Suppose a cup of coffee is at 100 degrees Celsius at time t = 0, it is at 70 degrees
at t = 10 minutes, and it is at 50 degrees at t = 20 minutes. Compute the ambient temperature.
So in the book I was given Newton's Heat equation.
dT/dt=k(A-T) where T is temperature of the coffee. t is time. A is the ambient temperature.
This is a seperable equation so I isolated k to one side with dt and dT/(A-T)
When I took the integral of both sides I got
ln(A-T)=kt+C
which is then
A=e^(kt+C)+T
How do I find ambient temperature or k or C? I am very confused and cannot get past this step.
Thanks!
Well, finding the ambient temperature (A), the constant of proportionality (k), and the constant of integration (C) requires additional information. However, there's a simpler approach we can take using the given information.
Let's assume that the ambient temperature (A) is constant throughout this time period. Since we know the temperatures of the coffee at different times, we can create a system of equations to solve for A.
At t = 0, the coffee temperature is 100 degrees Celsius. Using the equation dT/dt = k(A - T), we can write it as:
dT/dt = k(A - 100)
At t = 10 minutes, the coffee temperature is 70 degrees Celsius. We can plug this into the equation:
dT/dt = k(A - 70)
Similarly, at t = 20 minutes, the coffee temperature is 50 degrees Celsius:
dT/dt = k(A - 50)
Now, by comparing the first and second equations, we can eliminate k:
k(A - 100) = k(A - 70)
Simplifying this expression, we get:
A - 100 = A - 70
From this, we can see that 100 = 70, which is not true. Therefore, there seems to be an inconsistency in the given information. It's likely that we need more data or there might be an error in the values provided.
Remember, when dealing with coffee, it's always important to stay grounded and remain as per-cup-tive as possible!
To find the ambient temperature, we need to use the given information about the coffee's temperature at different times. We can then substitute these values into the equation and solve for A.
Given:
At t = 0 minutes, the coffee temperature is 100 degrees Celsius.
At t = 10 minutes, the coffee temperature is 70 degrees Celsius.
At t = 20 minutes, the coffee temperature is 50 degrees Celsius.
Let's start by substituting the first set of values into the equation:
100 = A * e^(k * 0) + T
Since e^(k * 0) is equal to 1, we can simplify the equation to:
100 = A + T
Now, let's substitute the second set of values:
70 = A * e^(k * 10) + T
And finally, substitute the third set of values:
50 = A * e^(k * 20) + T
Now, we have a system of three equations with three unknowns (A, k, and T). We can solve this system to find the values.
Subtract the second equation from the first equation to eliminate T:
100 - 70 = (A + T) - (A * e^(k * 10) + T)
30 = -A * (e^(k * 10) - 1)
Subtract the third equation from the second equation to eliminate T:
70 - 50 = (A * e^(k * 10) + T) - (A * e^(k * 20) + T)
20 = A * (e^(k * 10) - e^(k * 20))
We now have a system of two equations:
30 = -A * (e^(k * 10) - 1)
20 = A * (e^(k * 10) - e^(k * 20))
To simplify, let's consider the term (e^(k * 10) - 1) as variable x, and (e^(k * 10) - e^(k * 20)) as variable y.
The equations now become:
30 = -A * x
20 = A * y
We can divide the two equations to eliminate A:
30/20 = -x/y
3/2 = -x/y
2x = -3y
Now we can solve for x in terms of y and substitute it back into one of the original equations:
2x = -3y
2(e^(k * 10) - 1) = -3(e^(k * 10) - e^(k * 20))
Expanding the equation:
2e^(k * 10) - 2 = -3e^(k * 10) + 3e^(k * 20)
Combine like terms:
5e^(k * 10) = 3e^(k * 20) - 2
We now have one equation with one unknown (k). To solve for k, you can use numerical methods or apply algebraic manipulation techniques. Keep in mind that the values of k might not have an exact analytical solution.
Once you have the value of k, you can substitute it back into one of the original equations to find A. The ambient temperature (A) is equal to the calculated value minus the temperature of the coffee at time t = 0.
I hope this helps! Let me know if you have any further questions.
To find the ambient temperature, as well as the values of k and C in the equation dT/dt = k(A - T), we can use the given information about the coffee's temperature at different times.
Let's start with the equation you derived: ln(A - T) = kt + C.
At t = 0, the coffee temperature is 100 degrees Celsius, so we have:
ln(A - 100) = 0k + C -- (Equation 1)
At t = 10 minutes, the coffee temperature is 70 degrees Celsius, so we have:
ln(A - 70) = 10k + C -- (Equation 2)
At t = 20 minutes, the coffee temperature is 50 degrees Celsius, so we have:
ln(A - 50) = 20k + C -- (Equation 3)
We now have a system of three equations (Equations 1, 2, and 3) with three unknowns: A, k, and C.
To solve this system of equations, we can use the elimination or substitution method. However, since the equations involve logarithms, it is more straightforward to use the properties of logarithms to combine the equations.
First, let's subtract Equation 1 from Equation 2:
ln(A - 70) - ln(A - 100) = 10k + C - C
Using the logarithmic property ln(x) - ln(y) = ln(x/y), we have:
ln((A - 70)/(A - 100)) = 10k
Next, let's subtract Equation 1 from Equation 3:
ln(A - 50) - ln(A - 100) = 20k + C - C
Using the logarithmic property ln(x) - ln(y) = ln(x/y), we have:
ln((A - 50)/(A - 100)) = 20k
Now we have two equations involving only k:
ln((A - 70)/(A - 100)) = 10k -- (Equation 4)
ln((A - 50)/(A - 100)) = 20k -- (Equation 5)
We can solve these two equations simultaneously to find the value of k. Once we have k, we can substitute it back into Equation 1 or 2 or 3 to find C. Finally, using the known value of T at any time (e.g., t = 0), we can find the ambient temperature A.
I always like to subsume the e^C into a new C so the function is
Ce^(kt)+T
A(0) = 100
so, C + T = 100
A(10) = 70
so, Ce^10k+T = 70
also, use the 3rd point, and then you can solve for the constants.