An object is heated to 100 degree C It is left to cool in a

An object is heated to 100 degree C. It is left to cool in a room that has a temperature of 30 degree C. After 5 minutes, the temperature of the object is 80 degree C. a. Use Newton\'s Law of Cooling to find a model for the temperature of the object, T, after t minutes. b. What is the temperature of the object after 20 minutes? c. When will the temperature of the object be 35 degree C? Throughout this chapter, we have been working with models that were given. However we can create functions that model data by observing patterns in scatter plots Figure 3.24 shows scatter plots for data that are exponential, logarithmic.

Solution

the formula for newtons law of cooling is given by

T(t) = Ts + (Ts - To ) e^-kt

where Ts = temperature of surrounding

To = initial temperature of body

t = time taken

k = constant

finding the value of k

80 = 30 + ( 30 - 100 ) e^-k(5)

80 = 30 + (-70)e^-5k

80 - 30 = - 70 e^-5k

-50 =  - 70 e^-5k

e^-5k = 5/7

-5k = -.33647

k = 0.06729

hence, the function becomes

T(t) = 30 + (30 - 100 ) e^-( 0.06729 )t

b) temperature of object after 20 minutes

T(t) = Ts + (Ts - To ) e^-( 0.06729 )*20

T(t) = 30 + ( 30 - 100 ) e^-( 0.06729 )*20

= 30 - 18.2214

= 11.78 degree celcius

c) 35 = 30 + (30 - 100 ) e^-( 0.06729 )t

35 = 30 + (-70 )e^-( 0.06729 )t

5 = -70 )e^-( 0.06729 )t

5/-70 = e^-( 0.06729 )t

-.0714285 = e^-( 0.06729 )t

t = 1.0615

hence, after 1.06 minutes temperature will be 35 degree celcius