Newton’s law of cooling

Newton's law of cooling

Newton's law of cooling states that the rate at which the temperature of an object decreases is proportional to the difference between its temperature and the temperature of the surrounding environment.

This law is modeled using an exponential function. Let:

• T(t) be the temperature of the object at time t
• $T_s$ be the constant surrounding temperature

Then the rate of change of temperature is given by:

$$\frac{dT}{dt} = -k(T - T_s)$$

Separating variables and integrating:

$$\int \frac{1}{T - T_s} \, dT = -k \int dt$$ $$\ln|T - T_s| = -kt + C$$

Exponentiating both sides:

$$T - T_s = Ce^{-kt}$$

Letting $T(0) = T_0$, we get:

$$T_0 - T_s = C \Rightarrow C = T_0 - T_s$$

So the temperature as a function of time is:

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

Example

Suppose the temperature of an object falls from $T_0 = 240°C$ to $T = 180°C$ in 50 minutes, and the surrounding temperature is $T_s = 20°C$. Find the time it takes to reach $T = 100°C$, correct to three significant figures.

We use:

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

Substitute known values:

$$180 = 20 + (240 - 20)e^{-50k}$$ $$160 = 220e^{-50k}$$

Divide both sides:

$$\frac{160}{220} = e^{-50k} \Rightarrow e^{-50k} = 0.72727$$

Take natural logarithm:

$$-50k = \ln(0.72727) \Rightarrow k = -\frac{\ln(0.72727)}{50} \approx 0.006369$$

Now solve for time t when $T(t) = 100°C$:

$$100 = 20 + 220e^{-0.006369t} \Rightarrow 80 = 220e^{-0.006369t} \Rightarrow \frac{80}{220} = e^{-0.006369t}$$

Take logarithm:

$$-0.006369t = \ln\left(\frac{80}{220}\right) \Rightarrow t = \frac{\ln(80/220)}{-0.006369} \approx 158.83 \text{ minutes}$$

Answer: It will take approximately 158 minutes and 50 seconds for the object to reach 100°C.

Example

A rod of iron is heated to $T_0 = 1080°C$ and plunged into water at $T_s = 25°C$. After 25 seconds, the temperature drops to $T = 82.5°C$. Find the cooling rate $k$.

Use the formula:

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

Substitute:

$$82.5 = 25 + (1080 - 25)e^{-25k} \Rightarrow 57.5 = 1055e^{-25k} \Rightarrow \frac{57.5}{1055} = e^{-25k}$$

Take logarithm:

$$-25k = \ln\left(\frac{57.5}{1055}\right) \Rightarrow k = -\frac{\ln(0.0545)}{25} \approx 0.1164$$

Answer: The cooling rate is approximately 0.1164 per second.