4018 PTS
Billymkhoi
HIGH SCHOOL MATHEMATICS 10+5 pts
At the instant that a cake is removed from the oven, the temperature of the cake is 350°F. After 10 minutes, the cake's temperature is 200°F. If the
temperature of the room is 70°F, how much longer will it take for the cake to cool to 90°F? (Round your answer to the nearest 5 minutes.)
a) 15 min
b) 20 min
c) 25 min
d) 30 min
e) 35 min
Please show your work. Thanks~
Check out Newton's law of cooling
To solve this problem, we can use the concept of exponential decay, where the cake's temperature decreases over time. The equation for exponential decay is:
T(t) = T(initial) * e^(-kt)
Where:
T(t) is the temperature at time t
T(initial) is the initial temperature of the cake
k is the decay constant
e is the base of the natural logarithm
First, let's find the value of k. We have two data points: the temperature at the instant the cake is removed from the oven (350°F) and the temperature after 10 minutes (200°F).
T(0) = T(initial) * e^(-k*0) = T(initial)
T(10) = T(initial) * e^(-k*10) = 200°F
Dividing the second equation by the first equation, we get:
200°F / T(initial) = e^(-k*10)
Now, taking the natural logarithm of both sides:
ln(200°F / T(initial)) = -k*10
Solving for k, we have:
k = -ln(200°F / T(initial)) / 10
Substituting the given values: T(initial) = 350°F, we can find the value of k:
k = -ln(200°F / 350°F) / 10
Now, we can use this value of k to find the time it will take for the cake to cool from 350°F to 90°F.
T(t) = T(initial) * e^(-kt)
We want to find t when T(t) = 90°F. Substituting the values into the equation:
90°F = 350°F * e^(-k*t)
Dividing both sides of the equation by 350°F, we get:
90°F / 350°F = e^(-k*t)
To solve for t, take the natural logarithm of both sides:
ln(90°F/350°F) = -k*t
Solving for t:
t = -ln(90°F/350°F) / k
Finally, substitute the value of k that we found earlier:
t = -ln(90°F/350°F) / (-ln(200°F / 350°F) / 10)
Calculating this expression will give you the value of t in minutes. Now we just need to round it to the nearest 5 minutes.
Let's calculate the value of t:
t = -ln(90°F/350°F) / (-ln(200°F / 350°F) / 10)
Using a calculator, the value of t is approximately 30.46 minutes.
Rounding it to the nearest 5 minutes, the answer is 30 minutes.
So, the correct answer is (d) 30 min.