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
To solve this problem, we can use the given information and the equation dT/dt = -k(T - A) to determine the time it takes for the water to cool to a specific temperature.
Given:
Initial temperature, T₀ = 90°C
Final temperature, T = 60°C
Room temperature, A = 30°C
Time, t = unknown (to be solved)
We know that the rate of change of temperature with respect to time is proportional to the temperature difference between the water and the room temperature.
Using the given information, we can set up the following equation:
dT/dt = -k(T - A)
Substituting the values:
dT/dt = -k(T - A)
(85 - 30)/(1 minute) = -k(90 - 30)/(t minutes)
Simplifying further:
55 = -k(60)/(t)
To find the value of t, we need to solve for k. We can rearrange the equation:
k = (55t)/(60)
Now, substitute the value of k in the original equation:
dT/dt = -(55t)/(60)(T - A)
We need to integrate this equation to find t. Integrating both sides gives:
∫(dT/(T - 30)) = -55/60 ∫dt
Integrating the left side gives:
ln|T - 30| = -(55/60)t + C1
where C1 is the constant of integration.
Now, let's find C1 using the initial condition provided:
ln|90 - 30| = -(55/60)(1 minute) + C1
ln|60| = -55/60 + C1
ln|60| + 55/60 = C1
C1 ≈ 3.138
Substituting the value of C1 back into the equation:
ln|T - 30| = -(55/60)t + 3.138
Next, we can substitute the final temperature T = 60 and solve for t:
ln|60 - 30| = -(55/60)t + 3.138
ln|30| = -(55/60)t + 3.138
Now, isolate t:
-(55/60)t = ln|30| - 3.138
t = (ln|30| - 3.138) * (60/55)
Calculating the value gives approximately t ≈ 5.24.
Rounding to the nearest minute, it will take the water about 5 minutes to cool to 60°C. Therefore, the closest answer is 5.
To find the time it will take the water to cool to 60°C, we can use the given information and the equation dT/dt = -k(T-A).
Given:
Initial temperature (T0) = 90°C
Final temperature (Tf) = 60°C
Room temperature (A) = 30°C
First, we need to find the value of k. We can use the fact that the water cools from 90°C to 85°C in 1 minute. Plugging these values into the equation, we have:
dT/dt = -k(T - A)
85 - 90 = -k(90 - 30)
-5 = -k(60)
Next, we can rearrange the equation to solve for k:
k = 5/60
k = 1/12
Now, we can find the time it will take for the water to cool from 90°C to 60°C. Plugging the values into the equation and rearranging, we have:
dT/dt = -k(T - A)
dT = -k(T - A)dt
(Tf - T0) = -k∫dt
Tf - T0 = -k(t - t0)
60 - 90 = -1/12(t - 0)
-30 = -1/12t
Finally, we can solve for t:
t = (-30)/(1/12)
t = -360
Since time cannot be negative, we discard the negative sign and take the absolute value:
t = 360 minutes
Therefore, it will take approximately 360 minutes (or 6 hours) for the water to cool to 60°C.
So, the closest answer is 8.