What is Newton’s Law of Cooling?

Newton’s Law of Cooling states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings.

Newton’s law of cooling states that the rate at which an exposed body changes temperature through radiation is roughly proportional to the temperature difference between the object and its environment, given that the difference is small.

According to Newton’s Law of Cooling, the rate of heat loss from an object is directly proportional to the difference between the temperature of the object and its environment.

Table of Contents:

• Newton’s Law of Cooling Formula
• Newton’s Law of Cooling Derivation
• Limitations of Newton’s Law of Cooling
• Solved Examples

Newton’s law of cooling states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature: dT/dt = k(T_t - T_s)

T_t = temperature of the body at time t

T_s = temperature of the surrounding,

k is a positive constant that depends on the area and type of surface of the body in question.

Newton’s Law of Cooling Formula

The greater the difference in temperature between a system and its surroundings, the more quickly heat is transferred and, therefore, the more quickly the body temperature of the system changes. This is expressed by Newton’s law of cooling formula.

T(t) = T_s + (T_o - T_s)e^-kt

t = time

T(t) = Temperature of the given body at time t

T_s = surrounding temperature,

Initial Temperature of the Body (T_o)

k = constant

Newton’s Law of Cooling Derivation

The rate of cooling of a body is directly proportional to the temperature difference and the surface area exposed when the temperature difference between the body and its surrounding is small.

$dQ/dt\propto (q-q_s)$, where q and q_s are temperatures corresponding to the object and its surroundings.

From the expression above, we can see that $$\frac{dQ}{dt} = -k[Q - Q_s] . . . . . . . . (1)$$

This expression represents Newton’s law of cooling. It can be derived directly from Stefan’s law, which gives,

k = [4eσ×θ³o/mc²]^A

Now, $$\frac{d\theta}{dt} = -k[\theta - \theta_0]$$

$$⇒ (\int_{\theta_1}^{\theta_2}\frac{d\theta}{(\theta-\theta_o)} = -k\int_{0}^{1}dt)$$

Where,

Qⁱ = Initial Temperature of Object

Q^{F = Final Temperature of Object.}

$$ln \left(\frac{q_f - q_0}{q_i - q_0}\right) = kt$$

$$(q_f - q_0) = (q_i - q_0) \cdot e^{-kt}$$

$$q_f = q_0 + (q_i – q_0) \mathrm{e}^{-kt} ……….(3).$$

⇒ Check: Heat Transfer by Conduction

Applying Newton’s Law of Cooling

We can sometimes approximate values from Newton’s law by assuming a constant rate of cooling which is equal to the rate of cooling corresponding to the average temperature of the body over the given interval.

i.e. $$\frac{d\theta}{dt} = k(q - q_0) ……….. (4)$$

If q_i and q_f are the initial and final temperatures of the body, then

q = (q_i + q_f)/2 . . . . . (5)

Remember that equation (5) is only an approximation and equation (1) must be used for exact values.

Limitations of Newton’s Law of Cooling

The difference in temperature between the body and its surroundings must be minimal.

The loss of heat from the body should only be through radiation.

The major limitation of Newton’s law of cooling is that the temperature of the surrounding environment must remain constant while the body is cooling.

Solved Examples

Example 1: A body at temperature 40ºC is kept in a surrounding of constant temperature 20ºC. It is observed that its temperature falls to 35ºC in 10 minutes. How much more time will it take for the body to attain a temperature of 30ºC?

Solution:

Q_f = Q_i * e^(-kt)

Now, for the interval where the temperature is between 40ºC and 35ºC.

(35 - 20) = 15 and (40 - 20) = 20; therefore, 20 - 15 = 5

e^-10k = 3/4

$$k = \frac{\ln\frac{4}{3}}{10} …….. (a)$$

Now, for the next interval;

30 - 20 = 15 and 35 - 20 = 15

$$e^{-kt} = \frac{2}{3}$$

kt = ln(3/2) (b)

From equations (a) and (b);

t = 10 × [ln(3/2)/ln(4/3)] = 14.096 min.

Aliter (approximate method):

For the interval in which temperature falls from 40ºC to 35ºC

q = (40 + 35)/2 = 37.5ºC

From equation (4):

dθ/dt = k(q₀ - q)

(35 - 40)/10 = k(37.5 - 20)

k = (35 - 40)/(37.5 - 20)

$$k = 1/32 min^{-1}$$

Now, for the interval in which temperature falls from 35ºC to 30ºC.

$$( \frac{35 + 30}{2} = 32.5^\circ C )$$

From equation (4);

(30 - 35) / 5 = (32.5 - 20)

Therefore, t = 14 min, where t is the required time and 5/12.5 × 35 = 14 min.

Example 2: The oil is heated to 70ºC. It cools to 50ºC after 6 minutes. Calculate the time taken by the oil to cool from 50ºC to 40ºC given the surrounding temperature Ts = 25ºC.

The time taken by the oil to cool from 50ºC to 40ºC is approximately 5 minutes, given the surrounding temperature T_s = 25ºC.

Solution:

Given:

Temperature of oil after 6 minutes, T(t) = 50ºC

T_s = 25ºC

T_o = 70ºC

t = 6 min

Substituting the given data into Newton’s law of cooling formula yields:

T(t) = T_s + (T_s – T_o) e^-kt

[T(t) - T_s] / [T_o - T_s] = e^-kt

$$-k\large\frac{ln\ T(t) - Ts}{To - Ts}$$

-kt = [ln 50 - 25]/70 - 25 = ln 0.555

$$-kt = \frac{[\ln 50 - 25]}{70 - 25} = \ln 0.555$$

k = 0.092 (when -0.555/6 is simplified)

If T(t) = 45ºC (average temperature as the temperature decreases from 50ºC to 40ºC)

[ln T(t) - Ts] / [T_o - T_s] = kt

-t = ln(45 - 25) / (70 - 25)

-0.092t = -0.597

t = -0.597/-0.092 = 6.543

t = -0.597 / -0.092 = 6.489 min.

Example 3: Water is heated to 80ºC for 10 min. How much would be the temperature if k = 0.56 per min and the surrounding temperature is 25ºC?

Solution:

T_s = 25ºC

To = 80ºC

t = 10 min

k = 0.56

Now, substituting the above data into Newton’s law of cooling formula,

T(t) = T_s + (T_o - T_s) e^-kt

25 + (80 - 25) × e^-0.56 = 25 + [55 × e^-0.56] = 56.35 ºC

The temperature cooled from 80ºC to 56.35ºC over 10 minutes.

Frequently Asked Questions (FAQs)

What is Newton’s law of cooling?

Newton’s Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature.

Why is Newton’s law of cooling important?

Newton’s law of cooling is important because it explains the rate at which an object cools down in relation to the temperature difference between the object and its surroundings. This law helps us to understand the behavior of heat transfer and can be used to predict the temperature of an object over time.

The formula of Newton’s law of cooling is: $$\frac{dT}{dt} = -k(T - T_m)$$ where T is the temperature of the object, T_m is the temperature of the surrounding medium, and k is the cooling constant.

What is the formula of Newton’s law of cooling?

Newton’s law of cooling states that the rate of change of temperature of a body is proportional to the difference between its own temperature and the ambient temperature, and is given by the formula:

T(t) = T_s + (T_o - T_s)e^-kt