Answer:
To determine the time it takes for the coffee to cool to 60°C, 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 denote:
- T(t) as the temperature of the coffee at time t
- T_r as the room temperature (21°C)
- k as the cooling constant
According to Newton's Law of Cooling, we can write the differential equation:
dT/dt = -k(T - T_r)
To solve this differential equation, we need an initial condition. In this case, we know that at t = 0 (when the coffee is poured into the mug), the temperature of the coffee is T(0) = 90°C.
Now we can solve the differential equation to find the time when the coffee temperature reaches 60°C.
Separating variables and integrating, we get:
∫(1 / (T - T_r)) dT = -∫k dt
ln|T - T_r| = -kt + C
Taking the exponential of both sides:
T - T_r = Ce^(-kt)
Applying the initial condition T(0) = 90°C, we have:
90 - 21 = Ce^(0) => C = 69
Therefore, the equation becomes:
T - 21 = 69e^(-kt)
To find the value of k, we can use the information given that after 1 minute, the coffee temperature is 85°C:
85 - 21 = 69e^(-k * 1)
64 = 69e^(-k)
Dividing both sides by 69:
e^(-k) = 64/69
Taking the natural logarithm of both sides:
-k = ln(64/69)
Solving for k:
k ≈ -0.065
Now we can plug in the values into the equation T - 21 = 69e^(-kt) and solve for the time t when the temperature T equals 60°C:
60 - 21 = 69e^(-0.065t)
39 = 69e^(-0.065t)
Dividing both sides by 69:
e^(-0.065t) = 39/69
Taking the natural logarithm of both sides:
-0.065t = ln(39/69)
Solving for t:
t ≈ -ln(39/69) / 0.065
Using a calculator, we find that t ≈ 4.44 minutes.
Therefore, it will take approximately 4.44 minutes before the coffee temperature reaches 60°C and becomes safe to drink.