Answer:
A) It will get to a temperature of 125°F at 9:19 PM
B) It will get to a temperature of 150°F at 9:16 PM
C) as time passes temperature approaches the initial temperature of 450°F
Step-by-step explanation:
We are given;
Initial temperature; T_i = 450°F
Room temperature; T_r = 70°F
From Newton's law of cooling, temperature after time (t) is given as;
T(t) = T_r + (T_i - T_r)e^(-kt)
Where k is cooling rate and t is time after the initial temperature.
Now, we are told that After 5 minutes, the temperature is 300°F.
Thus;
300 = 70 + (450 - 70)e^(-5k)
300 - 70 = 380e^(-5k)
230/380 = e^(-5k)
e^(-5k) = 0.6053
-5k = In 0.6053
-5k = -0.502
k = 0.502/5
k = 0.1004 /min
A) Thus, at temperature of 125°F, we can find the time from;
125 = 70 + (450 - 70)e^(-0.1004t)
125 - 70 = 380e^(-0.1004t)
55/380 = e^(-0.1004t)
In (55/380) = -0.1004t
-0.1004t = -1.9328
t = 1.9328/0.1018
t ≈ 19 minutes
Thus, it will get to a temperature of 125°F at 9:19 PM
B) Thus, at temperature of 150°F, we can find the time from;
150 = 70 + (450 - 70)e^(-0.1004t)
150 - 70 = 380e^(-0.1004t)
80/380 = e^(-0.1004t)
In (80/380) = -0.1004t
-0.1004t = -1.5581
t = 1.5581/0.1004
t ≈ 16 minutes.
Thus, it will get to a temperature of 150°F at 9:16 PM
C) As time passes which means as it approaches to infinity, it means that e^(-kt) gets to 1.
Thus,we have;
T(t) = T_r + (T_i - T_r)
T_r will cancel out to give;
T(t) = T_i
Thus, as time passes temperature approaches the initial temperature of 450°F
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