# Newton’s Law of Cooling

Newton’s Law of Cooling is the fundamental law that describes the rate of heat transfer by a body to its surrounding through radiation. This law state that the rate at which the body radiate heats is directly proportional to the difference in the temperature of the body from its surrounding, given that the difference in temperature is low. i.e. the higher the difference between the temperature of the body and its surrounding the more heat is lost and the lower the temperature the less heat is lost. Newton’s Law of Cooling is a special case of Stefan-Boltzmann’s Law.

In this article, we will learn about, Newton’s Law of Cooling, Newton’s Law of Cooling Formula, its Derivation, Examples, and others in detail.

## Newton’s Law of Cooling Definition

**Newton **was the first to study the relationship between the heat lost by a body to its surrounding. He states that the more difference in the temperature between the object and its surrounding the more heat is radiated by the body.

**Newton’s Law of Cooling** states that

“The rate of heat loss from a body is directly proportional to the difference in temperature between the body and its surroundings, given that the temperature difference is not large.”

This law is used to explain, why hot water or milk left on a table cools faster than a little warm milk or water left on the table. Newton’s law of cooling helps us to teel the temperature of anybody without actually measuring it, given the initial temperature of the body and the temperature of the surrounding.

**Newton’s Law of Cooling Formula**

Newton’s Law of Cooling Formula is a formula for calculating the temperature of a material as it loses heat to its surrounding through radiation.

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

The rate of loss of heat (**– dQ/dt)** of the body is directly proportional to the difference in temperature** [ΔT = (T _{2 }– T_{1})]** of the body and the surroundings.

We can represent it as,

– dQ/dt ∝ (T_{2 }– T_{1})

– dQ/dt = k(T_{2}– T_{1})where,

kis a proportionality constant

Solving the above differential equation we get,

T(t) = T_{s}+ (T_{o}– T_{s}) e^{-kt}where,

tis the timeT(t)is the temperature of the Body at time tTis the surrounding temperature_{s}Tis the Initial temperature of the body_{o}kis the proportionality constant

**Derivation of Newton’s Law of Cooling**

Newton’s Law of Cooling formula can be derived using the solution of the differential equation. Let a body of mass m, with specific heat capacity s, be at temperature T_{2} and T_{1} is the temperature of the surroundings.

If the temperature falls by a small amount **dT _{2}** in time

**dt**, then the amount of heat lost is,

dQ = ms dT_{2}

Rate of loss of heat is given by,

dQ/dt = ms (dT_{2}/dt)

According to Newton’s law of cooling,

– dQ/dt = k(T_{2}– T_{1})Comparing the above equation

– ms (dT_{2}/dt) = k (T_{2 }– T_{1})

dT_{2}/(T_{2}–T_{1}) = – (k / ms) dt

dT_{2 }/(T_{2 }– T_{1}) = – Kdtwhere,

K = k/m sIntegrating the above equation

log_{e}(T_{2}– T_{1}) = – K t + c

T_{2}= T_{1}+ C’ e^{–Kt}where,

C’ = e^{c}

The relation between the drop in temperature of the body and the time is shown using the cooling graph. The slope of this graph shows the rate of fall of the temperature.

The cooling curve is a graph that shows the relationship between body temperature and time. The rate of temperature fall is determined by the slope of the tangent to the curve at any point. The image added below shows the Temperature drop and time relation.

In general,

T(t) = T_{A}+(T_{H}-T_{A})e^{-kt}where

T(t)is the Temperature at time tTis the Ambient temperature or temp of the surroundings_{A}Tis the temperature of the hot object_{H }kis the positive constant and t is the time

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

At a constant rate of cooling, the rate of cooling is related to the average temperature of the body during the interval then we can calculate the approximate value, using Newton’s Law of Cooling

dθ/dt = k(q – q_{s})where,

qis the temperature of the bodyqis the temperature of the surrounding_{s}

Now if the average temperature of the body is q, where,

q = (q_{i}+ q_{f})/2

**Verification of Newton’s Law of Cooling**

We can easily verify Newton’s Law of Cooling by the experiment described below:

In the experiment, we take a double-walled vessel (V) with water in between the two walls. Inside the double-walled vessel, we take a copper calorimeter (C) containing hot water.

We use two thermometers T_{2} to measure the temperatures of the water in the calorimeter and T_{1 }to measure the temperature_{ }of the hot water between the double walls. After equal intervals of time, both temperatures are noted and a graph between the log_{e} (T_{2}–T_{1}) and time (t) is plotted that appears as a straight line with a negative slope.

## Newton’s Law of Cooling Graph

The graph of Newton’s Law of Cooling is added below, in this graph the log of difference between the two temperatures and the time is shown.

**Limitations of Newton’s Law of Cooling**

Various limitations of Newton’s Law of Cooling are,

- Newton’s Law of Cooling holds true if the temperature difference between the body and the environment is small.
- The heat loss by the body is only in form of the Radiation.
- The temperature of the surroundings must remain constant during the cooling of the body, if not then Newton’s Law of Cooling does not holds true.

**Applications of Newton’s Law of Cooling**

Various applications of Newton’s Law of Cooling are,

- To estimate how long a warm object will take to cool down to a specific temperature.
- To determine the temperature of a drink in a refrigerator after a particular length of time has passed.
- It helps to indicate the time of death by looking at the possible body temperature at the time of death and the current body temperature.

**Read More,**

## Solved Examples **Newton’s Law of Cooling**

**Example 1: A pan filled with hot food cools from 94 °C to 86 °C in 2 minutes when the room temperature is at 20 °C. How long will it take to cool from 71 °C to 69 °C?**

**Solution:**

Average of 94 °C and 86 °C is 90 °C,

- T
_{2}= 90 °C- T
_{1}= 20 °CDrop in tem. of food is 8 °C in 2 minutes.

According to Newton’s law of cooling,

– dQ/dt = k(T_{2 }–T_{1})

8 °C /2 min = k(90 – 20)

4 = k(70)………(1)Average of 69 °C and 71 °C is 70 °C

- T
_{2}= 70 °C- T
_{1}= 20 °CAccording to Newton’s law of cooling,

2 °C /dt = k(70 – 20)……(2)From equation (1) and (2),

Change in time = 0.7 min = =42 secThus, the food will take 42 sec to cool from 71 °C to 69 °C.

**Example 2: A body at a temperature of 40ºC is kept in a surrounding of constant temperature of 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:**

Given,

- q
_{i}= (40 – 20)ºC- q
_{f}= (35 – 20)ºCAccording to Newtons law of cooling

q_{f }= q_{i }e^{-kt}Now, for the interval in which temperature falls from 40 ºC to 35 ºC.

(35 – 20) = (40 – 20) e

^{-(10k)}e

^{-10k}= 3/4-10k = (ln 4/3)

k = 0.2876/10

k = 0.02876

Now using Newon’s Formula again,

(30 – 20) = (35 – 20)e

^{-kt}10 = 15e

^{-kt}e

^{-kt }= 2/3-kt = ln(2/3)

t = 0.40546/k

Using the value of the k,

t = 0.40546/0.02876

t = 14.098 min

Thus, the time taken by body to reach the temp of 30ºC is 14.098 min

**Example 3: The oil is heated to 70 ºC. It cools to 50 ºC after 6 minutes. Calculate the time taken by the oil to cool from 50 ºC to 40 ºC given the surrounding temperature T _{s} = 25 ºC**

**Solution:**

Given,

Temperature of oil after 6 min i.e. T(t) is equal to 50 ºC

- Ambient Temperature T
_{s}= 25 ºC- Temperature of Oil, T
_{o}= 70 ºC- Time to Cool to 50ºC = 6 min
According to Newton’s law of cooling,

T(t) = T

_{s}+ (T_{0}– T_{s}) e^{-kt}{T(t) – T

_{s}}/(T_{o}– T_{s}) = e^{-kt}-kt = ln[(T(t) – T

_{s})/(T_{o}– T_{s})] ………(1)Substitute the values

-kt = ln[(50 – 25)/(70 – 25)]

-k = (ln 0.55556)/6

k = 0.09796

Average Temperature from 50 ºC to 40 ºC is equal to 45 ºC

Againg using Newton’s Law of cooling

-(0.09796)t = ln[(45 – 25)/(70 – 25)]

-0.09796t = ln(0.44444)

0.09796t = 0.81093

t = 0.09796/0.58778 =

8.278 minThus, the time take by oil to cool from 50 ºC to 40 ºC is

8.278 min

**Example 4: Water is heated to 80 ºC for 10 min. How much would be its temperature in degrees Celsius, if k = 0.056 per min and the surrounding temperature is 25 ºC?**

**Solution:**

Given,

- Ambient Temperature T
_{s}= 25 ºC- Temperature of water T
_{0}= 80 ºC- Time for which Water is heated (t) = 10 min
- Value of constant k = 0.056.
According to Newton’s law of cooling,

T(t) = T

_{s}+ (T_{0}– T_{s}) e^{-kt}Substituting the value

T(t)= 25 + (80 – 25)e

^{-(0.056×10)}T(t) = 25 + 55 e

^{-(0.056×10)}T(t) = 25 + 31.42

T(t) = 56.42

After 10 min the temperature of water would be

56.42 ºC.

**FAQs on Newton’s Law of Cooling**

**Q1: What is Newton’s Law of Cooling?**

**Answer:**

Newton’s Law of Cooling states that, “the rate of heat loss by a body is directly proportional to the difference in temperature between the body and its surroundings.”

### Q2: What is Newton’s Law of Cooling Formula?

**Answer:**

The Newton’s Law of Cooling formula states that,

T(t) = T_{s}+ (T_{o}– T_{s}) e^{-kt}

### Q3: What is k in Newton’s Law of Cooling?

**Answer:**

The

kin Newton’s Law of Cooling formula is the constant that depends on the material, i.e. changing the material changes thekin Newton’s Law of Cooling.

**Q4: Why Hot Milk is easier to drink from a Bowl than from a Glass?**

**Answer:**

Bowl has a greater surface area than glass therefore more heat loses to its surroundings in the form of heat radiation through the bowl and thus it is easier for us to drink hot milk from the bowl.

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