Explanation:
To solve the problem, we'll use the information given to find the values of A and k in the equation T = 20 + Ae^(-kt), where T is the temperature in degrees Celsius at time t.
a. Finding the value of A:
We're given that the initial temperature of the tea was 70℃. Substituting this into the equation, we get:
70 = 20 + Ae^(0) (since e^0 = 1)
70 - 20 = A
A = 50
So the value of A is 50.
b. Finding the value of k:
We're told that it takes 4 minutes for the tea to decrease in temperature from 70℃ to 50℃. We can use this information to set up an equation and solve for k.
Substituting T = 70 and t = 4 into the equation, we have:
70 = 20 + 50e^(-4k)
Subtracting 20 from both sides:
50 = 50e^(-4k)
Dividing both sides by 50:
1 = e^(-4k)
Taking the natural logarithm (ln) of both sides:
ln(1) = ln(e^(-4k))
0 = -4k
Dividing both sides by -4:
0 = k
However, we need to check if this solution satisfies the condition for the tea to decrease in temperature from 70℃ to 50℃ in 4 minutes.
Substituting T = 50 and t = 4 into the equation, we have:
50 = 20 + 50e^(-4(0))
50 = 20 + 50e^0
50 = 20 + 50(1)
50 = 20 + 50
50 = 70
Since 50 is not equal to 70, the value of k = 0 does not satisfy the given condition.
Let's try another approach:
Substituting T = 50 and t = 4 into the equation, we have:
50 = 20 + 50e^(-4k)
Subtracting 20 from both sides:
30 = 50e^(-4k)
Dividing both sides by 50:
0.6 = e^(-4k)
Taking the natural logarithm (ln) of both sides:
ln(0.6) = ln(e^(-4k))
ln(0.6) = -4k ln(e)
ln(0.6) = -4k(1)
ln(0.6) = -4k
Simplifying further:
k = ln(0.6) / -4
Using a calculator or computer to evaluate ln(3/5) / -4, we get approximately:
k = ln(5/3) / 4
To recap:
a. The value of A is 50.
b. The value of k is approximately k = ln(5/3) / 4
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Note: The calculations provided are based on the given information and assumptions. If there are any additional details or specific conditions provided, please let me know, and I'll be happy to assist you further.