# Newton's Law of Cooling - Formulas, Limitations, Examples

## What is Newton’s Law of Cooling?

Newton’s law of cooling describes the rate at which an exposed body changes temperature through radiation which is approximately proportional to the difference between the object’s temperature and its surroundings, provided the difference is small.

Definition: According to Newton’s law of cooling, the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surroundings.

### Table of Content:

Newton’s law of cooling is given by, dT/dt = k(Tt – Ts)

Where,

• Tt = temperature at time t and
• Ts = temperature of the surrounding,
• k = Positive constant that depends on the area and nature of the surface of the body under consideration.

### Newton’s Law of Cooling Formula

Greater the difference in temperature between the system and surrounding, more rapidly the heat is transferred i.e. more rapidly the body temperature of body changes. Newton’s law of cooling formula is expressed by,

T(t) = Ts + (To – Ts) e-kt

Where,

• t = time,
• T(t) = temperature of the given body at time t,
• Ts = surrounding temperature,
• To = initial temperature of the body,
• k = constant.

## Newton’s Law of Cooling Derivation

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

dQ/dt ∝ (q – qs)], where q and qs are temperature corresponding to object and surroundings.

From above expression , dQ/dt = -k[q – qs)] . . . . . . . . (1)

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

k = [4eσ×θ3o/mc] A . . . . . (2)

Now, dθ/dt = -k[θ – θo]

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

Newton’s Law of Cooling – Temperature vs Time

where,

qi = initial temperature of object,

qf = final temperature of object.

ln (qf – q0)/(qi – q0) = kt

(qf – q0) = (qi – q0) e-kt

qf = q0 + (qi – q0) e -kt . . . . . . (3).

⇒ Check: Heat transfer by conduction

### Methods to Apply Newton’s Law of Cooling

Sometime when we need only approximate values from Newton’s law, we can assume a constant rate of cooling, which is equal to the rate of cooling corresponding to the average temperature of the body during the interval.

i.e. dθ\dt = k(<q> – q0) . . . . . . . (4)

If qi and qf be the initial and final temperature of the body then,

<q> = (q+ qf)/2 . . . . . (5)

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

## Limitations of Newtons Law of Cooling

• The difference in temperature between the body and surroundings must be small,
• The loss of heat from the body should be by radiation only,
• The major limitation of Newton’s law of cooling is that the temperature of surroundings must remain constant during the cooling of the body.

## 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. Find how much more time will it take for the body to attain a temperature of 30ºC.

Solution:

From Newtons law of cooling, qf = qi e-kt

Now, for the interval in which temperature falls from 40 to 35oC.

(35 – 20) = (40 – 20) e-k.10

e-10k = 3/4

k = [ln 4/3]/10 . . . . (a)

Now, for the next interval;

(30 – 20) = (35 – 20)e-kt

e-kt = 2/3

kt = ln 3/2 . . . . (b)

From equation (a) and (b);

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

Aliter : (by approximate method)

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

<q> = (40 + 35)/2 = 37.5ºC

From equation (4);

dθ/dt = k(<q> – q0)

(35 – 40)/10 = k(37.5 – 20)

k = 1/32 min-1

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

<q> = (35 + 30)/2 = 32.5oC

From equation (4);

(30 – 35)/t = (32.5 – 20)

Therefore, the required time t = 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.

Solution:

Given:

Temperature of oil after 10 min = 50oC,

• Ts = 25oC,
• To = 70oC,
• t = 6 min

On substituting the given data in Newton’s law of cooling formula, we get;

T(t) = Ts + (Ts – To) e-kt

[T(t) – Ts]/[To – Ts] = e-kt

-kt ln = [ln T(t) – Ts]/To – Ts

-kt = [ln 50 – 25]/70 – 25 = ln 0.555

k = – (-0.555/6) = 0.092

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

Time taken is -kt ln e = [ln T(t) – Ts]/[To – Ts]

-(0.092) t = ln 45 – 25/[70 – 25]

-0.092 t = -0.597

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 = 0.056 per min and the surrounding temperature is 25oC?

Solution:

Given:

• Ts = 25oC,
• To = 80oC,
• t = 10 min,
• k = 0.056

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

T(t) = Ts + (To – Ts) × e-kt

= 25 + (80 – 25) × e-0.56 = 25 + [55 × 0.57] = 45.6 oC

Temperature cools down from 80oC to 45.6oC after 10 min.