**You might be thinking**, It’s so difficult to derive the formula for Newton’s law of cooling, right?

(As there are so many difficult mathematical steps involved in the derivation of newton’s law of cooling)

Maybe yes, I don’t know.

**Don’t worry,**

I’ll show you the **exact steps** to derive the above mentioned Newton’s law of cooling formula.

Newton’s law of cooling formula

Before moving onto the Newton’s law of cooling formula,

Let’s take a quick look on the statement of Newton’s law of cooling.

**According to Newton’s law of cooling,**

*“The rate of heat lost by a body is **directly proportional** to temperature difference of a body and its surroundings”*

Therefore,

**– dQ / dt ∝ ∆T**

– dQ / dt = k ∆T

– dQ / dt = k (T_{2} – T_{1})

dQ / dt = – k (T_{2} – T_{1})

Where,

dQ / dt = Rate of heat lost by a body

∆T = (T_{2} – T_{1}) = Temperature difference between the body and its surroundings

T_{1} = Temperature of the surroundings

T_{2} = Temperature of the body

k = positive constant which depends upon the area and nature of the surface of the body

### Newton’s law of cooling derivation

Newton’s law of cooling formula can be stated as:

**T _{(t)}**

**= T**

_{s}**+ (T**

_{0}**-T**

_{s}**) e**

^{-Kt}Let’s consider one example in order to derive this above mentioned Newton’s law of cooling formula.

**Example:**

A body having an initial temperature of **T _{0}** cools down gradually when it is left on the table. If the temperature of the surrounding is

**T**, what will be the temperature of the body after time

_{s}**t**?

**Solution:****Given data:**

Initial temperature of the body = T

_{0}

Temperature of the surroundings = T

_{s}

Temperature of the body after time t, T

_{(t)}=?

**According to newton’s law of cooling,**

**– dQ / dt ∝ ∆T**

– dQ / dt = k ∆T

– dQ / dt = k (T

_{2}– T

_{1})

**…………… Equation (1)**

Let,

m = mass of the body at temperature T

_{2}

s = specific heat capacity of the body at temperature T

_{2}

T

_{1}= temperature of the surroundings

If the temperature of the body falls down by a small amount dT

_{2}in time dt then,

**Amount of heat lost can be given by,**

∆Q = ms ∆T

dQ = ms dT

_{2}

**Rate of loss of heat is given by,**

dQ / dt = ms (dT

_{2}/ dt)

**………..…. Equation (2)**

Substituting the value of Equation (2) in Equation (1) we get,

– ms (dT

_{2}/ dt) = k (T

_{2}– T

_{1})

dT

_{2}/ (T

_{2}– T

_{1}) = – k dt / ms

**dT**

_{2}/ (T_{2}– T_{1}) = – K dt**………..…. Equation (3)**

Where,

K = k / ms

T

_{2}= temperature of the body

T

_{1}= temperature of the surroundings

But,

**According to our question,**

T = temperature of the body

T

_{s}= temperature of the surroundings

Substituting the above terms in Equation (3) we get,

dT / (T – T

_{s}) = – K dt

**………..…. Equation (4)**

Now,

The initial temperature of the body is T

_{0}and the temperature of the body after time t is T

_{(t)}.

Therefore, the following limits should be considered:

**Limit for L.H.S:**T

_{0}to T

_{(t)}

**Limit for R.H.S:**0 to t

Therefore,

On integrating Equation (4) with proper limits we get,

On integrating the above equation we get,

On solving the above equation with proper limits we get,

ln { (T_{(t)} – T_{s}) / (T_{0} – T_{s}) } = – K (t – 0)

ln { (T_{(t)} – T_{s}) / (T_{0} – T_{s}) } = – Kt

(T_{(t)} – T_{s}) / (T_{0} – T_{s}) = e^{-Kt}

(T_{(t)} – T_{s}) = (T_{0} – T_{s}) e^{-Kt}

**T _{(t)} = T_{s} + (T_{0} -T_{s}) e^{-Kt}**

Where,

T_{(t)} = Temperature of the body at time **t**

T_{s} = Temperature of the surroundings

T_{0} = Temperature of the body

This formula can be used to calculate the temperature of the body at a given time.

### Newton’s law of cooling problems

**Numerical 1:**

A metal sphere is heated up to 80 °C. In 3 minutes, the sphere cools down up to 66 °C. If the temperature of the surroundings is 20 °C, how much time will the metal sphere take to cool down from 50 °C to 46 °C?

**Solution:****Case 1: Metal sphere cools down from 80 °C to 60 °C**

The average temperature of 80 °C and 66 °C is 73 °C, which is 53 °C above the room temperature.

(i.e. the metal sphere cools up to 14 °C in 3 minutes)

Now,

As per the Newton’s law of cooling,

dT_{2} / (T_{2} – T_{1}) = – K dt

dT_{2} / dt = – K (T_{2} – T_{1})

So we can write as,

Change in temperature / time = K T

14 °C / 3 min = K (53 °C) **…………… Equation (1)****Case 2: Metal sphere cools down from 50 °C to 46 °C**

The average temperature of 50 °C to 46 °C is 48 °C, which is 28 °C above the room temperature.

(i.e. the soup cools up to 4 °C in **t** Time)**Note:** K remains the same in both the cases

Therefore,

Change in temperature / time = K T

4 °C / Time = K (28 °C) **…………… Equation (2)**

On dividing Equation (1) and Equation (2) we get,

(14 °C / 3 min) × (Time / 4 °C) = K (53 °C) / K (28 °C)

(14 × Time) / (3 × 4) = 53 / 28

(7 × Time) / 6 = 53 / 28

Time = (53 × 6) / (28 × 7)

Time = 318 / 196

Time = 318 / 196

Time = 1.6 **minutes**

Time = 96 **seconds**

Therefore, the metal sphere takes **1.6 minutes** to cool down from 50 °C to 46 °C.

**Numerical 2:**

When a hot copper rod having a temperature of 66 °C is placed in a surrounding temperature of 25 °C, it’s temperature falls down to 50 °C in 2 minutes. Calculate the time taken by the copper rod to cool down from 45 °C to 41 °C.

**Solution:****Case 1: Copper rod cools down from 66 °C to 50 °C**

The average temperature of 66 °C and 50 °C is 58 °C, which is 33 °C above the room temperature.

(i.e. copper rod cools up to 16 °C in 2 minutes)

Now,

Using Newton’s law of cooling,

dT

_{2}/ (T

_{2}– T

_{1}) = – K dt

dT

_{2}/ dt = – K (T

_{2}– T

_{1})

So we can write as,

Change in temperature / time = K T

16 °C / 2 min = K (33 °C)

**…………… Equation (1)**

**Case 2:**

**Copper rod cools down from 45 °C to 41 °C**The average temperature of 45 °C to 41 °C is 43 °C, which is 18 °C above the room temperature.

(i.e.copper rod cools up to 4 °C in

**t**Time)

**Note:**K remains the same in both the cases

Therefore,

Change in temperature / time = K T

4 °C / Time = K (18 °C)

**…………… Equation (2)**

On dividing Equation (1) and Equation (2) we get,

(16 °C / 2 min) × (Time / 4 °C) = K (33 °C) / K (18 °C)

(2 × Time) = 33 / 18

Time = 33 / 36

Time = 0.9

**minutes**

Time = 54

**seconds**

Therefore, the copper rod takes

**0.9 minutes**to cool down from 45 °C to 41 °C.

**........**Don’t you think it is easy to remember the formula of Newton’s law of cooling?

(Let me know by leaving a **comment**)

If you want to **read more** about the Newton’s laws,

**You can check here:**

**Newton’s law of cooling**

**Newton’s first law of motion**

Newton’s first law example

**Newton’s second law of motion**

Newton’s second law example

Newton’s second law equation

Definition of newton’s second law

**Newton’s third law of motion**

Newton’s third law example

**Newton’s laws of motion**

How many newton’s laws are there

**Newton’s law of inertia**

Newton’s law of inertia examples

**Newton’s universal law of gravitation**