The busy Mother pours herself a coffee into a paper cup before making her way to the amusement park. The coffee temperature is 35 C when the cup is placed on the kitchen counter. The Mother needs to tend to her son and her coffee is forgotten. When the Mother finally returns to her coffee 35 minutes later, the temperature is now 21 C.
a.what type of function best models the cooling of a hot liquid?
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 the mother have had to return in order to enjoy her cup of coffee?
I need to solve this question using math (advanced functions), not physics.
a. The cooling of a hot liquid can be modeled using an exponential decay function.
b. The mathematical model for this situation can be represented by the equation:
T(t) = T0 * e^(-kt)
Where:
T(t) is the temperature of the liquid at time t
T0 is the initial temperature of the liquid
k is the cooling constant
t is the time that has elapsed
c. To find the time when the coffee would have reached the optimal temperature of 28°C, we can set T(t) = 28 and solve for t:
28 = 35 * e^(-35k)
Divide both sides by 35:
0.8 = e^(-35k)
Take the natural logarithm of both sides:
ln(0.8) = -35k
Solve for k:
k = -ln(0.8) / 35
Now we can substitute this value of k in the equation T(t) = 28, and solve for t:
28 = 35 * e^((-ln(0.8) / 35) * t)
Divide both sides by 35:
0.8 = e^((-ln(0.8) / 35) * t)
Take the natural logarithm of both sides:
ln(0.8) = (-ln(0.8) / 35) * t
Solve for t:
t = ln(0.8) / (-ln(0.8) / 35)
t ≈ 35.22 minutes
Therefore, the mother would have to return to her coffee at approximately 35.22 minutes in order to enjoy her cup of coffee at the optimal temperature of 28°C.
a. The cooling of a hot liquid can be modeled using an exponential decay function.
b. The mathematical model for this situation can be represented by the equation: T(t) = T0 * e^(-kt), where T(t) is the temperature of the liquid at time t, T0 is the initial temperature of the liquid, e is the base of natural logarithm (approximately 2.71828), k is a constant representing the rate of cooling, and t is the time elapsed.
c. In order to find the time at which the mother would have to return to enjoy her coffee at the optimal temperature, we need to determine the value of t that corresponds to T(t) = 28°C. However, since the initial temperature of the coffee is 35°C, we need to adjust the equation accordingly.
The adjusted equation would be: 28 = 35 * e^(-kt)
To solve for t, you would need to rearrange the equation and isolate t. Here are the steps to do that:
1. Divide both sides of the equation by 35: (28/35) = e^(-kt)
2. Take the natural logarithm of both sides of the equation to remove the exponential: ln(28/35) = -kt.
3. Finally, solve for t by dividing both sides of the equation by -k: t = ln(28/35)/(-k).
To find the value of t, you would need to know the specific value of the constant k, which represents the rate of cooling. Unfortunately, the problem statement does not provide that information.
recall Newton's Law of Cooling states that
T(t) = T0 - (T0-Ta)e^(-kt)
Unfortunately, you have not specified Ta (the ambient temperature), so all we have so far is
T(t) =35 - (35-Ta)e^(-kt)
and you know that T(35)=21
You can see that the cooler the room is, the faster the coffee will cool.
On the other hand, maybe you are using a linear cooling model, in which case
T(t) = 35 - (35-21)/35 t = 35 - 2/5 t
If so, then you want t such that T(t) = 28