔSₗₐₖₑ
Entropy change of the lake
calculated · kJ/K
Formulas
Q
=
m
×
c
×
(
T₁
−
T₀
)
292.5 = 2 × 0.45 × (623 − 298)
ΔSblock
=
m
×
c
×
ln(
T₀
T₁
)
−0.6637
=
2
×
0.45
×
ln(
298
623
)
ΔSₗₐₖₑ
=
Q
T₀
0.9815
=
292.5
298
Sgen = ΔSblock + ΔSₗₐₖₑ
0.3178 = −0.6637 + 0.9815
Assumptions
•
The block is a solid with constant specific heat c; it cools until it reaches the lake’s temperature.
•
The lake is so large that its temperature T₀ stays the same: it takes heat Q at T₀.
•
Temperatures in kelvins; no heat is lost elsewhere.
Step-by-step
WE KNOW
m = 2 kg, c = 0.45 kJ/(kg·K), T₁ = 623 K, T₀ = 298 K
FIND
heat given to the lake (Q), entropy change of the block (ΔSblock), entropy change of the lake (ΔSₗₐₖₑ), entropy generated (Sgen)
Step 1 · Find heat given to the lake (Q)
Use Q = m × c × (T₁ − T₀)
The
heat
the
block
gives
up
as
it
cools
from
T₁
to
T₀
all
goes
into
the
lake.
Q
=
2
×
0.45
×
(623
−
298)
Q
=
2
×
0.45
×
325
Q
=
0.9
×
325
Q
=
292.5
kJ
Step 2 · Find entropy change of the block (ΔSblock)
Use ΔSblock = m × c × ln(T₀ ÷ T₁)
Add
up
dQ
T
as
the
block
cools:
∫mc
dT
T
is
mc
ln(
T₀
T₁
),
negative
as
it
cools.
ΔSblock
=
2
×
0.45
×
ln(
298
623
)
ΔSblock = 2 × 0.45 × ln(0.4783)
ΔSblock = 2 × 0.45 × (−0.7375)
ΔSblock = 0.9 × (−0.7375)
ΔSblock = −0.6637 kJ/K
Step 3 · Find entropy change of the lake (ΔSₗₐₖₑ)
Use ΔSₗₐₖₑ = Q ÷ T₀
The
lake
takes
in
Q
at
one
fixed
temperature,
so
its
entropy
rises
by
Q
T₀
.
ΔSₗₐₖₑ
=
292.5
298
ΔSₗₐₖₑ = 0.9815 kJ/K
Step 4 · Find entropy generated (Sgen)
Use Sgen = ΔSblock + ΔSₗₐₖₑ
Block and lake together are isolated: the entropy generated is the sum of their changes.
Sgen = −0.6637 + 0.9815
Sgen = 0.3178 kJ/K
Check
Put the numbers back in. Both sides should match.
292.5 = 2 × 0.45 × (623 − 298) ✓
−0.6637
=
2
×
0.45
×
ln(
298
623
)
✓
0.9815
=
292.5
298
✓
0.3178 = −0.6637 + 0.9815 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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