Understanding Newton’s Law of Cooling

In the year 1701, Sir Isaac Newton discovered Newton’s Law of cooling. It has been observed that liquids, such as tea, coffee, or boiled water, that were initially at a higher temperature, cooled down after some time. In other words, a hot body loses heat until it reaches the temperature of its surroundings. Newton’s Law of cooling provides an explanation for this phenomenon. The law states that the rate at which a body loses heat to its surroundings varies directly to the temperature difference that exists between it and its surrounding environment. In other words, heat is lost more rapidly when there is a large temperature difference between a body and its surrounding environment. 

Formula

According to Newton’s law of cooling, at a given time (t), the temperature of an object (T) is given by,

Tt=Ts+T0–Tse-kt

In this equation,

k is the proportionality constant

T0 is the body’s original temperature

Ts is the surrounding temperature

Deriving the equation

As we know, the rate at which a body loses heat to its surrounding dTdt varies directly to the temperature difference between it and its surrounding environment T-Ts. This can be written as:

dTdtT-Ts

To remove the proportionality symbol, we add a proportionality constant k. Therefore,

dTdt=-kT-Ts

(Since the rate of temperature change decreases with time, therefore, a negative sign has been added here).

Now, let T0 denote the body’s initial temperature, and Tf denote the body’s temperature at any given time t. 

Let us integrate both sides of the above equation as follows:

TfT0dTT-Ts=-kdt

Integrating LHS and RHS, we get,

ln Tf–TsT0–Ts =-kt

Taking exponent on both sides, we get,

Tf–TsT0–Ts=e-kt

On cross multiplication, we get,

Tf–Ts=T0–Tse-kt

Taking Ts to the other side, we get,

Tf=Ts+T0–Tse-kt

The above equation can also be written as

Tt=Ts+T0–Tse-kt

Applications

The law can be applied to the following areas:

• It is used in indicating the time of death of a person. This is done by taking the current body temperature and the temperature that the body had at the time of death. 
• It is used to estimate the time it will take for a hot body to come down to a certain temperature. 
• It is used to estimate the body’s temperature at any particular time after some time has passed by taking into account its original temperature and the temperature of its surroundings.

Limitations 

This law takes certain assumptions or conditions into account. If those conditions are not fulfilled, the law does not stand true. Those conditions are as follows: 

• Heat should only be lost via radiation.
• The difference between a body’s temperature and its surroundings should be small.
• When an object loses heat to its surrounding, or in other words, when an object cools, the temperature of its surrounding should remain constant.

Problems

Example 1:

Suppose the original temperature of a body is 40℃, and it is kept at a surrounding that has a temperature of 20℃. If, in 10 minutes, the body’s temperature falls to 35℃, how much time will the body take to reach a temperature of 30℃?

Solution:

Given:

T0=40℃ (Original temperature) 

TS=20℃ (Temperature of the surrounding) 

T(10)=35℃ (Temperature after 10 minutes)

We know that,

Tt=Ts+T0–Tse-kt

Substituting the given values, we get,

35=20+40-20e-k∙10

Simplifying the above equation, we get,

35=20+20e-k∙10

Subtracting 20 from both sides, we get,

15=20e-k∙10

Dividing both sides by 20, we get,

3/4=e-k∙10

Taking ln on both sides

ln⁡3/4=lne-k∙10

Using lnex=xlne, we get,

ln3/4=-10kln(e)

Since, ln e =1, therefore,

ln3/4=-10k

Solving for k, we get,

k=-ln3/410

Now, for Tt=30℃ and T0=35℃

30=20+35-20e-kt

On further simplification, we get,

e-kt=2/3

Taking ln on both sides and simplifying, we get,

kt=-ln2/3

Substituting the value of k, we get,

t=10ln2/3ln3/4

On simplification, we get,

t=14.09

Therefore, to reach a temperature of 30℃, the body will take 14.09, or approximately 14 minutes.

Example 2:

Suppose the temperature of a body is 80℃. If the surrounding temperature is 25℃, and the proportionality constant is 0.056 per minute, find the body’s temperature after 10 minutes.

Solution:

Given:

T0=80℃ (Initial temperature) 

TS=25℃ (Surrounding temperature) 

k=0.056 (Proportionality constant)

To find:

 T(10)=? (Temperature after 10 minutes)

We know that,

Tt=Ts+T0–Tse-kt

Substituting the given values, we get,

T(10)=25+80-25e-(0.056)(10)

On simplification, we get

T(10)=25+55e-0.56

On substituting the value of e-0.56 and simplifying further, we get,

T(10)=56.42

Therefore, after 10 minutes, the temperature of the body will be 56.42℃.

Conclusion 

Newton’s Law of cooling was given by Sir Isaac Newton in 1701. The law states that the rate at which a body loses heat varies directly to the difference in temperature that exists between it and its surroundings. 

The law can be applied to estimate the time of death of a body, the time taken by an object to reach a certain temperature, the temperature of a body after a certain time, etc. One of the limitations of the law is that it can be applied only when the temperature difference is small and when the temperature of the surroundings remains constant.

According to Newton’s law of cooling, at a given time (t), the temperature of an object (T) is given by

Tt=Ts+T0–Tse-kt

In this equation,

k is the proportionality constant

T0 is the body’s original temperature

Ts is the surrounding temperature.