To tackle this problem, we need to understand the relationship between the moisture content of the sheet and the time it takes to lose that moisture. The statement indicates that the rate of moisture loss is proportional to the current moisture content, which is a classic example of exponential decay. Let's break this down step by step.
Understanding the Moisture Loss
When a wet porous substance loses moisture at a rate proportional to its moisture content, we can express this mathematically. If we let \( M(t) \) represent the moisture content at time \( t \), the relationship can be described by the differential equation:
dm/dt = -kM
Here, \( k \) is a positive constant that represents the rate of moisture loss. This equation indicates that as moisture decreases, the rate of loss also decreases.
Exponential Decay Model
The solution to this differential equation is:
M(t) = M_0 e^{-kt}
where \( M_0 \) is the initial moisture content. This equation shows that moisture decreases exponentially over time.
Applying the Given Conditions
From the problem, we know two key points:
- In the first hour, the sheet loses half of its moisture.
- It loses 99% of its moisture in time \( t \).
Let's analyze the first condition. If the moisture content is halved in the first hour, we can set up the equation:
M(1) = M_0 / 2
Substituting into the exponential decay formula gives:
M_0 e^{-k \cdot 1} = M_0 / 2
Dividing both sides by \( M_0 \) (assuming \( M_0 \neq 0 \)) leads to:
e^{-k} = 1/2
Taking the natural logarithm of both sides, we find:
-k = ln(1/2)
Thus, we have:
k = -ln(1/2) = ln(2)
Finding Time \( t \)
Now, let's use the second condition where the sheet loses 99% of its moisture. This means that only 1% of the original moisture remains:
M(t) = 0.01M_0
Substituting into the decay formula gives:
M_0 e^{-kt} = 0.01M_0
Again, dividing by \( M_0 \) leads to:
e^{-kt} = 0.01
Taking the natural logarithm of both sides results in:
-kt = ln(0.01)
Substituting \( k = ln(2) \) into the equation gives:
-ln(2) \cdot t = ln(0.01)
Solving for \( t \) yields:
t = -\frac{ln(0.01)}{ln(2)}
Calculating \( 2^{(t/2)} \)
Now, we need to find the value of \( 2^{(t/2)} \). We can express \( t \) in terms of base 2:
t = -\frac{ln(0.01)}{ln(2)} = -\frac{ln(10^{-2})}{ln(2)} = \frac{2ln(10)}{ln(2)}
Thus, \( t/2 \) becomes:
\frac{t}{2} = \frac{ln(10)}{ln(2)}
Now, substituting this into \( 2^{(t/2)} \) gives:
2^{(t/2)} = 2^{\frac{ln(10)}{ln(2)}} = 10
Final Result
Therefore, the value of \( 2^{(t/2)} \) is 10. This illustrates how we can use the principles of exponential decay to solve real-world problems involving moisture loss.
