A freshly brewed cup of coffee has temperature 95°C in a 20°C room. When its temperature is 71°C, it is cooling at a rate of 1°C per minute. When does this occur?
To determine when the coffee reaches a temperature of 71°C, we can use the formula for cooling:
ΔT = (T_initial - T_final) * e^(-kt)
where:
ΔT = change in temperature over time
T_initial = initial temperature
T_final = final temperature
k = cooling constant
t = time (in minutes)
e = base of natural logarithm (approximately 2.71828)
From the given information, we can plug in the values:
ΔT = 95°C - 71°C = 24°C
T_initial = 95°C
T_final = 71°C
k = unknown
t = unknown
Now, let's solve for the cooling constant (k):
24 = (95 - 71) * e^(-k*t)
Divide both sides by 24:
1 = (95 - 71) / 24 * e^(-k*t)
Simplify:
1 = 2.5 * e^(-k*t)
Take the natural logarithm (ln) of both sides to isolate the exponential term:
ln(1) = ln(2.5 * e^(-k*t))
ln(1) = ln(2.5) + ln(e^(-k*t))
Since ln(1) = 0 and ln(e^(-k*t)) = -k*t, we have:
0 = ln(2.5) - k*t
Now, we can substitute the given information:
ln(2.5) = -k*1
Divide both sides by -k:
t = ln(2.5) / -k
Now we need to solve for k. We know that when the coffee is cooling at a rate of 1°C per minute, its temperature is 71°C, so we can substitute these values:
1 = (95 - 71) * e^(-k * t)
1 = 24 * e^(-k * 1)
Divide both sides by 24:
1/24 = e^(-k)
Taking the natural logarithm (ln) of both sides:
ln(1/24) = ln(e^(-k))
ln(1/24) = -k
Now we have a value for k:
k = -ln(1/24)
Using this value for k, we can now substitute it back into the equation for t:
t = ln(2.5) / -k
Plug in the value of k:
t = ln(2.5) / -(-ln(1/24))
Simplify:
t = ln(2.5) / ln(1/24)
Using a calculator, we can calculate the value of t:
t ≈ 1.902
Therefore, the coffee reaches a temperature of 71°C at approximately 1.902 minutes.
To find when the coffee reaches a temperature of 71°C, we can use the concept of rate of change of temperature. Since the coffee is cooling at a rate of 1°C per minute, we know that the difference in temperature between any two points in time will also be 1°C if cooling continues at a constant rate.
The initial temperature of the coffee is 95°C, and we want to find out when it reaches a temperature of 71°C. We can calculate the time it takes by finding the difference in temperature between the initial and target temperatures, and dividing it by the cooling rate.
Temperature difference = 95°C - 71°C = 24°C
Cooling rate = 1°C per minute
Time = Temperature difference / Cooling rate
Time = 24°C / 1°C per minute
Time = 24 minutes
Therefore, it will take 24 minutes for the coffee to cool from 95°C to 71°C at a rate of 1°C per minute.
Recall the differential equation:
dT/dt = -k(T-A)
So, you want
-1 = -k(71-20)
k = 1/(71-20) = 1/51
Now you know that
T(t) = 20 + (95-20)e^-t/51 = 71
t = 20