dT/dt = -0.25(T - 20), T(0) = 100, when is T = 30 Step by Step

Step by step

The question

Question: find the time t when T = 30, if dT/dt = −0.25(T − 20) and T(0) = 100.

dTdt = −0.25(T − 20)

T(0) = 100

1 Read the equation

T is the temperature, t is the time, and 20 is the room temperature.

The bigger the gap T − 20, the faster it cools.

the rate = −0.25 × (T − 20): T − 20 is the gap to the room

2 Separate and integrate

This is a separable equation.

Divide by T − 20, times dt.

dTT − 20 = −0.25 dt

Integrate both sides, as in the separable lesson.

ln |T − 20| = −0.25t + C

Use e on both sides.

e^C is just a number: call it A.

T − 20 = Ae^(−0.25t)

Add 20 to both sides.

T = 20 + Ae^(−0.25t)

3 Use the starting value

Put in t = 0 and T = 100.

e⁰ = 1.

100 = 20 + Ae^(0)

So A = 80: the gap at the start.

A = 100 − 20 = 80

The formula: T = 20 + 80e^(−0.25t).

T = 20 + 80e^(−0.25t)

Careful!

Don't forget the room temperature.

Only the gap, 80, shrinks; T gets close to 20, not to 0.

Careful: not T = 100e^(−0.25t) ✗

4 Answer the question

Put in T = 30.

Now t is the unknown.

30 = 20 + 80e^(−0.25t)

Take away 20 from both sides.

30 − 20 = 80e^(−0.25t)

Divide both sides by 80.

18 = e^(−0.25t)

Take ln of both sides.

ln(18) = −0.25t

Divide.

So t ≈ 8.32.

t = ln(18)−0.25 ≈ 8.32

5 Check

Let's check.

At t = 0: 100, the start.

Right!

T(0) = 20 + 80 × 1 = 100

Put t ≈ 8.32 back in: about 29.99.

Right!

T(8.32) ≈ 29.99

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer