The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT/dt=-k(T-A), where T is the water temperature, A is the room temperature, and k is a positive constant.
If the water cools from 90°C to 85°C in 1 minute at a room temperature of 30°C, how long (to the nearest minute) will it take the water to cool to 60°C?
3
4
5 <- my answer
8
Oops.
ln(T-A) = -kt + ln(c)
T-A = c*e^(-kt)
T = A + c*e^(-kt)
T(0) = 90, so
30 + c = 90
c = 60
T(t) = 30 + 60e^(-kt)
T(1) = 85, so
30 + 60e^-k = 85
e^-k = 55/60
-k = ln(55/60)
k = 0.087
T(t) = 30 + 60e^(-.087t)
we want when T=60:
30 + 60e^(-.087t) = 60
e^(-.087t) = 1/2
-.087t = -ln2
t = ln2/.087 = 7.96
Looks like 8 to me. How did you arrive at 5?
dT/dt = -k(T-A)
dT/(T-A) = -k dt
ln(T-A) = -kt
T-A = e^(-kt)
T = A + e^(-kt)
Now just plug in your numbers to find k and then solve for when T=60
To solve this problem, we can use the differential equation provided by Newton's Law of Cooling:
dT/dt = -k(T - A)
Given that the water cools from 90°C to 85°C in 1 minute at a room temperature of 30°C, we can substitute the known values into the equation:
dT/dt = -k(T - A)
dT/dt = -k(90 - 30)
dT/dt = -60k
Now, we can solve the differential equation using separation of variables. Let's integrate both sides:
∫ 1 dT = ∫ -60k dt
ln|T| = -60kt + C
Next, we need to find the value of the constant C. We can use the initial condition where T = 90°C at t = 0. Substituting these values into the equation:
ln|90| = -60k(0) + C
ln|90| = C
So, the equation becomes:
ln|T| = -60kt + ln|90|
Now, let's solve for k. We can use the information that the water cools from 90°C to 85°C in 1 minute:
ln|85| = -60k(1) + ln|90|
Using a calculator, we can evaluate the natural logarithms:
4.442651 = -60k + 4.49981
Simplifying the equation:
-60k = 4.442651 - 4.49981
-60k = -0.057159
k = 0.000952652
Now that we know the value of k, we can find the time it takes for the water to cool to 60°C. Let's substitute T = 60 and solve for t:
ln|60| = -60(0.000952652)t + ln|90|
Using a calculator again:
4.094345 = -0.057159t + 4.49981
Simplifying:
-0.057159t = 0.405465
t = -0.405465 / -0.057159
t ≈ 7.115
Rounding to the nearest minute, it will take approximately 7 minutes for the water to cool to 60°C.
Therefore, the correct answer is option 8.
To find out how long it will take for the water to cool to 60°C, we need to use the differential equation given by Newton's Law of Cooling:
dT/dt = -k(T - A)
Given:
Initial temperature (T0) = 90°C
Final temperature (T) = 60°C
Room temperature (A) = 30°C
Let's first find the value of k. We can use the information given in the problem to calculate it.
dT/dt = -k(T - A)
Rearranging the equation, we get:
dT/(T - A) = -k dt
Integrating both sides of the equation, we have:
∫ dT/(T - A) = -∫ k dt
ln|T - A| = -kt + C
Using the initial condition T0 = 90°C at time t = 0 minutes:
ln|90 - 30| = -k(0) + C
ln(60) = C
Substituting this value of C back into the equation, we have:
ln|T - A| = -kt + ln(60)
Now, let's use the information that the water cools from 90°C to 85°C in 1 minute to find the value of k.
At t = 0 minutes, T = 90°C:
ln|90 - 30| = -k(0) + ln(60)
ln(60) = ln(60)
At t = 1 minute, T = 85°C:
ln|85 - 30| = -k(1) + ln(60)
ln(55) = -k + ln(60)
Now we can solve these two equations to find the value of k:
ln(55) = -k + ln(60)
k = ln(60) - ln(55)
Now that we have the value of k, we can solve for the time it will take for the water to cool to 60°C.
Using the equation ln|T - A| = -kt + ln(60), we substitute T = 60°C, A = 30°C, and the calculated value of k:
ln|60 - 30| = -k t + ln(60)
ln(30) = -(ln(60) - ln(55)) t + ln(60)
Now, we can solve for t by isolating it on one side of the equation:
ln(30) - ln(60) = -(ln(60) - ln(55)) t
Now, we can calculate t by rearranging the equation as follows:
t = (ln(60) - ln(55))⁻¹ * (ln(30) - ln(60))
Calculating the value of t, we get:
t ≈ 4 minutes
Therefore, it will take approximately 4 minutes for the water to cool from 90°C to 60°C at a room temperature of 30°C.