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A large room has a temperature of 70 degrees Fahrenheit. When a cup of tea is brought into the room, the tea has a temperature of 120 degrees Fahrenheit. The function [tex] f(t) = C e^{-kt} + 70 [/tex] represents the situation, where [tex] t [/tex] is time in minutes, [tex] C [/tex] is a constant, and [tex] k [/tex] is a constant.

After 3 minutes, the cup of tea has a temperature of 100 degrees. What will be the temperature of the tea, in degrees Fahrenheit, after 5 minutes? Round your answer to the nearest tenth, and do not include units.



Answer :

GinnyAnswer
Sure! Let's find the temperature of the tea after 5 minutes step-by-step.

1. Understand the given function:
The function [tex]\( f(t) = C \cdot e^{-kt} + 70 \)[/tex] describes how the temperature of the tea changes over time, where:
- [tex]\( t \)[/tex] is time in minutes,
- [tex]\( C \)[/tex] and [tex]\( k \)[/tex] are constants,
- The room temperature is 70 degrees Fahrenheit.

2. Initial conditions:
- At [tex]\( t = 0 \)[/tex] minutes, the temperature of the tea is 120 degrees Fahrenheit:
[tex]\( f(0) = 120 = C \cdot e^{-k \cdot 0} + 70 \)[/tex].
Since [tex]\( e^0 = 1 \)[/tex], this simplifies to [tex]\( 120 = C + 70 \)[/tex].
Solving for [tex]\( C \)[/tex]:
[tex]\[ C = 120 - 70 = 50. \][/tex]

3. Using the temperature after 3 minutes to find [tex]\( k \)[/tex]:
- At [tex]\( t = 3 \)[/tex] minutes, the temperature of the tea is 100 degrees Fahrenheit:
[tex]\( f(3) = 100 = 50 \cdot e^{-3k} + 70 \)[/tex].
Isolate the exponential term:
[tex]\[ 100 - 70 = 50 \cdot e^{-3k} \implies 30 = 50 \cdot e^{-3k} \][/tex]
Divide both sides by 50:
[tex]\[ e^{-3k} = \frac{30}{50} = 0.6 \][/tex]
Take the natural logarithm of both sides to solve for [tex]\( k \)[/tex]:
[tex]\[ -3k = \ln(0.6) \][/tex]
Therefore:
[tex]\[ k = -\frac{\ln(0.6)}{3} \][/tex]

4. Use [tex]\( k \)[/tex] to determine the temperature after 5 minutes:
- Now we want to find the temperature after 5 minutes, [tex]\( f(5) \)[/tex]:
[tex]\[ f(5) = 50 \cdot e^{-k \cdot 5} + 70 \][/tex]
Substituting [tex]\( k = -\frac{\ln(0.6)}{3} \)[/tex]:
[tex]\[ f(5) = 50 \cdot e^{\ln(0.6)/3 \cdot (-5)} + 70 \][/tex]
Simplify the exponent:
[tex]\[ f(5) = 50 \cdot e^{\ln(0.6) \cdot (-5/3)} + 70 \][/tex]
Simplifying further inside the exponent:
[tex]\[ e^{\ln(0.6) \cdot (-5/3)} = e^{\ln((0.6)^{-5/3})} = (0.6)^{-5/3} \][/tex]
Compute the value inside:
[tex]\[ f(5) = 50 \cdot (0.6)^{-5/3} + 70 \approx 91.3 \][/tex]

Thus, the temperature of the tea after 5 minutes will be approximately 91.3 (rounded to the nearest tenth).

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