**Simply think, **

What happens when a cup of **hot tea** is left on the table?

The answer is simple.

It cools down gradually and attains the temperature of surroundings.

In other words,

When a cup of hot tea is left on the table,

It gradually loses its temperature and reaches the temperature of surroundings.

See this,

As you can see,

The temperature of the hot tea gradually decreases and reaches the temperature of surroundings.

**Now the question is,**

Is there any way to know **how much time **the cup of hot tea takes to reach the temperature of surroundings?

This question can be understood by Newton’s law of cooling.

In the 17^{th} century,

Isaac Newton studied the cooling of the bodies and formulated the law of cooling.

**Newton’s law of cooling states that:**

*“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})

By this formula of Newton’s law of cooling, different numericals can be solved. (Which we’ll see later)

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 experiment

You already know,

When a cup of **hot tea** is left on the table, it cools down slowly.

(According to newton’s law of cooling)

Now the question is,

**How does the body cool down** by exchanging heat with the surroundings?

To understand this question, let’s perform one simple experiment.

- Take some water (say 300 ml) in a calorimeter.

- Cover the calorimeter with a lid having two holes.

- Fix the thermometer inside the calorimeter, such that the bulb of the thermometer is dipped inside the water.

- Mark the reading of the thermometer as T
_{1}. (Which is the temperature of surroundings)

- Heat the water filled inside the calorimeter above the surrounding temperature.

- After some time, remove the heat source.

- Note the readings of the thermometer by mixing the water after every one minute.

- Take readings until the temperature of water T
_{2}reaches about 5 °C. (Approx value)

Finally,

- Plot the graph of
**∆T**(along the y-axis) and**t**(along the x-axis).

From this graph,

You will understand how the cooling of hot water depends on the difference of its temperature from that of surroundings.

You’ll notice that, *cooling rate of the body keeps on decreasing as the temperature drops down.*

The above activity shows that a hot body loses heat to its surroundings in the form of **heat radiation.**

Therefore,

**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”*

## Newton’s law of cooling – Assumptions

The following assumptions are made in newton’s law of cooling:

- Temperature difference between the body and its surroundings must be
**small.**

- The loss of heat from a body must occur through
**radiation**only.

- The temperature of the surroundings must remain
**constant**while the cooling of a body.

## Newton’s law of cooling – Limitations

The following are the limitations for newton’s law of cooling:

- Newton’s law of cooling is not applicable for
**large temperature differences.**

- Newton’s law of cooling is not applicable for bodies having
**high temperatures.**

## Verification of Newton’s law of cooling

Newton’s law of cooling can be verified by using the experimental setup shown below.

The setup consists of,

- Double walled vessel containing water in between the two walls.

- One copper calorimeter containing hot water is placed inside the double walled vessel.

- Two thermometers are used to measure the temperatures of water.

- Mark
**T**as the temperature of water kept in the calorimeter._{2}

- Mark
**T**as the temperature of hot water in between the double walls._{1}

Temperature of hot water inside the calorimeter is noted at equal intervals of time.

## Newton’s law of cooling graph

By measuring the temperature of water at equal intervals of time,

You’ll get the following graph.

As you can see,

When a graph is plotted between **log**_{e}** (T**_{2}** – T**_{1}**)** and time **t**,

A straight line graph having a negative slope is obtained as shown in the figure.

## Newton’s law of cooling differential equation

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**

Where,

dQ / dt = Rate of heat lost by a body

∆T = (T_{2} – T_{1}) = Temperature difference

Using above relation,

– dQ / dt ∝ ∆T

– dQ / dt = k ∆T

– dQ / dt = k (T_{2} – T_{1}) **…………… Equation (1)**

Where,

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

Let,

m = mass of a body at temperature T_{2}

s = specific heat capacity at temperatures T_{2}

T_{1} = temperature of the surroundings

If the temperature falls by a small amount dT_{2} in time dt, then the amount of heat lost,

∆Q = ms ∆T

dQ = ms dT_{2}

Therefore, 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**; where k / ms = K

Integrating on both sides we get,

log_{e} (T_{2} – T_{1}) = – Kt + c

T_{2} – T_{1} = e^{(- Kt +c)}

T_{2} – T_{1} = e^{-Kt} e^{c}

T_{2} – T_{1} = e^{-Kt} C^{‘}

**T _{2}**

**= T**

_{1}**+ C**

^{‘}**e**; where C

^{-Kt}^{‘}= e

^{c}

By this equation of Newton’s law of cooling,

You can calculate the time of cooling of a body in a particular range of temperature.

## Newton’s law of cooling problems

**Numerical 1:**

A hot cup of soup cools down from 90 °C to 82 °C in 4 minutes when placed on the table. If the temperature of the surroundings is 25 °C, How much time will the soup take to cool down from 65 °C to 61 °C?

**Solution:****Case 1: When soup cools down from 90 °C to 82 °C**

The average temperature of 90 °C and 82 °C is 86 °C, which is 61 °C above the room temperature.

(i.e. the soup cools up to 8 °C in 4 minutes)

Now,

According to the formula of 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

8 °C / 4 min = K (61 °C) **…………… Equation (1)****Case 2: When soup cools down from 65 °C to 61 °C**

The average temperature of 65 °C to 61 °C is 63 °C, which is 38 °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 (38 °C) **…………… Equation (2)**

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

(8 °C / 4 min) × (Time / 4 °C) = K (61 °C) / K (38 °C)

Time / 2 = 61 / 38

Time = (61 × 2) / 38

Time = 122 / 38

Time = 3.2 **minutes**

Time = 192 **seconds**

Therefore, the hot cup of soup takes **3.2 minutes** to cool down from 65 °C to 61 °C.

**Numerical 2:**

A hot oil drum cools down from 80 °C to 70 °C in 2 minutes. How much time will it take to cool from 60 °C to 54 °C if the surrounding area has a temperature of 20 °C?

**Solution:****Case 1: When the oil drum cools down from 80 °C to 70 °C**

The average temperature of 80 °C and 70 °C is 75 °C, which is 55 °C above the room temperature.

(i.e. the oil drum cools up to 10 °C in 2 minutes)

Now,

According to Newton’s law of cooling equation,

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

10 °C / 2 min = K (55 °C)

**…………… Equation (1)**

**Case 2: When**

**the oil drum cools down from 60 °C to 54 °C**The average temperature of 60 °C to 54 °C is 57 °C, which is 37 °C above the room temperature.

(i.e. the oil drum cools up to 6 °C in

**t**Time)

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

Therefore,

Change in temperature / time = K T

6 °C / Time = K (37 °C)

**…………… Equation (2)**

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

(10 °C / 2 min) × (Time / 6 °C) = K (55 °C) / K (37 °C)

(5 × Time) / 6 = 55 / 37

Time = (55 × 6) / (37 × 5)

Time = 330 / 185

Time = 1.7

**minutes**

Time = 102

**seconds**

Therefore, the oil drum takes

**1.7 minutes**to cool down from 60 °C to 54 °C.

**Numerical 3:**

A hot metal rod cools down from 95 °C to 81 °C in 3 minutes. How much time will it take to cool from 76 °C to 70 °C if the temperature of the surrounding area is 24 °C?

**Solution:****Case 1: When the metal rod cools down from 95 °C to 81 °C**

The average temperature of 95 °C and 81 °C is 88 °C, which is 64 °C above the room temperature.

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

Now,

According to Newton’s law of cooling equation,

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 (64 °C)

**…………… Equation (1)**

**Case 2: When**

**the****metal rod cools down from 76 °C to 70 °C**The average temperature of 76 °C to 70 °C is 73 °C, which is 49 °C above the room temperature.

(i.e. metal rod cools up to 6 °C in

**t**Time)

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

Therefore,

Change in temperature / time = K T

6 °C / Time = K (49 °C)

**…………… Equation (2)**

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

(14 °C / 3 min) × (Time / 6 °C) = K (64 °C) / K (49 °C)

(14 × Time) / 18 = 64 / 49

Time = (64 × 18) / (49 × 14)

Time = 1152 / 686

Time = 1.6

**minutes**

Time = 96

**seconds**

Therefore, a hot metal rod takes

**1.6 minutes**to cool down from 76 °C to 70 °C.

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

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

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