SME
Differential Equations
First-order ODEs
Use this for “Soup at 85 °C cools in a 22 °C kitchen with k = 0.05 per minute. What is its temperature after 15 minutes?”
Chart
Horizontal asymptote T = 20
T(10) ≈ 45.75
This is an example. Type your own values to start a new problem.
T(t)
=
Tₐ (°C)
+
(
T₀ (°C)
−
Tₐ (°C)
)
e
−
t
T(
t (min)
)
=
T (°C)
Formulas
T
=
Tₐ
+
(
T₀
−
Tₐ)
×
e
−
k
×
t
45.7516
=
20
+
(90
−
20)
×
e
−0.1 × 10
Assumptions
•
Newton’s law of cooling: the temperature changes at a rate proportional to its gap from the room, dT/dt = −k(T − Tₐ), with k > 0.
•
The equation is separable: ∫ dT ÷ (T − Tₐ) = −∫ k dt gives ln|T − Tₐ| = −kt + C, so T = Tₐ + (T₀ − Tₐ)e−kt.
•
The room stays at Tₐ the whole time, and T approaches Tₐ without ever crossing it.
•
Temperatures are in °C, time in minutes and k per minute.
Step-by-step
WE KNOW
Tₐ = 20 °C, T₀ = 90 °C, k = 0.1 min⁻¹, t = 10 min
FIND
temperature at time t (T)
Step 1 · Find temperature at time t (T)
Use T = Tₐ + (T₀ − Tₐ) × e−k × t
The
gap
T₀
−
Tₐ
shrinks
by
the
factor
e
−
kt
.
Add
what
is
left
of
it
to
the
room
temperature.
dT/dt
=
−0.1(
T
−
20)
T
(
t
)
=
20
+
70
e
−0.1
t
T
(10)
=
20
+
70e⁻¹
=
20
+
70
×
0.367879
=
45.7516
T
=
45.7516
°C
Check
Put the numbers back in. Both sides should match.
45.7516
=
20
+
(90
−
20)
×
e
−0.1 × 10
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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