dT/dt = -0.2(T - 25), T(0) = 95, when is T = 40 Step by Step

Step by step

The question

Question: find the time t when T = 40, if dT/dt = −0.2(T − 25) and T(0) = 95.

dTdt = −0.2(T − 25)

T(0) = 95

1 Read the equation

T is the temperature, t is the time, and 25 is the room temperature.

The bigger the gap T − 25, the faster it cools.

the rate = −0.2 × (T − 25): T − 25 is the gap to the room

2 Separate and integrate

This is a separable equation.

Divide by T − 25, times dt.

dTT − 25 = −0.2 dt

Integrate both sides, as in the separable lesson.

ln |T − 25| = −0.2t + C

Use e on both sides.

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

T − 25 = Ae^(−0.2t)

Add 25 to both sides.

T = 25 + Ae^(−0.2t)

3 Use the starting value

Put in t = 0 and T = 95.

e⁰ = 1.

95 = 25 + Ae^(0)

So A = 70: the gap at the start.

A = 95 − 25 = 70

The formula: T = 25 + 70e^(−0.2t).

T = 25 + 70e^(−0.2t)

Careful!

Don't forget the room temperature.

Only the gap, 70, shrinks; T gets close to 25, not to 0.

Careful: not T = 95e^(−0.2t) ✗

4 Answer the question

Put in T = 40.

Now t is the unknown.

40 = 25 + 70e^(−0.2t)

Take away 25 from both sides.

40 − 25 = 70e^(−0.2t)

Divide both sides by 70.

314 = e^(−0.2t)

Take ln of both sides.

ln(314) = −0.2t

Divide.

So t ≈ 7.7.

t = ln(314)−0.2 ≈ 7.7

5 Check

Let's check.

At t = 0: 95, the start.

Right!

T(0) = 25 + 70 × 1 = 95

Put t ≈ 7.7 back in: about 40.01.

Right!

T(7.7) ≈ 40.01

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer