Newton's Law of Cooling

When you take a hot cup of coffee and set it on a table, it cools down. But it does not cool at a constant rate — it cools faster at first (when the temperature difference from the room is large) and slower later (when it is closer to room temperature). This behavior is captured precisely by Newton’s Law of Cooling, one of the most elegant applications of exponential functions.

The Model

Newton’s Law of Cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and the surrounding (ambient) temperature. The resulting formula is:

$T(t) = T_s + (T_0 - T_s)e^{-kt}$

where:

• $T(t)$ = temperature of the object at time $t$
• $T_s$ = surrounding (ambient) temperature
• $T_0$ = initial temperature of the object (at $t = 0$)
• $k$ = cooling constant ($k > 0$)
• $t$ = time

Understanding the Formula

At $t = 0$: $T(0) = T_s + (T_0 - T_s)e^0 = T_s + T_0 - T_s = T_0$ — the initial temperature.

As $t \to \infty$: $e^{-kt} \to 0$, so $T \to T_s$ — the object approaches room temperature.

The term $(T_0 - T_s)e^{-kt}$ represents the temperature excess (or deficit) above (or below) the ambient temperature, and it decays exponentially.

Cooling: When $T_0 > T_s$ (object hotter than surroundings), the temperature decreases toward $T_s$.

Warming: When $T_0 < T_s$ (object cooler than surroundings), the formula still works — the temperature increases toward $T_s$. For example, a cold drink warming up in a hot room.

Finding the Cooling Constant $k$

The constant $k$ depends on the physical properties of the object and its environment (material, surface area, air circulation). You find $k$ from a data point: if you know the temperature at some specific time, you can solve for $k$.

Example 1: A cup of coffee is brewed at 200 degrees F and placed in a 72-degree room. After 5 minutes, the coffee is 180 degrees. Find $k$ and the complete model.

Given: $T_0 = 200$, $T_s = 72$, $T(5) = 180$.

Setup:

$180 = 72 + (200 - 72)e^{-5k}$

$180 = 72 + 128e^{-5k}$

$108 = 128e^{-5k}$

$e^{-5k} = \frac{108}{128} = 0.84375$

$-5k = \ln(0.84375) = -0.16990$

$k = 0.03398$

Model: $T(t) = 72 + 128e^{-0.03398t}$

Verification: $T(5) = 72 + 128e^{-0.170} = 72 + 128 \times 0.8438 = 72 + 108 = 180$ degrees F. Confirmed.

Solving for Time

Once you have the model, you can solve for when the object reaches a specific temperature.

Example 2 (Continuing from Example 1): When will the coffee cool to 120 degrees?

$120 = 72 + 128e^{-0.03398t}$

$48 = 128e^{-0.03398t}$

$e^{-0.03398t} = 0.375$

$-0.03398t = \ln(0.375) = -0.9808$

$t = \frac{0.9808}{0.03398} = 28.9 \text{ minutes}$

The coffee reaches 120 degrees after about 29 minutes.

Example 3: When will it be within 1 degree of room temperature (73 degrees)?

$73 = 72 + 128e^{-0.03398t}$

$1 = 128e^{-0.03398t}$

$e^{-0.03398t} = \frac{1}{128} = 0.007813$

$-0.03398t = \ln(0.007813) = -4.852$

$t = \frac{4.852}{0.03398} = 142.8 \text{ minutes}$

It takes nearly 2.5 hours to get within 1 degree of room temperature. The last few degrees take a very long time — the exponential tail.

Application: Forensic Time of Death

Forensic investigators use Newton’s Law of Cooling to estimate the time of death. Normal body temperature is 98.6 degrees F. After death, the body cools toward the ambient temperature.

Example 4: A body is found at 8:00 PM in a 65-degree room with a temperature of 82 degrees. One hour later, the temperature is 78.5 degrees. Estimate the time of death.

Step 1 — Find $k$: Let $t = 0$ be 8:00 PM, so $T(0) = 82$ and $T_s = 65$.

$78.5 = 65 + (82 - 65)e^{-k}$

$13.5 = 17e^{-k}$

$e^{-k} = \frac{13.5}{17} = 0.7941$

$k = -\ln(0.7941) = 0.2305$

Step 2 — Find when the body was 98.6 degrees: Now we work backward from the time the body was found. We need the model starting from death, but since we defined $t = 0$ at 8:00 PM with temperature 82, we need to find when (before 8:00 PM) the temperature was 98.6.

Reset: let $T_0 = 98.6$ (at death), $T_s = 65$. The model from the time of death is:

$T(\tau) = 65 + 33.6e^{-0.2305\tau}$

At 8:00 PM ($\tau$ hours after death), $T = 82$:

$82 = 65 + 33.6e^{-0.2305\tau}$

$17 = 33.6e^{-0.2305\tau}$

$e^{-0.2305\tau} = 0.5060$

$-0.2305\tau = \ln(0.5060) = -0.6812$

$\tau = 2.96 \text{ hours}$

The person died approximately 3 hours before 8:00 PM — around 5:00 PM.

Application: Cooking (Warming Model)

Newton’s Law also models warming. When you put a cold item in a hot environment, the temperature rises toward the ambient temperature.

Example 5: A frozen turkey at 25 degrees F is placed in a 325-degree oven. After 1 hour it is 75 degrees. When will it reach 165 degrees (safe internal temperature)?

Given: $T_0 = 25$, $T_s = 325$, $T(1) = 75$.

Note: $T_0 < T_s$, so $(T_0 - T_s) = 25 - 325 = -300$. The model is:

$T(t) = 325 - 300e^{-kt}$

Find $k$:

$75 = 325 - 300e^{-k}$

$-250 = -300e^{-k}$

$e^{-k} = \frac{250}{300} = 0.8333$

$k = -\ln(0.8333) = 0.1823$

Find $t$ for $T = 165$:

$165 = 325 - 300e^{-0.1823t}$

$-160 = -300e^{-0.1823t}$

$e^{-0.1823t} = \frac{160}{300} = 0.5333$

$-0.1823t = \ln(0.5333) = -0.6286$

$t = \frac{0.6286}{0.1823} = 3.45 \text{ hours}$

The turkey reaches 165 degrees after about 3 hours and 27 minutes.

Note: Real cooking is more complex — ovens have radiation and convection effects that Newton’s Law simplifies. But the model gives a useful first approximation.

Application: HVAC Engineering

HVAC engineers use Newton’s Law of Cooling to estimate how quickly a building heats or cools after the thermostat is adjusted.

Example 6: An office building is at 68 degrees F when the heating system fails at midnight. The outside temperature is 20 degrees F. The cooling constant for the building is $k = 0.04$ per hour. When will the interior reach 55 degrees?

$T(t) = 20 + (68 - 20)e^{-0.04t} = 20 + 48e^{-0.04t}$

$55 = 20 + 48e^{-0.04t}$

$35 = 48e^{-0.04t}$

$e^{-0.04t} = 0.7292$

$-0.04t = \ln(0.7292) = -0.3158$

$t = 7.9 \text{ hours}$

The building reaches 55 degrees around 7:54 AM — before workers typically arrive. The HVAC engineer can use this to plan backup heating or insulation improvements.

Nursing Application: IV Fluid Temperature

In nursing, intravenous fluids sometimes need to reach body temperature before administration. If an IV bag is removed from a 40-degree F refrigerator and placed in a 72-degree room, Newton’s Law models the warming:

$T(t) = 72 + (40 - 72)e^{-kt} = 72 - 32e^{-kt}$

With a typical $k \approx 0.1$ per minute for a standard IV bag, the fluid reaches 65 degrees in:

$65 = 72 - 32e^{-0.1t} \implies 7 = 32e^{-0.1t} \implies t = \frac{-\ln(7/32)}{0.1} = \frac{1.520}{0.1} = 15.2 \text{ min}$

Limitations of the Model

Newton’s Law of Cooling assumes:

• The ambient temperature is constant (if the room heats up from the hot object, the model breaks down)
• The object has uniform temperature (no hot spots or cold core)
• Heat transfer is primarily by convection (the model is less accurate for radiation-dominated scenarios)
• The temperature difference is not extreme (the law is an approximation that works well for moderate temperature ranges)

For precision applications, more sophisticated models (like lumped capacitance or finite element analysis) are used. But for everyday problems — coffee cooling, time of death, building heat loss — Newton’s Law is remarkably accurate.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Newton’s Law of Cooling: $T(t) = T_s + (T_0 - T_s)e^{-kt}$ — temperature approaches ambient exponentially
• The formula works for both cooling ($T_0 > T_s$) and warming ($T_0 < T_s$)
• The constant $k$ is found from a known temperature at a known time — it depends on material properties and environment
• A larger $k$ means faster temperature change (poor insulation); smaller $k$ means slower change (good insulation)
• Applications include forensics (time of death), cooking, HVAC (building heat loss), and nursing (fluid warming)
• The model assumes constant ambient temperature and uniform object temperature

Return to College Algebra for more topics in this section.