You place a cup of 205oF coffee on a table in a room that is 72oF, and 10 minutes later, it is 195oF. Approximately how long will it be before the coffee is 180oF? Use Newton's law of cooling:
1 hour
25 minutes
15 minutes
45 minutes
As I recall, our model is
T(t) = 205 + (72-205)e^-kt
So, solve for k in T(10) = 195
Then use that k to solve for t in T(t) = 180
What do you get?
To determine the approximate time it will take for the coffee to reach 180oF, we can use Newton's law of cooling. This law states that the rate of change of temperature of an object is directly proportional to the difference between its temperature and the surrounding temperature.
Given:
Initial temperature of the coffee (T0) = 205oF
Temperature of the room (Troom) = 72oF
Temperature of the coffee after 10 minutes (T10) = 195oF
Temperature to be reached (T) = 180oF
We can use the formula:
(T - Troom) = (T0 - Troom) * e^(-kt)
where k is a constant and t is the time in minutes.
Rearranging the formula to solve for t:
t = ln((T - Troom) / (T0 - Troom)) / -k
Let's calculate the value of k first:
(T10 - Troom) = (T0 - Troom) * e^(-k * 10)
(195 - 72) = (205 - 72) * e^(-k * 10)
123 = 133 * e^(-10k)
e^(-10k) = 123 / 133
-10k = ln(123 / 133)
k ≈ -0.0754
Now we can substitute the values into the formula:
t = ln((T - Troom) / (T0 - Troom)) / -k
t = ln((180 - 72) / (205 - 72)) / -0.0754
t ≈ ln(108 / 133) / 0.0754
t ≈ 0.4312 / 0.0754
t ≈ 5.71 hours
Approximately, it will take around 5.71 hours for the coffee to reach 180oF. Therefore, the closest option is 1 hour.
To answer this question, we can use Newton's law of cooling, which describes the rate of temperature change for an object in a different environment. According to this law, the rate of change of temperature is proportional to the difference between the temperature of the object and the temperature of its surroundings.
The general formula for Newton's law of cooling is:
dT/dt = -k(T - Ts)
Where:
- dT/dt is the rate of change of temperature with respect to time,
- k is the cooling constant,
- T is the temperature of the object,
- Ts is the temperature of the surroundings.
First, we need to find the cooling constant (k) for the cup of coffee. To do this, we can use the initial temperature of the coffee (205°F), the final temperature after 10 minutes (195°F), and the temperature of the room (72°F).
Using the formula above, we can rearrange it to solve for the cooling constant (k):
k = -(dT/dt) / (T - Ts)
k = -(195°F - 205°F) / (10 minutes)
k = -10°F / 10 minutes
k = -1°F/minute
Now that we have the cooling constant, we can use it to find approximately how long it will take for the coffee to reach 180°F.
Using the same formula, but this time solving for time (t):
t = -(1/k) * ln((Tf - Ts) / (Ti - Ts))
Where:
- t is the time it takes for the object to cool,
- Tf is the final temperature we want (180°F),
- Ti is the initial temperature (205°F),
- Ts is the temperature of the surroundings (72°F).
Plugging in the values:
t = -(1/(-1°F/minute)) * ln((180°F - 72°F) / (205°F - 72°F))
t ≈ 41.8 minutes
So approximately, it will take about 41.8 minutes for the coffee to reach 180°F. Since this is not one of the given answer choices, the closest option is 45 minutes.