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Lesson 4: Newton's Law of Cooling

Estimated time: 30-40 minutes

Learning Objectives

By the end of this lesson, you will be able to:

State Newton's Law of Cooling as a differential equation
Solve dT/dt = k(T - T_env) and interpret the solution
Determine k from temperature measurements
Apply the model to forensic time-of-death estimation
Apply the model to cooking and engineering problems

The Law

Newton's Law of Cooling states that the rate of temperature change of an object is proportional to the difference between the object's temperature and the ambient (environmental) temperature.

Newton's Law of Cooling: dT/dt = k(T - T_env), where T(t) is the object's temperature, T_env is the constant ambient temperature, and k < 0 (cooling) or k > 0 (warming toward a hotter environment).

Note: When an object is hotter than its surroundings, T - T_env > 0 and the object cools (dT/dt < 0), so k must be negative. Some textbooks write dT/dt = -k(T - T_env) with k > 0; both formulations are equivalent.

Solving the Equation

Let u = T - T_env. Then du/dt = dT/dt = ku, so u(t) = u₀e^kt.

General Solution: T(t) = T_env + (T₀ - T_env)e^kt, where T₀ = T(0) is the initial temperature.

As t → ∞, the exponential term decays to 0 (since k < 0 for cooling), and T → T_env. The object approaches ambient temperature asymptotically -- it never quite reaches it.

Determining k from Data

Example 1: Finding the Cooling Constant

A cup of coffee at 90°C is placed in a 20°C room. After 10 minutes, the coffee is 70°C. Find k.

Step 1: T(t) = 20 + (90 - 20)e^kt = 20 + 70e^kt.

Step 2: T(10) = 70: 70 = 20 + 70e^10k ⇒ 50 = 70e^10k ⇒ e^10k = 5/7.

Step 3: k = ln(5/7)/10 = (ln 5 - ln 7)/10 ≈ -0.03365 per minute.

Model: T(t) = 20 + 70e^-0.03365t.

Example 2: When Will the Coffee Be Drinkable?

Using the model above, when does the coffee reach 50°C?

Solution: 50 = 20 + 70e^-0.03365t ⇒ 30 = 70e^-0.03365t ⇒ e^-0.03365t = 3/7.

t = -ln(3/7)/0.03365 = ln(7/3)/0.03365 ≈ 25.2 minutes.

Forensic Application: Time of Death

A key application of Newton's Law is estimating time of death. Normal body temperature is about 37°C (98.6°F). After death, the body cools toward ambient temperature following this law.

Example 3: Time of Death Estimation

A body is found at 8:00 PM with temperature 29°C. The room is 20°C. At 9:00 PM, the body is 27°C. Estimate the time of death (assuming normal body temp of 37°C at death).

Step 1: Let t = 0 be 8:00 PM. T(0) = 29, T(1) = 27, T_env = 20.

Step 2: T(t) = 20 + 9e^kt. From T(1) = 27: 27 = 20 + 9e^k ⇒ e^k = 7/9 ⇒ k = ln(7/9) ≈ -0.2513.

Step 3: At the time of death (t = t_d), body was 37°C. Working backwards from 8:00 PM:

37 = 20 + (37 - 20)e^kt_d where t_d is measured from death to 8:00 PM.

Actually, let us set t = 0 at time of death. Then T(0) = 37, and at time t₁ (the time from death to 8 PM), T(t₁) = 29.

29 = 20 + 17e^kt₁ ⇒ e^kt₁ = 9/17.

Also T(t₁ + 1) = 27: 27 = 20 + 17e^k(t₁+1) ⇒ e^k(t₁+1) = 7/17.

Dividing: e^k = 7/9 ⇒ k = ln(7/9) ≈ -0.2513 (consistent).

t₁ = ln(9/17)/k = ln(9/17)/ln(7/9) ≈ 0.6376/0.2513 ≈ 2.54 hours.

Conclusion: Death occurred about 2.5 hours before 8:00 PM, i.e., around 5:30 PM.

Warming Problems

The same equation applies when an object warms up (cold object in a warm environment). Here T₀ < T_env, and the object's temperature increases toward T_env.

Example 4: Warming a Cold Object

A thermometer reading 5°C is brought into a 25°C room. After 3 minutes it reads 15°C. When will it read 24°C?

Step 1: T(t) = 25 + (5 - 25)e^kt = 25 - 20e^kt.

Step 2: T(3) = 15: 15 = 25 - 20e^3k ⇒ 20e^3k = 10 ⇒ e^3k = 1/2 ⇒ k = -ln 2/3 ≈ -0.2310.

Step 3: Find t when T = 24: 24 = 25 - 20e^kt ⇒ 20e^kt = 1 ⇒ e^kt = 1/20.

t = ln(1/20)/k = ln(20) · 3/ln(2) ≈ 12.97 minutes.

Limitations of the Model

Assumes constant ambient temperature T_env.
Works best when the temperature difference is moderate (not extreme).
Assumes the object has uniform temperature (no significant internal temperature gradients).
For very hot objects (radiation is significant), more complex models are needed.

Check Your Understanding

1. A hot plate at 150°C cools in a 25°C room. Write the general solution for T(t).

Answer: T(t) = 25 + (150 - 25)e^kt = 25 + 125e^kt where k < 0 is the cooling constant.

2. As t → ∞, what temperature does the hot plate approach?

Answer: T → 25°C (the ambient temperature). The exponential term decays to zero.

3. A body at 35°C is in a 22°C room. One hour later it is 31°C. Find k.

Answer: 31 = 22 + 13e^k ⇒ 9 = 13e^k ⇒ k = ln(9/13) ≈ -0.3677 per hour.

4. Using Newton's cooling with T_env = 20, T₀ = 80, and k = -0.1, find the temperature after 5 minutes.

Answer: T(5) = 20 + 60e^-0.1(5) = 20 + 60e^-0.5 ≈ 20 + 60(0.6065) = 20 + 36.39 = 56.39°C.

Key Takeaways

Newton's Law: dT/dt = k(T - T_env) with solution T(t) = T_env + (T₀ - T_env)e^kt.
The object temperature approaches ambient temperature asymptotically (never quite reaches it).
Find k by using two temperature measurements at known times.
Key applications: forensic time-of-death estimation, cooking times, engineering heat transfer.
The model assumes constant T_env, uniform object temperature, and moderate temperature differences.

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