Newton’s law of Cooling Calculator, Formula, Calculation With Example

Newton’s law of Cooling Calculator:


Check out the detailed steps to calculate the temperature of an object using Newton’s law of cooling. Follow these rules and guidelines to get the result easily.

Get the ambient temperature, cooling constant, initial temperature and time from the query. Find the negative product exponential by multiplying the cooling coefficient by the cooling time. There are three main mechanisms of heat transfer. They are heat conduction, convection and radiation. Newton’s law of cooling applies to heat conduction, convection.

The cooling time of an item depends on two factors. One of the factors is the difference between the temperature of an object and its surroundings. The greater the difference, the faster the cooling. The second factor is the cooling coefficient, It depends on the engine and the amount of heat transfer.

Newton’s Law of Cooling Formula:

Newton’s law of cooling formula is the temperature of the object at the time T(oc) in degree celsius is equal to the Tambient is the surrounding temperature Tambient(oc) in degree Celsius and addition the T initial(oc) in degree Celsius   and minus the ambient is the surrounding temperature Tambient(oc)  in degree Celsius and multiply the  cooling coefficient e-kt.Hence the newton’s law of cooling formula can be written as

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T(oc)=Tambient(oc) +( T initial(oc) –Tambient(oc) ) * e-kt

K= h * A /C

Where

K=is the cooling coefficient

H=is the heat transfer coefficient

A=is the area of the heat exchange

C=is the heat capacity

Newton’s law of cooling states that the rate of change of temperature of an object is directly proportional to the difference between body temperature and its surroundings.

The Newton’s law of cooling formula is along the lines

T=Tambient +( T initial –Tambient ) * e-kt

Where,

T=is the temperature of the object at the time t

Tambient is the surrounding temperature

T initial is the object temperature

K=is the cooling coefficient

T=is the time of cooling

Example:1

Water is heated to 80c for 18 min .How much would be the temperature if k=0.56 per min and the surrounding temperature is 50c?

Answer:

Given that

Surrounding temperature Tambient =50c

Water temperature T initial = 80c

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Time t =18 min

Cooling coefficient k=0.56

Newton ‘s law of cooling formula is

T=Tambient +( T initial –Tambient ) * e-kt

T=50 +(80-50) * e-0.56

T=50 +(30) *0.57

=50+17.1

=67.1 c

Temperature cools down form  80c to 67.1 c after 18 minutes

Example:2

Water is heated to 70c for 20 min .How much would be the temperature if k=0.86 per min and the surrounding temperature is 30c?

Answer:

Given that

Surrounding temperature Tambient =30c

Water temperature T initial = 70c

Time t =20 min

Cooling coefficient k=0.86

Newton ‘s law of cooling formula is

T=Tambient +( T initial –Tambient ) * e-kt

T=30 +(70-30) * e-0.86

T=30+(40) *0.87

=30+34.8

=64.8c

Temperature cools down form  70c to 64.8 c after 20  minutes

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