Any help will be greatly appreciated, especially with D. I think I can handle the other ones

Question

I'm starting a new more complex course and this is a more difficult them my previous problems.Can you assist me with solving these? mainly D

Common sense tells us that the coffee will be cooling off and its temperature will decrease and approach the ambient temperature of the room, 69 degrees.

So, the temperature difference between the coffee temperature and the room temperature will decrease to 0.

We will fit the temperature difference data (Table 2) to an exponential curve of the form y = A e^-btbt.

Notice that as t gets large, y will get closer and closer to 0, which is what the temperature difference will do.
So, we want to analyze the data where t = time elapsed and y = C - 69, the temperature difference between the coffee temperature and the room temperature.



y = 89.976 e ^- 0.023 t where t = Time Elapsed (minutes) and y = Temperature Difference (in degrees)

(a) Use the exponential function to estimate the temperature difference y when 25 minutes have elapsed. Report your estimated temperature difference to the nearest tenth of a degree.




(b) Since y = C - 69, we have coffee temperature C = y + 69. Take your difference estimate from part (a) and add 69 degrees. Interpret the result by filling in the blank:

When 25 minutes have elapsed, the estimated coffee temperature is ________ degrees.

(c) Suppose the coffee temperature C is 100 degrees. Then y = C - 69 = ____ degrees is the temperature difference between the coffee and room temperatures.

(d) Consider the equation _____ = 89.976 e^-0.023t where the ____ is filled in with your answer from part (c).

Answer 1

y= 89.976e^-.023t

 

a.  at t =25 mins

y= 89.976e^-.023(25)

  = 50.6 degrees

 

b.  50.6 + 69 = 119.6 degrees

     This is the coffee temperature after 25 mins

 

c.  If c= 100 degrees

     y=c-69= 100-69 =31 degrees

 

d.  31 = 89.976e^-.023t

     31/89.976 = e^-.023t

     ln(31/89.976) = ln(e^-.023t)

     -.023t = ln(31/89.976)

     t = [ln(31/89.976)]/(-.023)

     t = 46.3 min

 

at elapsed time of 46.3 mins the temperature

difference between the room and the coffee

will equal 31 degrees.

 

Hope this helps.  Let me know if you

have any questions.