A cold metal bar at -30 degrees Celsius is submerged in a pool maintained at a temperature of 40 degrees Celsius. Half a minute later, the temperature of the bar is 20 degrees Celsius. How long will it take the bar to attain a temperature of 30 degrees Celsius?
I know that dy/dt=k(y-b) and y(t)=b+Ce^kt
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To find the time it will take for the cold metal bar to attain a temperature of 30 degrees Celsius, we need to determine the value of the constant k and find the appropriate value of t.
Given:
Initial temperature of the bar, y(0) = -30 degrees Celsius
Temperature of the pool, b = 40 degrees Celsius
Temperature of the bar after half a minute, y(0.5 minutes) = 20 degrees Celsius
To find the value of k, we can use the formula dy/dt = k(y - b).
Substituting the values, we have:
dy/dt = k(y - b)
20 - (-30) = k(20 - 40)
50 = -20k
Solving for k, we get:
k = -50/20
k = -2.5
Now, using the equation y(t) = b + Ce^(kt), we can substitute the values of b, C, and k to find the value of t when the bar reaches a temperature of 30 degrees Celsius.
30 = 40 + Ce^(-2.5t)
Subtracting 40 from both sides, we have:
-10 = Ce^(-2.5t)
To determine the value of C, we need to use the initial condition, which states that the temperature of the bar 0.5 minutes later (t=0.5) is 20 degrees Celsius.
20 = 40 + Ce^(-2.5 * 0.5)
20 = 40 + Ce^(-1.25)
Subtracting 40 from both sides, we have:
-20 = Ce^(-1.25)
Now we can substitute this value of C into the equation for t when the temperature is 30 degrees Celsius.
-10 = (-20)e^(-2.5t)
Dividing both sides by -20, we get:
0.5 = e^(-2.5t)
Taking the natural log of both sides, we have:
ln(0.5) = ln(e^(-2.5t))
ln(0.5) = -2.5t
Solving for t, we get:
t = ln(0.5)/-2.5
Using a calculator, we can find the value of t:
t ≈ 0.277
Therefore, it will take approximately 0.277 minutes (or 16.6 seconds) for the bar to attain a temperature of 30 degrees Celsius.
To find out how long it will take for the metal bar to attain a temperature of 30 degrees Celsius, we can use the differential equation dy/dt = k(y - b), where y represents the temperature of the bar at time t, k is a constant, and b is the temperature of the environment (40 degrees Celsius in this case).
To solve this differential equation, we'll first substitute the given initial condition into the general solution of the equation. In this case, we know that the temperature of the bar half a minute later is 20 degrees Celsius. Therefore, we have y(0.5) = 20.
Using the general solution formula y(t) = b + Ce^(kt), we can substitute the values y(0.5) = 20 and b = 40 into the equation and solve for C:
20 = 40 + Ce^(0.5k)
Subtracting 40 from both sides gives us:
-20 = Ce^(0.5k)
Dividing both sides by C and taking natural logarithm:
ln(-20/C) = 0.5k
Now, we can use the information that the temperature of the bar will be 30 degrees Celsius to find the time it takes for this temperature to be reached. Let's substitute y(t) = 30 and b = 40 into the general solution formula:
30 = 40 + Ce^(kt)
Subtracting 40 from both sides gives us:
-10 = Ce^(kt)
Dividing both sides by C and taking natural logarithm:
ln(-10/C) = kt
To find the value of k, we can divide the second equation by the first equation:
ln(-10/C) / ln(-20/C) = kt / 0.5k
Simplifying the equation:
ln(-10/C) / ln(-20/C) = 2t
Since the left side of the equation is a constant, we can solve this equation for t. Divide both sides by 2:
t = ln(-10/C) / ln(-20/C) / 2
Plug in the known values and use a calculator to find the value of t. Keep in mind that negative values inside the logarithms are not defined, so the values of C that yield negative arguments for the logarithm cannot be used.
Now, you can calculate the value of t to find out how long it will take for the metal bar to attain a temperature of 30 degrees Celsius.
The solution will be of the form
T - 40 = (-70)*exp(-kt)
Note that this satisfies T = -30 when t = 0 and T-> 40 as t -> infinity
Solve for k by using the fact that
T = 20 at t = 30 s.
-20 = -70*exp(-30 t)
exp(-30k) = 0.2857
-30k = -1.253
k = 4.18*10-2
(1/k) = 23.9 seconds
Now solve for t when T = 30 C