Question 1129730: can someone please assist me in understanding what I am doing wrong:
A pot of boiling soup with an internal temperature of 100° Fahrenheit was taken off the stove to cool in a 65°F room. After fifteen minutes, the internal temperature of the soup was 95°F.
Use Newton's Law of Cooling to write a formula
T(t) that models this situation, where T is the temperature of the soup in degrees Fahrenheit and t is time in minutes.
Here is my work:
For the pot of boiling soup;
A=100-65=35
T(t)= 35 e^ ( ( kt ) ) +65
*From there I isolated the variables, took the ln of each equations and then divided the ln of 30/35
by 15.
The formula that models the cooling of the soup is
T(t)= 35 e^( (-0.0103 ) ) +65
This however keeps being labeled incorrect
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Here are some thoughts as to why you may have the wrong answer
- Scenario #1: You forgot to put a lowercase t in the exponent for the term e^( (-0.0103 ) )
- Scenario #2: You rounded the decimal value for k in the incorrect format (eg: you rounded to 4 decimal places but maybe your teacher wanted 6 instead)
Assuming that only scenario #1 applies, then the temperature T(t) function is approximately:
T(t)= 35*e^(-0.0103*t)+65
You do not need two sets of parenthesis. The asterisk symbols indicate multiplication which may be omitted depending on the computer system your teacher uses.
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As a way to check the answer, let's plug in t = 0 and see what happens
T(t)= 35*e^(-0.0103*t) + 65
T(0)= 35*e^(-0.0103*0) + 65 .... replace every lowercase t with 0; use PEMDAS to simplify
T(0)= 35*e^(0) + 65
T(0)= 35*1 + 65
T(0)= 35 + 65
T(0)= 100
As expected, we get a result of 100. So when the time is t = 0 minutes (aka the starting point), the temperature is T = 100 degrees Fahrenheit. This is part 1 of confirming the answer.
Onto part 2: Plug in t = 15. We should get T = 95
T(t)= 35*e^(-0.0103*t) + 65
T(15)= 35*e^(-0.0103*15) + 65 ... replace every t with 15
T(15)= 35*e^(-0.1545) + 65
T(15)= 35*0.856843492142097 + 65 .... this is approximate
T(15)= 29.9895222249734 + 65
T(15)= 94.9895222249734
If you round to the nearest tenth or the nearest whole number, then you'll get approximately T(15) = 95. Due to rounding error, the 94.9895222249734 should be a lot closer to 95 (or close enough that the calculator can't determine the difference). This concludes showing how after t = 15 minutes, the temperature is now approximately T = 95 degrees F.
These two parts help verify that the formula T(t)= 35*e^(-0.0103*t)+65 is the correct function your teacher wants. Or at least, it's a good approximation of the function.
Side Note: it's unfortunate that T shows up as the temperature and t is the time, which in my opinion is a bit confusing. It would be more helpful to only use t once. If t is the time, then H could represent the temperature (H for heat maybe) making the function to be H(t)= 35*e^(-0.0103*t) + 65. However, your teacher is using uppercase T for temperature so it's best to stick with that so you don't lose points.
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