The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus 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?
dT/dt = -kT-A
dT/(kT+A) = -k dt
1/k ln(kT+A) = -kt + C
T(0) = 90 so 1/k ln(90k+30) = C
T(1) = 85 so 1/k ln(85k+30) = C-k
Solve for C and k, then evaluate t when
1/k ln(60k+30) = C-kt
dT/dt = -k (T-A) yes
try
T = A + (To - A) e^-kt
then
dT/dt = -k(To-A) e^-kt
but T-A = (To-A)e^-kt
so dT/dt = -k(T-A) it works
so
85 = 30 + (90-30)e^-k(1)
55 = 60 e^-k
ln(55/60) = -k solve for k
then use that k
60 = 30 +60 e^-k t
.5 = e^-kt
ln .5 = -k t solve for t
dT/dt = -kT-A
I think not
dT/dt = -k(T-A)
To solve this problem, we can use Newton's Law of Cooling equation to find the time it takes for the water to cool from 90°C to 85°C. Then, we can use this time to extrapolate and find out how long it will take for the water to cool to 60°C.
Let's break down the problem and solve it step by step:
1. Determine the known values:
- Initial temperature of the water (T₀): 90°C
- Final temperature of the water (T): 85°C
- Room temperature (A): 30°C
2. Apply Newton's Law of Cooling equation:
dT/dt = -k * (T - A)
3. Integrate both sides of the equation:
∫(1/(T - A)) dT = -k * ∫dt
The left side integrates to:
ln|T - A| = -kt + C₁,
where C₁ is the constant of integration.
4. Apply initial condition:
At t = 0, T = T₀:
ln|T₀ - A| = C₁
Therefore, the equation becomes:
ln|T - A| = -kt + ln|T₀ - A|
5. Solve for k:
Given T₀ = 90°C and T = 85°C, we substitute these values into the equation. Also, k is a positive constant, so we take the absolute value.
ln|85 - 30| = -k(1) + ln|90 - 30|
ln(55) = -k + ln(60)
Rearranging the equation:
k = ln(60) - ln(55)
6. Now, we have the value of k, so we can use it to find the time it takes for the water to cool from 90°C to 85°C.
Plugging in the values:
ln|T - A| = -(ln(60) - ln(55)) * t + ln|90 - 30|
Since we're looking for the time it takes to cool to 60°C, we set T = 60°C.
ln|60 - 30| = -(ln(60) - ln(55)) * t + ln|90 - 30|
Simplifying further:
ln(30) = -(ln(60) - ln(55)) * t + ln(60)
7. Solve for t:
Rearrange the equation to solve for t:
-(ln(60) - ln(55)) * t = ln(30) - ln(60)
Divide both sides by -(ln(60) - ln(55)):
t = (ln(30) - ln(60))/(ln(55) - ln(60))
8. Plug in the values and calculate the time:
Substitute the values into the equation:
t = (ln(30) - ln(60))/(ln(55) - ln(60))
Using a calculator, we can evaluate the expression.
t ≈ 2.64 minutes
Therefore, it will take approximately 2.64 minutes (to the nearest minute) for the water to cool to 60°C.