# Newton’s law of cooling formula

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.

Contents

## 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 (T2 – T1)

dQ / dt = – k (T2 – T1)

Where,

dQ / dt = Rate of heat lost by a body

∆T = (T2 – T1) = Temperature difference between the body and its surroundings

T1 = Temperature of the surroundings

T2 = 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

T(t) = Ts + (T0 -Ts) e-Kt

Let’s consider one example in order to derive this above mentioned Newton’s law of cooling formula.

Example 1:
A body having an initial temperature of T0 cools down gradually when it is left on the table. If the temperature of the surrounding is Ts, what will be the temperature of the body after time t?

Solution:

Given data:

Initial temperature of the body = T0

Temperature of the surroundings = Ts

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 (T2 – T1) …………… Equation (1)

Let,

m = mass of the body at temperature T2

s = specific heat capacity of the body at temperature T2

T1 = temperature of the surroundings

If the temperature of the body falls down by a small amount dT2 in time dt then,

Amount of heat lost can be given by,

∆Q = ms ∆T

dQ = ms dT2

Rate of loss of heat is given by,

dQ / dt = ms (dT2 / dt) ………..…. Equation (2)

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

– ms (dT2 / dt) = k (T2 – T1)

dT2 / (T2 – T1) = – k dt / ms

dT2 / (T2 – T1) = – K dt ………..…. Equation (3)

Where,

K = k / ms

T2 = temperature of the body

T1 = temperature of the surroundings

But,

According to our question,

T = temperature of the body

Ts = temperature of the surroundings

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

dT / (T – Ts) = – K dt ………..…. Equation (4)

Now,

The initial temperature of the body is T0 and the temperature of the body after time t is T(t).

Therefore, the following limits should be considered:

Limit for L.H.S: T0 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) – Ts) / (T0 – Ts) } = – K (t – 0)

ln { (T(t) – Ts) / (T0 – Ts) } = – Kt

(T(t) – Ts) / (T0 – Ts) = e-Kt

(T(t) – Ts) = (T0 – Ts) e-Kt

T(t) = Ts + (T0 -Ts) e-Kt

Where,

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

Ts = Temperature of the surroundings

T0 = 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,

dT2 / (T2 – T1) = – K dt

dT2 / dt = – K (T2 – T1)

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,

dT2 / (T2 – T1) = – K dt

dT2 / dt = – K (T2 – T1)

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)