If a hot cup of coffee, initially at 190◦F, cools to 125◦F in 5 minutes when places in a room with a constant temperature of 75◦F, how long will it take the coffee to reach 100◦F.
I do not know how to set up the problem correctly, I have tried multiple times but always end up with the wrong answer. I know the formula but not sure about what value goes where
so, which formula are you using? You expect help, but do not show what you have tried so far. If using Newton's law of cooling, then you have
T(t) = 75+(190-75)e^(-t/r) = 75 + 115e^(-t/r)
You need to find r. Since T(5) = 125,
75 + 115e^(-5/r) = 125
so r = 5/log(23/10) = 6
That means T(t) = 75 + 115e^(-t/6)
and T will bee 100 when
75 + 115e^(-t/6) = 100
t = 6log(23/5) = 9.156 minutes
google can provide further explanations and examples of this formula.
Assuming that you will be given the formula based on Newton's Law of Cooling, which is
T(t) = c e^(-kt) + Ta, where T is temperature, t is time, and Ta is the ambiant tempereature, c and k are constants.
(you titled your post as Pre Calculus, so I assume you don't have to arrive at this formula)
given:
Ta = 75
T(0) = 190
T(5) = 125
we want T(t) = 100
T(0) = c e^0 + Ta
190 = c + 75
c = 115
so T(t) = 115 e^(-kt) + 75
when t = 5, T(5) = 125
125 = 115 e^(-5k) + 75
e(-5k) = 50/115 = .43478..
using ln and log rules
-5k = ln .43478...
k = .16658...
T(t) = 115 e^(-.16658..t) + 75
100 = 115 e^(-.16658..t) + 75
e^(-.16658..t) = 25/115 = .21739...
solving this for t , once again using natural logs, I got
t = 9.16
check my calculations
ahhh, I agree with oobleck, it must be correct
To solve this problem, we can use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its current temperature and the surrounding temperature.
Let's define some variables:
- To: Initial temperature of the coffee (190◦F).
- T: Current temperature of the coffee.
- Tr: Room temperature (75◦F).
- k: The constant of proportionality.
Using Newton's Law of Cooling, we can express the rate of change of temperature as follows:
ΔT/Δt = -k * (T - Tr),
where ΔT/Δt represents the rate of cooling (change in temperature with respect to time).
Now, let's use the information given in the problem to set up an equation:
When the coffee cools from 190◦F to 125◦F in 5 minutes:
ΔT = 125 - 190 = -65,
Δt = 5.
Substituting these values into the equation, we get:
-65/5 = -k * (125 - 75).
Simplifying:
-13 = -k * 50.
Divide both sides by -50:
k = 13/50.
Now that we have the value of k, we can find the time it takes for the coffee to reach 100◦F. We'll let T represent the current temperature (100◦F) and solve for t (time):
ΔT/Δt = -k * (T - Tr).
Substituting in the values:
(100 - 75)/t = -13/50 * (100 - 75).
Simplifying:
25/t = -13/50 * 25.
Dividing by -13/50:
t = -25 / (-13/50).
Simplifying further, we get:
t = 25 * (50/13).
Evaluating this expression, we find that t ≈ 96.15 minutes.
Therefore, it will take approximately 96.15 minutes for the coffee to cool from 190◦F to reach 100◦F.