SME
University Physics I: Mechanics · Problem type
The final temperature of a mixture
Use this for “A 0.2 kg aluminium block at 90 °C goes into 0.5 kg of water at 20 °C. Find the final temperature.”
Chart
This is an example. Type your own values to start a new problem.
m₁ (
)
×
c₁ (J/(kg·K))
×
(
T₁ (°C)
−
Tf (°C)
)
=
m₂ (
)
×
c₂ (J/(kg·K))
×
(
Tf (°C)
−
T₂ (°C)
)
The left side is Q₁, the heat the hot body gives (the first bar); the right side is Q₂, the heat the cold body takes (the second).
Q₁
Heat the hot body gives
worked out (not typed)
Q₂
Heat the cold body takes
worked out (not typed)
Formulas
m₁
×
c₁
×
(
T₁
−
Tf)
=
m₂
×
c₂
×
(Tf
−
T₂
)
0.2 × 900 × (90 − 25.5433) = 0.5 × 4,186 × (25.5433 − 20)
Q₁
=
m₁
×
c₁
×
m₂
×
c₂
×
T₁
−
T₂
m₁
×
c₁
+
m₂
×
c₂
11,602
=
0.2
×
900
×
0.5
×
4,186
×
90 − 20
0.2 × 900 + 0.5 × 4,186
Q₂
=
Q₁
11,602 = 11,602
Assumptions
•
No heat is lost to the cup or the air: all the heat the hot body gives, the cold body takes.
•
Neither body melts, boils or reacts, and each specific heat stays the same over the range.
•
Water: c = 4186 J/(kg·K); aluminium: 900 J/(kg·K). A change of 1 °C is a change of 1 K.
Step-by-step
WE KNOW
m₁ = 0.2 kg, c₁ = 900 J/(kg·K), T₁ = 90 °C, m₂ = 0.5 kg, c₂ = 4,186 J/(kg·K), T₂ = 20 °C
FIND
final temperature (Tf), heat the hot body gives (Q₁), heat the cold body takes (Q₂)
Step 1 · Find final temperature (Tf)
Use m₁ × c₁ × (T₁ − Tf) = m₂ × c₂ × (Tf − T₂)
Heat lost equals heat gained; gather the Tf terms on one side. Tf is the average of the two temperatures weighted by each mc.
Tf
=
(
m₁
×
c₁
×
T₁
+
m₂
×
c₂
×
T₂
)
÷
(
m₁
×
c₁
+
m₂
×
c₂
)
Tf = (0.2 × 900 × 90 + 0.5 × 4,186 × 20) ÷ (0.2 × 900 + 0.5 × 4,186)
Tf
=
180 × 90 + 2,093 × 20
180 + 2,093
Tf
=
16,200 + 41,860
2,273
Tf
=
58,060
2,273
Tf = 25.5433 °C
Step 2 · Find heat the hot body gives (Q₁)
Use Q₁ = m₁ × c₁ × m₂ × c₂ × (T₁ − T₂) ÷ (m₁ × c₁ + m₂ × c₂)
Put
Tf
into
Q₁
=
m₁
c₁
(
T₁
−
Tf):
the
heat
depends
only
on
the
starting
difference.
Q₁
=
0.2
×
900
×
0.5
×
4,186
×
90 − 20
0.2 × 900 + 0.5 × 4,186
Q₁
=
0.2
×
900
×
0.5
×
4,186
×
70
180 + 2,093
Q₁
=
0.2
×
900
×
0.5
×
4,186
×
70
2,273
Q₁
=
180
×
0.5
×
4,186
×
70
2,273
Q₁
=
90
×
4,186
×
70
2,273
Q₁
=
376,740
×
70
2,273
Q₁
=
26,371,800
2,273
Q₁
=
11,602
J
Step 3 · Find heat the cold body takes (Q₂)
Use Q₂ = Q₁
No heat is lost, so the cold body takes all the heat the hot body gives.
Q₂
=
11,602
J
Check
Put the numbers back in. Both sides should match.
0.2 × 900 × (90 − 25.5433) = 0.5 × 4,186 × (25.5433 − 20) ✓
11,602
=
0.2
×
900
×
0.5
×
4,186
×
90 − 20
0.2 × 900 + 0.5 × 4,186
✓
11,602 = 11,602 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.