Newton’s law of cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. So the rate of cooling for a bottle of lemonade at a room temperature of 75°F which is placed into a refrigerator with temperature of 38°F can be modeled by dT dt equals k times the quantity T minus 38 where T(t) is the temperature of the lemonade after t minutes and T(0) = 75. After 30 minutes the lemonade has cooled to 60°F, so T(30) = 60.
A) To the nearest degree, what is the temperature of the lemonade after an additional 30 minutes?
B) To the nearest minute, how long does it take for the lemonade to cool to 55°F?
Using your numbers, you can make the function
T(t) = 75+(38-75)e^(-kt) = 75 - 37e^(-kt)
Now just plug in the point (30,60) to find k.
Then you can find t such that T(t) = 55
To solve these problems, we will use the differential equation given by Newton's law of cooling:
dT/dt = k(T - 38)
We also have an initial condition T(0) = 75, which gives us the initial temperature of the lemonade.
A) To find the temperature of the lemonade after an additional 30 minutes (t = 60 minutes), we need to solve the differential equation.
Let's separate the variables and integrate:
(dT / (T - 38)) = k dt
Integrating both sides gives:
ln(abs(T - 38)) = kt + C
where C is the constant of integration.
To find C, we can use the initial condition T(0) = 75:
ln(abs(75 - 38)) = k(0) + C
ln(37) = C
Now we have the equation:
ln(abs(T - 38)) = kt + ln(37)
Substituting the values T = 60 and t = 30 into the equation, we can solve for k:
ln(abs(60 - 38)) = k(30) + ln(37)
ln(22) = 30k + ln(37)
30k = ln(22) - ln(37)
k = (ln(22) - ln(37)) / 30
Now, using the value of k, we can find the temperature of the lemonade after an additional 30 minutes:
ln(abs(T - 38)) = ((ln(22) - ln(37)) / 30) * 60 + ln(37)
ln(abs(T - 38)) = (ln(22) - ln(37)) * 2 + ln(37)
ln(abs(T - 38)) = ln[(22 - 37)^2 * 37]
abs(T - 38) = (22 - 37)^2 * 37
abs(T - 38) = 15^2 * 37
abs(T - 38) = 675
T - 38 = 675 or T - 38 = -675
T = 713 or T = -637
We discard the negative solution since temperature cannot be negative. Therefore, the temperature of the lemonade after an additional 30 minutes (t = 60 minutes) is approximately 713°F (rounded to the nearest degree).
B) To find how long it takes for the lemonade to cool to 55°F, we can use the same differential equation and initial condition.
We need to solve the differential equation for t when T = 55:
(dT / (T - 38)) = k dt
Let's separate the variables and integrate:
ln(abs(T - 38)) = kt + C
Applying the initial condition T(0) = 75, we can find C:
ln(abs(75 - 38)) = k(0) + C
ln(37) = C
Now we have the equation:
ln(abs(T - 38)) = kt + ln(37)
Substituting T = 55, we can solve for t:
ln(abs(55 - 38)) = k t + ln(37)
ln(17) = 55k + ln(37)
55k = ln(17) - ln(37)
k = (ln(17) - ln(37)) / 55
Now, we can find the time t when T = 55:
ln(abs(T - 38)) = ((ln(17) - ln(37)) / 55) * t + ln(37)
ln(abs(55 - 38)) = ((ln(17) - ln(37)) / 55) * t + ln(37)
ln(17) = ((ln(17) - ln(37)) / 55) * t + ln(37)
ln(17) - ln(37) = ((ln(17) - ln(37)) / 55) * t
t = (55 * (ln(17) - ln(37))) / (ln(17) - ln(37))
Calculating this expression gives t ≈ 179.98 minutes. Rounded to the nearest minute, it takes approximately 180 minutes for the lemonade to cool to 55°F.