NEWTON'S LAW OF COOLING PROJECT

CIVIL ENGINEERING MATHEMATICS 2 (BFC 13903)

NEWTON’S LAW OF COOLING

LECTURER NAME : DR AMANI FIRZAH BT MOHD KAMIL

1. 2. 3. 4. 5. 6.

GROUP MEMBERS MUHAMMAD FARID BIN MUSTAFA NURHAFIDZ BIN ABDUL RAHAMAN AMYRAH SYAHYERAH BINTI KEIRUDIN NURUL HIKMAH BINTI ARIS WAN FATIN AFIFAH BINTI WAN MOHD FAUZI NURUL DIANA SYAKILA BINTI MOHAMAD ROSLI

MATRIC NO. DF140075 DF140104 DF140087 DF140022 DF140024 DF140099

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 over time which can be applied to many phenomena in science and engineering, including the discharge of a capacitor and the decay in radioactivity. The graph drawn between the temperature of the body and time is known as cooling curve. The slope of the tangent to the curve at any point gives the rate of fall of temperature. We 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) TS To k t

Temperature at time t Ambient temperature (temp of surroundings) Temperature of hot object at time 0 Positive constant Time

QUESTION  The oil is heated to 60oC. It cools to 50oC after 6 minutes. Find the time taken by the oil to cool from 50oC to 40oC (Surrounding temperature Ts = 25oC) SOLUTION  Given temperature of oil at 60°C T(6) = 50°C T s=25 ° C T o=60 ° C t=6 minutes

 The Newton’s Law of cooling formula is given by : T ( t )=T s + ( T o−T s ) e−kt T ( t ) −T s −kt =e T o−T s

e

−kt ln =ln

−kt=ln

T (t )−T s T o−T s

T ( t ) −T s T o −T s

 Substitute the given value into equation

– k ( 6 )=ln

50−25 60−25

−6 k=−0.336

– k ( t )=ln

T ( t )−T s T o−T s

k=

−0.336 −6

k =0.056

If

Tt

= 45°C (average temperature is considered as temperature decreases from 50°C to

40°C)

The time taken is

– 0.056 ( t )=ln

– k ( t )=ln

45−25 60−25

−0.0056t =−0.5596 t=9.993 minutes

T ( t )−T s T o−T s

CONCLUSION To find the time taken of the oil to cool down from 50°C to 40°C, we use Newton’s Law of Cooling formula : T ( t )=T s + ( T o−T s ) e−kt After finding the value of k, we substitute the value into the equation and got the answer which is 9.993 minutes was taken of the oil to cool down from 50°C to 40°C. REFERENCES   

http://vlab.amrita.edu/?sub=1&brch=194&sim=354&cnt=1 http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html http://www.math.ubc.ca/~cass/courses/m256-7b/heat.pdf

FACTS 



The metal core is heated till 1500oF, T o=1500 °C It cools to 1120oCF after 60 minutes, T(60) = 1120oC



t = 60 minute



Surrounding temperature, Ts = 80oC

IDEAS   

Using the related formula Draw the graph The slope of the tangent

LEARNING ISSUES 1) Set up equation by using all given information. the Newton’s Law of cooling formula : T ( t )=T s + ( T o−T s ) e−kt T ( t ) −T s −kt =e T o−T s

−kt ln e=ln

T (t )−T s T o−T s

−kt=ln

T ( t ) −T s T o −T s

2) Subtitute the given value into equation −kt=ln

T ( t ) −T s T o −T s

3) Finding the value of k and substitute the value into the equation to find t. FILA TABLE