a bottle of water with temperature 70 F is placed in a refrigerator with an internal temperature of 34 F. After 30 minutes, the temperature of the water has decreased to 52 F
a)Give the cooling constant k(6th decimal place), and find an expression for the temperature of the water as a function of time.
b)What is the temperature of the water after 40 minutes.
c)After what amount of time will the temperature of the water reach 38 F.
I just wanna know the steps of each part and what to plug in.
I think the first part(a) uses :
52-34=(70-34)e^(k*30)
Although, I think "52-34" is wrong.
please help
for part a I got k = -.023105, if this isn't correct, can someone explain what I did wrong?
how do I now find p(t)?
Sure, I'd be happy to help!
a) To find the cooling constant, we can use the formula for exponential decay:
T(t) = T_initial + (T_final - T_initial) * e^(-kt)
where T(t) is the temperature at time t, T_initial is the initial temperature, T_final is the final temperature, and k is the cooling constant.
To find k, we can rearrange the formula as:
(T(t) - T_initial) / (T_final - T_initial) = e^(-kt)
Let's plug in the given values:
T_initial = 70 F
T_final = 52 F
t = 30 minutes
Insert these values into the formula to get:
(52 - 70) / (52 - 34) = e^(-k * 30)
Simplifying it further:
-18 / 18 = e^(-k * 30)
-1 = e^(-k * 30)
Now, we can take the natural logarithm of both sides:
ln(-1) = ln(e^(-k * 30))
ln(e^(-k * 30)) = -k * 30
Using the property of logarithms, ln(e^x) = x:
-k * 30 = ln(-1)
Now, unfortunately, we can't take the natural logarithm of a negative number, so there seems to be a mistake in the original equation. The temperature difference in the numerator can't be negative.
Please double-check the given information or let me know if there are any corrections, and I'll be happy to assist further!
To find the cooling constant k and an expression for the temperature of the water as a function of time, we can use the formula for exponential decay:
T(t) = T0 + (T1 - T0) * e^(-kt)
where:
T(t) is the temperature of the water at time t,
T0 is the initial temperature of the water,
T1 is the final temperature of the water,
k is the cooling constant, and
e is Euler's number, approximately equal to 2.71828.
a) To find the cooling constant k, we can use the given information: T0 = 70 F, T1 = 52 F, and t = 30 minutes:
52 = 70 + (52 - 70) * e^(-k * 30)
Now, we can solve this equation to find the value of k.
b) To find the temperature of the water after 40 minutes, we can use the expression we derived in part a) and substitute t = 40 minutes:
T(40) = 70 + (52 - 70) * e^(-k * 40)
c) To find the amount of time it takes for the temperature of the water to reach 38 F, we can set T(t) = 38 F in the expression from part a):
38 = 70 + (52 - 70) * e^(-k * t)
Now, we can solve this equation to find the value of t.
Sure, let's break down the steps for each part:
a) To find the cooling constant, we can use the formula for exponential decay:
T(t) = T0 + (T1 - T0) * e^(-kt)
where T(t) is the temperature at time t, T0 is the initial temperature, T1 is the final temperature, and k is the cooling constant.
In this case, T0 = 70 F, T1 = 34 F, and t = 30 minutes. Plugging these values into the formula, we get:
52 = 70 + (34 - 70) * e^(-k*30)
We can simplify this equation further by subtracting 70 from both sides:
-18 = -36 * e^(-k*30)
Dividing both sides by -36, we get:
1/2 = e^(-k*30)
To solve for k, we can take the natural logarithm (ln) of both sides:
ln(1/2) = ln(e^(-k*30))
ln(1/2) = -k*30
Now, we can solve for k. Dividing both sides by -30:
k = ln(1/2) / -30
Calculating this value, we obtain the cooling constant k to be approximately -0.023025.
To find an expression for the temperature of the water as a function of time, we can substitute the obtained value of k back into the original equation:
T(t) = 70 + (34 - 70) * e^(-0.023025t)
b) To find the temperature after 40 minutes, we simply plug t = 40 into the equation:
T(40) = 70 + (34 - 70) * e^(-0.023025 * 40)
Evaluating this expression, we can determine the temperature after 40 minutes.
c) To find the time it takes for the temperature to reach 38 F, we set T(t) equal to 38 and then solve for t:
38 = 70 + (34 - 70) * e^(-0.023025t)
Simplifying:
-32 = -36 * e^(-0.023025t)
Dividing both sides by -36:
2/3 = e^(-0.023025t)
Taking the natural logarithm of both sides:
ln(2/3) = -0.023025t
Finally, we solve for t by dividing both sides by -0.023025:
t = ln(2/3) / -0.023025
By evaluating this expression, we can determine the time it takes for the water to reach 38 F.
I hope this explanation helps! Let me know if you have any further questions.
I strongly suggest you listen to the two videos by Khan
In the first he establishes the formula:
T(t) = c e^(-kt) + Ta , where Ta is the ambiant temperature.
your given data will be:
Ta = 34
T(0) = 70
T(30) = 52
your t will be in minutes
The first video results in the basic equation, and you may actually skip it
https://www.khanacademy.org/math/differential-equations/first-order-differential-equations/modeling-with-differential-equations/v/newtons-law-of-cooling
In the second video, it follows your problem quite well, all you have to do is plug in your values
https://www.khanacademy.org/math/differential-equations/first-order-differential-equations/modeling-with-differential-equations/v/applying-newtons-law-of-cooling-to-warm-oatmeal