A cup of coffee at 173 degrees is poured into a mug and left in a room at 72 degrees. After 6 minutes, the coffee is 137 degrees.
Assuming that Newton's Law of Cooling applies:
After how many minutes will the coffee be 100 degrees?
Isaac's law says,
T(t) = Ta + (Ts - Ta)e^kt
where Ta is the room temp
Ts is the temp of the object
and t is the time
T(6) = 72 + (173-72)e^(6k)
137 = 72 + 101e^6k
101e^6k = 65
e^6k = 65/101
6k = ln(65/101)
k = ln(65/101)/6 = -.073455541.. (put in calculator's memory)
so when T(t) = 100
100 = 72 + 101e^((-.073...)(t))
e^((-.073..)t) = 28/101
t = ln(28/101)/-.073..) = appr 17.5 minutes
check my arithmetic
saac's law says,
T(t) = Ta + (Ts - Ta)e^kt
where Ta is the room temp
Ts is the temp of the object
and t is the time
T(6) = 72 + (173-72)e^(6k)
137 = 72 + 101e^6k
101e^6k = 65
e^6k = 65/101
6k = ln(65/101)
k = ln(65/101)/6 = -.073455541.. (put in calculator's memory)
so when T(t) = 100
100 = 72 + 101e^((-.073...)(t))
e^((-.073..)t) = 28/101
t = ln(28/101)/-.073..) = appr 17.5 minutes
check my arithmetic
To determine the time it takes for the coffee to reach 100 degrees, we can use Newton's Law of Cooling, which states that the rate of change of temperature of an object is directly proportional to the difference between the object's temperature and the ambient temperature.
Let's denote:
Tc as the temperature of the coffee at any given time (in degrees)
Ta as the ambient temperature (in degrees)
k as the cooling constant (which depends on the specific conditions)
According to the problem, the initial temperature of the coffee (Tc0) is 173 degrees, the ambient temperature (Ta) is 72 degrees, and after 6 minutes, the coffee temperature (Tc6) is 137 degrees.
Using the given information, we can write the equation as:
(Tc6 - Ta) = (Tc0 - Ta) * e^(-k * 6)
Simplifying the equation, we have:
(137 - 72) = (173 - 72) * e^(-k * 6)
65 = 101 * e^(-6k)
To find the cooling constant (k), we can rearrange the equation as follows:
e^(-6k) = 65/101
Taking the natural logarithm (ln) of both sides of the equation, we have:
-6k = ln(65/101)
Now, we can solve for k:
k = ln(65/101) / -6
Using a scientific calculator or any online calculator, we can determine the value of k, which is approximately -0.017.
Now that we have the value of k, we can determine how long it will take for the coffee to reach 100 degrees. Again using Newton's Law of Cooling:
(100 - 72) = (173 - 72) * e^(-0.017 * t)
Simplifying the equation, we have:
28 = 101 * e^(-0.017 * t)
Dividing both sides of the equation by 101, we have:
0.277 = e^(-0.017 * t)
Taking the natural logarithm (ln) of both sides of the equation, we have:
-0.017 * t = ln(0.277)
Solving for t:
t = ln(0.277) / -0.017
Using a scientific calculator or any online calculator, we can determine the value of t, which is approximately 9.413.
Therefore, it will take approximately 9.413 minutes for the coffee to reach 100 degrees.
To solve this problem, we can use Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the surrounding temperature.
In this case, we know that the initial temperature of the coffee is 173 degrees, the surrounding temperature is 72 degrees, and after 6 minutes, the temperature of the coffee is 137 degrees.
Let's assume that the rate of cooling is given by the constant k. We can set up the following differential equation:
dT/dt = -k(T - Ts)
Where:
dT/dt represents the rate of change of temperature with respect to time,
T represents the temperature of the coffee at any given time,
Ts represents the surrounding temperature, and
k is the cooling rate constant.
To find the value of k, we can use the information given after 6 minutes:
137 = (173 - 72)e^(-6k)
To solve for k, we can divide both sides by 101 (173 - 72) and then take the natural logarithm of both sides:
ln(137/101) = -6k
k ≈ ln(137/101)/-6 ≈ 0.0204
Now that we have the value of k, we can use it to determine the time it takes for the coffee to reach 100 degrees.
Let's set up the equation with the known values:
100 = (173 - 72)e^(-0.0204t)
Divide both sides by 101 (173 - 72):
e^(-0.0204t) = 100/101
Taking the natural logarithm of both sides:
ln(e^(-0.0204t)) = ln(100/101)
-0.0204t = ln(100/101)
Finally, divide both sides by -0.0204 to solve for t:
t ≈ ln(100/101) / -0.0204
t ≈ 59.6 minutes
Therefore, the coffee will be approximately 100 degrees after about 59.6 minutes.