The temperature of a hard boiled egg varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the temperature of the egg, A is the room temperature, and k is a positive constant. If the egg cools from 100°C to 90°C in 1 minute at a room temperature of 70°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes.
To solve this problem, we need to use Newton's Law of Cooling and apply it to the scenario given.
Newton's Law of Cooling states that the rate of change of temperature of an object is directly proportional to the difference between its temperature and the surrounding temperature. Mathematically, it can be expressed as:
dT/dt = -k(T - A)
Where:
dT/dt represents the rate of change of temperature of the object.
k is a positive constant.
T is the temperature of the object.
A is the surrounding (room) temperature.
In this case, we are given that the egg cools from 100°C to 90°C in 1 minute at a room temperature of 70°C. Let's denote the temperature of the coffee as C (in Celsius) after 4 minutes.
We can set up a differential equation using Newton's Law of Cooling:
dT/dt = -k(T - A)
Given: T(0) = 100°C (initial temperature)
A = 70°C (room temperature)
T(1) = 90°C (temperature after 1 minute)
We can solve this differential equation to find the value of k:
dT/dt = -k(T - A)
dT = -k(T - A) dt
(1/(T - A)) dT = -k dt
Integrating both sides, we get:
ln|T - A| = -kt + C1
To find C1, we use the initial condition T(0) = 100°C:
ln|T(0) - A| = -k(0) + C1
ln|100 - 70| = C1
ln|30| = C1
C1 = ln(30)
Substituting the values back into the equation, we have:
ln|T - A| = -kt + ln(30)
To find k, we use the condition T(1) = 90°C:
ln|T(1) - A| = -k(1) + ln(30)
ln|90 - 70| = -k + ln(30)
ln|20| = -k + ln(30)
k = ln(30) - ln(20)
Now, we have the value of k. To find the temperature of the coffee after 4 minutes, we can use the equation:
ln|T - A| = -kt + ln(30)
Substituting in the given values:
A = 70°C (room temperature)
t = 4 minutes
ln|T - 70| = -(ln(30) - ln(20))(4) + ln(30)
Now, we can solve for T - the temperature of the coffee after 4 minutes - by taking the exponential of both sides:
e^(ln|T - 70|) = e^((ln(30) - ln(20))(4) + ln(30))
Simplifying further:
T - 70 = e^((ln(30) - ln(20))(4)) * e^(ln(30))
T - 70 = (30/20)^4 * 30
T - 70 = (3/2)^4 * 30
T - 70 = (81/16) * 30
T = (81/16) * 30 + 70
Calculating this, we find:
T ≈ 158.44°C
Therefore, the temperature of the coffee after 4 minutes is approximately 158.44°C (rounded to the nearest degree Celsius).