Mr. Currie pours himself some coffee into a paper cup before making his way to the amusement park. The coffee temperature is 350C when the cup is placed on the kitchen counter with room temperature of 20o C. Mr. Currie was called to the phone for last minute arrangements, and his coffee is forgotten. When he finally returns to his coffee 35 minutes later, the temperature had been decreasing by 1.2% per minute.
a) What type of function best models the cooling of a hot liquid?
Explain your choice.
b) What is the mathematical model for this situation? (i.e. - the equation)
c) If the optimal temperature for drinking a hot liquid is 280C, at what time would Mr. Currie have had to return in order to enjoy his cup of coffee
1 - .012 = .988 why did you do this?
According to the Newton's Law of Cooling we have that:
T(t) = Ts + (To - Ts)*e^(-k*t) ;
where:
t is the time in the preferred units (seconds, minutes, hours, etc.)
T(t) is the temperature of the object at time t
Ts is the sorrounding constant temperature
To is the initial temperature of the object
k is a constant to be found
so let's find k
Before I go any further, there is something very wrong with your data
since water boils at 100° C, there is no way that the coffee could have a temperature of 350°
nor can you drink coffee that has a temp of 280°, it would be very hot steam.
check your question and your typing.
its 35° and 20° sorry :$
Looking at this question again, I think they want this done in a much simpler way.
when t = 35, the T(35) the temperature has decreased by 1.2% each minute,
so temp after 35 minutes = 35(1.015^-35
= 20.7853°
to have a temp of 28°
we need
35(1.015)^-t = 28
1.015^-t = .8
take log of both sides
log (1.015^-t) = log .8
-t = log .8/log 1.015
-t = -1498
t = appr 15
let me know if we have to use Newton's Law of Cooking equation.
where did you get 1.015 from in the first equation?
also what kind of function is this?
sorry, that was a typo, should have been 1.012
for 1 + 1.2%
I am getting myself all messed up here, I noticed that I used 1.015 several times instead of 1.012
Almost too deep into the mess to recover, I should have done this:
reduce by 1.2% per minute ---- >1 - .012 = .988
35(.988)^35 = 22.9°
when does 35° become 28° ?
35(.988)^t = 28
.988^t = 28/35 = .8
t log .988 = log .8
t = log .8/log .988 = 18.48 minutes
does that make more sense?
You are dealing with an "exponential function"
a) The cooling of a hot liquid can be modeled using an exponential decay function. This is because the temperature of the coffee decreases over time with a constant rate of change, which is a characteristic of exponential decay.
b) The mathematical model for this situation can be represented by the equation:
\(T(t) = T_0 \cdot e^{kt}\)
Where:
- \(T(t)\) represents the temperature of the coffee at a given time \(t\).
- \(T_0\) is the initial temperature of the coffee.
- \(e\) is the base of the natural logarithm.
- \(k\) is the rate constant, equal to \(-0.012\), as the coffee temperature decreases by \(1.2\%\) per minute.
c) To determine at what time Mr. Currie would have had to return in order to enjoy his cup of coffee at the optimal temperature, we need to set \(T(t)\) equal to 280C and solve for \(t\).
\(280 = 350 \cdot e^{-0.012t}\)
To solve this equation, we can take the natural logarithm of both sides:
\(\ln(280) = \ln(350 \cdot e^{-0.012t})\)
Using the properties of logarithms, we can simplify this equation to:
\(\ln(280) = \ln(350) - 0.012t\ln(e)\)
Since \(\ln(e) = 1\), we have:
\(\ln(280) = \ln(350) - 0.012t\)
Now, we can solve for \(t\) by isolating it on one side of the equation:
\(0.012t = \ln(350) - \ln(280)\)
\(t = \frac{\ln(350) - \ln(280)}{0.012}\)
Evaluating this expression will give us the time at which Mr. Currie would have had to return to enjoy his cup of coffee at the optimal temperature of 280C.