Steply

Newton's law of cooling

Differential Equations · 20 problems, easy to harder

▶ Watch the first one solved

How we do it

  1. Read the equation
  2. Separate and integrate
  3. Use the starting value
  4. Find k
  5. Answer the question
  6. Check

The steps for dT/dt = -k(T - 20), T(0) = 5, T(10) = 12, find T(30). Every solution names its steps, so you always know what comes next.

Worked examples

dT/dt = -0.1(T - 20), T(0) = 90, find T(10) Easy

The question

Question: find T(10), if dT/dt = −0.1(T − 20) and T(0) = 90.

dTdt = −0.1(T − 20)
T(0) = 90

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 = −0.1 × (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 = −0.1 dt

Integrate both sides, as in the separable lesson.

ln |T − 20| = −0.1t + C

Use e on both sides.

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

T − 20 = Ae^(−0.1t)

Add 20 to both sides.

T = 20 + Ae^(−0.1t)

3 Use the starting value

Put in t = 0 and T = 90.

e⁰ = 1.

90 = 20 + Ae^(0)

So A = 70: the gap at the start.

A = 90 − 20 = 70

The formula: T = 20 + 70e^(−0.1t).

T = 20 + 70e^(−0.1t)

Careful!

Don't forget the room temperature.

Only the gap, 70, shrinks; T gets close to 20, not to 0.

Careful: not T = 90e^(−0.1t) ✗

4 Answer the question

Put in t = 10.

T(10) = 20 + 70e^(−0.1 × 10)

The power: −1.

= 20 + 70e^(−1)

A calculator: 70 × e^(−1) ≈ 25.75.

≈ 20 + 25.75

So T(10) ≈ 45.75.

≈ 45.75

5 Check

Let's check.

At t = 0: 90, the start.

Right!

T(0) = 20 + 70 × 1 = 90

And the derivative of the formula is −0.1 times the gap T − 20: the equation again.

Right!

T′ = 70 × −0.1e^(−0.1t) = −0.1 × (T − 20)

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer
≈ 45.75

▶ Watch it solved

dT/dt = -0.1(T - 20), T(0) = 90, when is T = 50 Medium

The question

Question: find the time t when T = 50, if dT/dt = −0.1(T − 20) and T(0) = 90.

dTdt = −0.1(T − 20)
T(0) = 90

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 = −0.1 × (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 = −0.1 dt

Integrate both sides, as in the separable lesson.

ln |T − 20| = −0.1t + C

Use e on both sides.

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

T − 20 = Ae^(−0.1t)

Add 20 to both sides.

T = 20 + Ae^(−0.1t)

3 Use the starting value

Put in t = 0 and T = 90.

e⁰ = 1.

90 = 20 + Ae^(0)

So A = 70: the gap at the start.

A = 90 − 20 = 70

The formula: T = 20 + 70e^(−0.1t).

T = 20 + 70e^(−0.1t)

Careful!

Don't forget the room temperature.

Only the gap, 70, shrinks; T gets close to 20, not to 0.

Careful: not T = 90e^(−0.1t) ✗

4 Answer the question

Put in T = 50.

Now t is the unknown.

50 = 20 + 70e^(−0.1t)

Take away 20 from both sides.

50 − 20 = 70e^(−0.1t)

Divide both sides by 70.

37 = e^(−0.1t)

Take ln of both sides.

ln(37) = −0.1t

Divide.

So t ≈ 8.47.

t = ln(37)−0.1 ≈ 8.47

5 Check

Let's check.

At t = 0: 90, the start.

Right!

T(0) = 20 + 70 × 1 = 90

Put t ≈ 8.47 back in: about 50.01.

Right!

T(8.47) ≈ 50.01

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer
t = ln(37)−0.1 ≈ 8.47

▶ Watch it solved

dT/dt = -k(T - 20), T(0) = 5, T(10) = 12, find T(30) Harder

The question

Question: find T(30), if dT/dt = −k(T − 20) and T(0) = 5, T(10) = 12.

dTdt = −k(T − 20)
T(0) = 5, T(10) = 12

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 warms.

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 = 5.

e⁰ = 1.

5 = 20 + Ae^(0)

So A = −15: the gap at the start.

A = 5 − 20 = −15

The formula: T = 20 − 15e^(−kt).

T = 20 − 15e^(−kt)

Careful!

Don't forget the room temperature.

Only the gap, -15, shrinks; T gets close to 20, not to 0.

Careful: not T = 5e^(−kt) ✗

4 Find k

Now the second value: t = 10, T = 12.

12 = 20 − 15e^(−10k)

Take away 20 from both sides.

12 − 20 = -15e^(−10k)

Divide both sides by −15.

815 = e^(−10k)

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

ln(815) = −10k

Divide by −10: k is about 0.0629.

k = ln(815)−10 ≈ 0.0629

So T ≈ 20 − 15e^(−0.0629t).

T ≈ 20 − 15e^(−0.0629t)

5 Answer the question

Put in t = 30.

T(30) = 20 − 15e^(−0.0629 × 30)

The power: −1.8858.

≈ 20 − 15e^(−1.8858)

A calculator: -15 × e^(−1.8858) ≈ −2.28.

≈ 20 + −2.28

So T(30) ≈ 17.72.

≈ 17.72

6 Check

Let's check.

At t = 0: 5, the start.

Right!

T(0) = 20 − 15 × 1 = 5

With this k, at t = 10 the formula gives 12: the second value.

Right!

T(10) ≈ 20 + -15e^(−0.0629 × 10) ≈ 12

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
≈ 17.72

▶ Watch it solved

Practice

Try each one, then open the answer. Stuck? Open the problem to watch every step.

Easy

  1. dT/dt = -0.1(T - 20), T(0) = 90, find T(10)
    Answer
    ≈ 45.75
  2. dT/dt = -0.2(T - 25), T(0) = 85, find T(5)
    Answer
    ≈ 47.07
  3. dT/dt = -0.05(T - 20), T(0) = 100, find T(20)
    Answer
    ≈ 49.43
  4. dT/dt = -0.3(T - 18), T(0) = 68, find T(2)
    Answer
    ≈ 45.44
  5. dT/dt = -0.1(T - 22), T(0) = 72, find T(15)
    Answer
    ≈ 33.16
  6. dT/dt = -0.2(T - 20), T(0) = 4, find T(3)
    Answer
    ≈ 11.22
  7. dT/dt = -0.15(T - 20), T(0) = 80, find T(8)
    Answer
    ≈ 38.07

Medium

  1. dT/dt = -0.1(T - 20), T(0) = 90, when is T = 50
    Answer
    t = ln(37)−0.1 ≈ 8.47
  2. dT/dt = -0.2(T - 25), T(0) = 95, when is T = 40
    Answer
    t = ln(314)−0.2 ≈ 7.7
  3. dT/dt = -k(T - 20), T(0) = 90, T(5) = 60, find k
    Answer
    k = ln(47)−5 ≈ 0.1119
  4. dT/dt = -k(T - 20), T(0) = 100, T(10) = 60, find T(20)
    Answer
    ≈ 40
  5. dT/dt = -k(T - 22), T(0) = 82, T(4) = 62, find T(12)
    Answer
    ≈ 39.78
  6. dT/dt = -0.05(T - 18), T(0) = 98, when is T = 38
    Answer
    t = ln(14)−0.05 ≈ 27.73
  7. dT/dt = -k(T - 25), T(0) = 95, T(3) = 75, when is T = 35
    Answer
    t = ln(17)−0.1122 ≈ 17.35

Harder

  1. dT/dt = -k(T - 20), T(0) = 5, T(10) = 12, find T(30)
    Answer
    ≈ 17.72
  2. dT/dt = -0.1(T - 4), T(0) = 20, find T(10)
    Answer
    ≈ 9.89
  3. dT/dt = -k(T - 70), T(0) = 350, T(10) = 250, find k
    Answer
    k = ln(914)−10 ≈ 0.0442
  4. dT/dt = 0.2(18 - T), T(0) = 90, find T(5)
    Answer
    ≈ 44.49
  5. dT/dt = -0.25(T - 20), T(0) = 100, when is T = 30
    Answer
    t = ln(18)−0.25 ≈ 8.32
  6. dT/dt = -k(T - 20), T(0) = 80, T(2) = 70, when is T = 25
    Answer
    t = ln(112)−0.0912 ≈ 27.26