Question

I need help in a bit of it so much

Answer 1

Given the table:

Let's compute the rate of change of temperature at all times in the table.

The time interval is 5 minutes.

To find the rate of change, apply the formula:

`(dT)/(dt)=(T(t+5)-T(t-5))/((t+5)-(t-5))`

Where t is the time.

We have the following:

• When t = 5:

`(dT)/(dt)=(T(5+5)-T(5-5))/((5+5)-(5-5))=(T(10)-T(0))/(10-0)`

Where:

T(0) = 260.0

T(10) = 234.7

Thus, we have:

`(dT)/(dt)=(234.7-260)/(10)=-2.53`

• When t = 10:

`(dT)/(dt)=(T(15)-T(5))/(10)=(228.9-244.2)/(10)=-1.53`

• When t = 15:

`(dT)/(dt)=(T(20)-T(10))/(20-10)=(225.4-234.7)/(10)=-0.93`

• When t = 20:

`(dT)/(dt)=(T(25)-T(15))/(25-15)=(223.3-228.9)/(10)=-0.56`

• When t = 25:

`(dT)/(dt)=(T(30)-T(20))/(30-20)=(222.0-225.4)/(10)=-0.34`

• When t = 30:

`(dT)/(dt)=(T(35)-T(25))/(35-25)=(221.2-223.3)/(10)=-0.21`

• When t = 35:

`(dT)/(dt)=(T(40)-T(30))/(40-30)=(220.7-222.0)/(10)=-0.13`

• When t = 40:

`(dT)/(dt)=(T(45)-T(35))/(45-35)=(220.4-221.2)/(10)=-0.08`

ANSWER:

AT t = 5: -2.53

At t = 10: -1.53

AT t = 15: -0.93

At t = 20: -0.56

At t = 25: -0.34

At t = 30: -0.21

At t = 35: -0.13

At t = 40: -0.08