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
8
minutes
To find out how long it will take the water to cool to 60°C, we need to use Newton's Law of Cooling and solve the differential equation dT/dt = -k(T - A).
Given information:
Initial temperature (T_0) = 90°C
Final temperature (T) = 60°C
Room temperature (A) = 30°C
We can rearrange the equation as follows:
dT/dt = -k(T - A)
dT/(T - A) = -k dt
We integrate both sides of the equation with respect to T and t:
∫ dT/(T - A) = -k ∫ dt
ln|T - A| = -kt + C
To find the constant of integration C, we can use the initial condition when t = 0 and T = 90°C:
ln|90 - 30| = -k(0) + C
ln|60| = C
C = ln|60|
Now we can substitute the final condition when t = t_1 and T = 60°C:
ln|60 - 30| = -k t_1 + ln|60|
ln|30| = -k t_1 + ln|60|
Simplifying further:
ln|30| - ln|60| = -k t_1
Using logarithm properties, we can rewrite the equation as:
ln(30/60) = -k t_1
ln(1/2) = -k t_1
- ln(2) = -k t_1
Now we solve for t_1:
t_1 = ln(2)/k
Since k is a positive constant, we can approximate ln(2) as 0.693.
t_1 ≈ 0.693/k
Now we need to determine the value of k. We can use the information that the water cools from 90°C to 85°C in 1 minute.
dT/dt = -k(T - A)
(85 - 90)/1 = -k(90 - 30)
-5 = -k(60)
k = -5/60
k = -1/12
Now we substitute the value of k into the equation for t_1:
t_1 ≈ 0.693/(-1/12)
t_1 ≈ -8.316 (approximately)
Since time cannot be negative, we take the absolute value:
t_1 ≈ 8.316
Rounding to the nearest minute, it will take the water approximately 8 minutes to cool to 60°C.
Therefore, the answer is 8.
To find out how long it will take for the water to cool to 60°C, we need to solve the differential equation using the given information.
The differential equation is given as:
dT/dt = -k(T - A)
We know that the initial temperature of the water, T(0) = 90°C, and after 1 minute, the temperature of the water has cooled to 85°C, i.e., T(1) = 85°C. The room temperature is A = 30°C.
To find the value of k, we can use the given information and solve for k:
dT/dt = -k(T - A)
=> (dT/dt)/(-k) = T - A
=> (1/(-k))(dT/dt) = T - A
Now, let's integrate both sides of the equation with respect to t:
∫(1/(-k))(dT/dt) dt = ∫(T - A) dt
=> (-1/k) ∫dT = ∫(T - A) dt
Integrating both sides:
(-1/k) * (T - A) = (1/2)T^2 - AT + C
where C is the constant of integration.
Let's substitute the values of T and t at the two given points:
At t = 0, T = 90°C:
(-1/k) * (90 - 30) = (1/2)(90^2) - (30)(90) + C
=> (-1/k) * 60 = 4050 - 2700 + C
=> (-1/k) * 60 = 1350 + C
At t = 1, T = 85°C:
(-1/k) * (85 - 30) = (1/2)(85^2) - (30)(85) + C
=> (-1/k) * 55 = 3612.5 - 2550 + C
=> (-1/k) * 55 = 1062.5 + C
Now, let's solve these two equations simultaneously to find the value of k:
(-1/k) * 60 = 1350 + C [Equation 1]
(-1/k) * 55 = 1062.5 + C [Equation 2]
Multiplying both sides of Equation 1 by 55, and Equation 2 by 60:
(-1/k) * 55 * 60 = 55 * 1350 + 55C [Equation 3]
(-1/k) * 55 * 60 = 60 * 1062.5 + 60C [Equation 4]
Simplifying these equations:
-3300/k = 74250 + 55C [Equation 3]
-3300/k = 63750 + 60C [Equation 4]
Setting the right-hand sides of Equations 3 and 4 equal to each other:
74250 + 55C = 63750 + 60C
Rearranging the equation:
5C = 10500
C = 2100
Now substituting the value of C back into Equation 1:
(-1/k) * 60 = 1350 + 2100
(-1/k) * 60 = 3450
-60/k = 3450
k = -60/3450
k = -2/115
So, the equation becomes:
dT/dt = (2/115)(T - 30)
To find out how long it will take for the water to cool to 60°C, we can set up a new differential equation as follows:
(2/115)(T - 30) = -k(T - A)
(2/115)(T - 30) = (2/115)(T - 30)
T - 30 = 60 - 30
T = 60
Now let's solve for t:
(2/115)(T - 30) * dt = dt
∫(T - 30) = ∫(2/115) dt
(T - 30) = (2/115) t
Substituting T = 60:
(60 - 30) = (2/115) t
30 = (2/115) t
t = 57.5
So, it will take approximately 57.5 minutes for the water to cool to 60°C. Rounding to the nearest minute, it will take 58 minutes.