When hot water or milk is left on a table, it gradually cools. It eventually reaches the temperature of the surroundings. The hot water or milk can cool down by exchanging heat with the environment.
Do you know why the tea in the cup gets colder after some time? During cooling it gives the heat to the environment. Also you may notice it cools faster when its temperature is high and cools slower when its temperature is lower. To describe such phenomena we need to learn about Newton’s law of cooling.
Table of contents
Newton’s law of cooling states that if the temperature of a body is not very different from the surroundings, then the rate of cooling is proportional to the temperature difference between the body and its surroundings.
According to Newton's Law of Cooling, the rate of heat loss through a body is directly proportional to the difference of temperatures between the body and environment.
Where, is Temperature difference between body and surrounding
Where, is the rate of cooling.
Let T be the temperature of the body and To bethe temperature of the surroundings. If the body is cooling down, that is, the heat is flowing out from the body, then the body is at a higher temperature than its surroundings.
Let the temperature difference between the body and its surroundings be small.
From Stefan’s law,we know that the rate of radiating heat is given as follows:
By substituting , we get the following
(where remains constant for a body)
The negative sign denotes that the temperature of the body decreases with time.
The greater the temperature difference between the body and its surrounding, the greater is the rate of cooling.
Therefore, a body can never be cooled to a temperature lesser than its surroundings by radiation.
If a body cools down by radiation from to , we get the following:
θ1 is the initial temperature of the body.
θ2 is the final temperature of the body.
θ0 is the temperature of the surroundings.
The law holds good for small differences in temperature.Let the body is at temperature at any time . Consider the initial temperature of the body and surrounding temperature be and . Let temperature of the body falls by in time .
Using the equation,
Integrating both sides,
Q. An object kept in a large room has a temperature of 25 °C. It takes 12 minutes to cool from 80 °C to 70 °C. How much time will be taken by the same object to cool from 70 °C to 60 °C?
Temperature of the surrounding, θ0=25 °C
Temperature of the body, θ1=80 °C
Time taken to cool, t = 12 minutes
Final temperature after 12 minutes, θ2=70 °C
Q. A liquid having mass is kept warm in a vessel by using an electric heater. The liquid is maintained at and the power supplied by the heater is and the surrounding temperature is . As the heater is switched off , the liquid starts cooling and it was observed that it takes for temperature to fall from to . Calculate the specific heat capacity of the liquid. Assume Newton's law of cooling to be applicable.
A liquid is maintained at by supply heat with power rating . when the switch of heater is turned off, it is observed that the temperature of liquid changes from the to in just 10 sec. If the mass of liquid is 250g and environment temperature is then find the specific heat capacity of liquid.
Let be specific heat, be mean temperature, be the surrounding temperature and be a constant.
From Newton's law of cooling, we know that :
From the question, we can say that power supplied by heater is equal to the rate of heat loss from the liquid to the surroundings [so as to maintain constant temperature]
After the heater is switched off the switch of heater is turn off, rate of loss of heat of liquid :
For fall in temperature from to in ,
Using , we get
Q. A bucket full of hot water cools from to in time , from to in time and from to in time . Then find the relation between , and .
From Newton's law of cooling, we know that,
where is the time taken for cooling.
From the data given in the question, we calculate in every case.
and is the same in all cases,
Therefore, we can say that,
Q. Mass of a liquid A is kept in a cup at a temperature of . When placed in a room having temperature of , it takes for the temperature of the liquid to drop to . Another liquid A having nearly the same density as A and of mass , kept in another identical cup at takes for its temperature to fall to when placed in a room having temperature . If the two liquids at and are mixed in a calorimeter where no heat is allowed to leak, find the final temperature of the mixture. Assume that Newton's law of cooling is applicable for the given temperature ranges.
Same density of two liquids means that they will occupy the same volume in the identical cups. Therefore, the surface area through which the heat leaks is the same for both.
For cooling of liquid A:
For cooling of liquid B, we have:
When the two liquids are mixed, let the final temperature of mixture be
Heat lost by A = Heat gained by B
Q. Newton's law of cooling is a special case of which law?
A. Newton's law of cooling is a special case of Stefan-Boltzmann's Law for small temperature differences.
Q. Why is Newton’s law of cooling important?
A. Newton’s law of cooling explains how fast a hot object can cool down. For example, the rate of cooling of hot water in pipes can be explained by Newton’s law of cooling.
Q. What does the rate of cooling of a body depend upon?
A. Nature of the surface of the body, the area of the body, the temperature difference between body and surroundings.
Q. Why does hot milk cool down faster in a bowl as compared to a glass?
A. Due to the larger surface area of the bowl, heat loss occurs faster in case of the bowl as compared to the glass.