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:
[tex]\frac{dT}{dt}=\frac{T(t+5)-T(t-5)}{(t+5)-(t-5)}[/tex]Where t is the time.
We have the following:
• When t = 5:
[tex]\frac{dT}{dt}=\frac{T(5+5)-T(5-5)}{(5+5)-(5-5)}=\frac{T(10)-T(0)}{10-0}[/tex]Where:
T(0) = 260.0
T(10) = 234.7
Thus, we have:
[tex]\frac{dT}{dt}=\frac{234.7-260}{10}=-2.53[/tex]• When t = 10:
[tex]\frac{dT}{dt}=\frac{T(15)-T(5)}{10}=\frac{228.9-244.2}{10}=-1.53[/tex]• When t = 15:
[tex]\frac{dT}{dt}=\frac{T(20)-T(10)}{20-10}=\frac{225.4-234.7}{10}=-0.93[/tex]• When t = 20:
[tex]\frac{dT}{dt}=\frac{T(25)-T(15)}{25-15}=\frac{223.3-228.9}{10}=-0.56[/tex]• When t = 25:
[tex]\frac{dT}{dt}=\frac{T(30)-T(20)}{30-20}=\frac{222.0-225.4}{10}=-0.34[/tex]• When t = 30:
[tex]\frac{dT}{dt}=\frac{T(35)-T(25)}{35-25}=\frac{221.2-223.3}{10}=-0.21[/tex]• When t = 35:
[tex]\frac{dT}{dt}=\frac{T(40)-T(30)}{40-30}=\frac{220.7-222.0}{10}=-0.13[/tex]• When t = 40:
[tex]\frac{dT}{dt}=\frac{T(45)-T(35)}{45-35}=\frac{220.4-221.2}{10}=-0.08[/tex]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
