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CIVIL ENGINEERING MATHEMATICS 2

(BFC 13903)

NEWTON’S LAW OF COOLING

  LECTURER NAME : DR AMANI FIRZAH BT MOHD KAMIL

GROUP MEMBERS MATRIC NO.

1. MUHAMMAD FARID BIN MUSTAFA DF100!"

2. NURHAFIDZ BIN ABDUL RAHAMAN DF1010

3. AM#RAH S#AH#ERAH BINTI KEIRUDIN DF100$!

. NURUL HIKMAH BINTI ARIS DF10022

". %AN FATIN AFIFAH BINTI %AN MOHD FAUZI DF1002

&. NURUL DIANA S#AKILA BINTI MOHAMAD ROSLI DF10099

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INTRODUCTION

Temperature difference in any situation results from energy flow into a system or energy flow

from a system to surroundings. The former leads to heating, whereas latter leads to cooling of 

an object. Newton’s Law of Cooling states that the rate of temperature of the body is proportional to the difference between the temperature of the body and that of the surrounding

medium. This statement leads to the classic equation of exponential decline oer time which

can be applied to many phenomena in science and engineering, including the discharge of a

capacitor and the decay in radioactiity.

The graph drawn between the temperature of the body and time is !nown as cooling cure.

The slope of the tangent to the cure at any point gies the rate of fall of temperature. 

"e can

say that the temperature of the body approaches that of its surroundings as time goes.

This equation represents Newton’s law of cooling.

T  (t )=T s+(T o−T s ) e−kt 

T#t$ Temperature at time t

T%  &mbient temperature #temp of surroundings$

To Temperature of hot object at time '! (ositie constant

t Time

 

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'UESTION

The oil is heated to )'oC. *t cools to +'oC after ) minutes. ind the time ta!en by the oil

to cool from +'oC to -'oC #%urrounding temperature Ts  /+oC$

SOLUTION

0ien temperature of oil at )'1C T#)$ +'1C

T s=25°C 

T o=60 ° C 

t =6minutes

The Newton’s Law of cooling formula is gien by 2

T  (t )=T s+(T o−T s ) e−kt 

T  ( t )−T s

T o−T s

=e−kt 

−kt lne=ln

T  (t )−T s

T o−T s

−kt =lnT  ( t )−T 

s

T o−T s

%ubstitute the gien alue into equation  – k ( t )=lnT ( t )−T s

T o−T s

 – k (6 )=ln 50−25

60−25

−6 k =−0.336

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k =−0.336

−6

k =0.056

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*fT 

t    -+1C #aerage temperature is considered as temperature decreases from +'1C to

-'1C$

The time ta!en is  – k ( t )=ln

T ( t )−T s

T o−T s

 – 0.056 ( t )=ln 45−25

60−25

−0.0056 t =−0.5596

t =

9.993minutes

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CONCLUSION

To find the time ta!en of the oil to cool down from +'1C to -'1C, we use Newton’s Law of 

Cooling formula 2

T  (t )=T s+(T o−T s )e−kt 

&fter finding the alue of !, we substitute the alue into the equation and got the answer 

which is 3.334 minutes was ta!en of the oil to cool down from +'1C to -'1C.

REFERENCES

• http255lab.amrita.edu56sub78brch73-8sim4+-8cnt7

• http255www.ugrad.math.ubc.ca5coursedoc5math7''5notes5diffeqs5cool.html

• http255www.math.ubc.ca59cass5courses5m/+):;b5heat.pdf 

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&CT% *<=&% L=&>N*N0 *%%?=%

• The metal core is heated

till 7+''o,

T o=1500 ° C 

• *t cools to 77/'oC after

)' minutes, T#)'$

77/'oC

• t )' minute

• %urrounding temperature,

Ts  @'oC

• ?sing the related formula

• <raw the graph

•The slope of the tangent

7$ %et up equation by using all gien

information. the Newton’s Law of

cooling formula 2

 T  (t )=T s+(T o−T s )e

−kt 

 

T  ( t )−T s

T o−T s=e

−kt 

  −kt ln e=lnT 

 (t 

)−T 

s

T o−T s

  −kt =lnT  ( t )−T 

s

T o−T s

/$ %ubtitute the gien alue into equation

 −

kt =

ln

T  ( t )−T s

T o−T s

4$ inding the alue of ! and substitute

the alue into the equation to find t.

FILA TABLE