dT/dt = -k(T - 20), T(0) = 80, T(2) = 70, when is T = 25 Step by Step

Step by step

The question

Question: find the time t when T = 25, if dT/dt = −k(T − 20) and T(0) = 80, T(2) = 70.

dTdt = −k(T − 20)

T(0) = 80, T(2) = 70

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 = −k × (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 = −k dt

Integrate both sides, as in the separable lesson.

ln |T − 20| = −kt + C

Use e on both sides.

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

T − 20 = Ae^(−kt)

Add 20 to both sides.

T = 20 + Ae^(−kt)

3 Use the starting value

Put in t = 0 and T = 80.

e⁰ = 1.

80 = 20 + Ae^(0)

So A = 60: the gap at the start.

A = 80 − 20 = 60

The formula: T = 20 + 60e^(−kt).

T = 20 + 60e^(−kt)

Careful!

Don't forget the room temperature.

Only the gap, 60, shrinks; T gets close to 20, not to 0.

Careful: not T = 80e^(−kt) ✗

4 Find k

Now the second value: t = 2, T = 70.

70 = 20 + 60e^(−2k)

Take away 20 from both sides.

70 − 20 = 60e^(−2k)

Divide both sides by 60.

56 = e^(−2k)

Take ln of both sides: ln and e cancel, so the power comes down.

ln(56) = −2k

Divide by −2: k is about 0.0912.

k = ln(56)−2 ≈ 0.0912

So T ≈ 20 + 60e^(−0.0912t).

T ≈ 20 + 60e^(−0.0912t)

5 Answer the question

Put in T = 25.

Now t is the unknown.

25 ≈ 20 + 60e^(−0.0912t)

Take away 20 from both sides.

25 − 20 ≈ 60e^(−0.0912t)

Divide both sides by 60.

112 ≈ e^(−0.0912t)

Take ln of both sides.

ln(112) ≈ −0.0912t

Divide.

So t ≈ 27.26.

t = ln(112)−0.0912 ≈ 27.26

6 Check

Let's check.

At t = 0: 80, the start.

Right!

T(0) = 20 + 60 × 1 = 80

Put t ≈ 27.26 back in: about 25.

Right!

T(27.26) ≈ 25

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: find k.

Then: answer the question.

Last: check.

Answer