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 f(t)=Ce(−kt)+70 represents the situation, where t is time in minutes, C is a constant, and k 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 :

Answer:

91.3

Step-by-step explanation:

The temperature of the tea, in degrees Fahrenheit, after 5 minutes is 91.40° F.

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A word problem is a verbal description of a problem situation. It consists of few sentences describing a 'real-life' scenario where a problem needs to be solved by way of a mathematical calculation.

For the given situation,

The function f(t) = Ce^(−kt)+70

[tex]f(t) = Ce^{(-kt)} +70[/tex]

At room temperature, the temperature of tea is 120 degrees

For t = 0,

⇒ [tex]120 = Ce^{(-k(0))} +70[/tex]

⇒ [tex]120 = Ce^{0}+70[/tex]

⇒ [tex]120 = C(1) +70[/tex]

⇒ [tex]C=120-70[/tex]

⇒ [tex]C=50[/tex]

At t = 3 minutes, the temperature of tea is 100 degrees

⇒ [tex]100 = 50e^{(-k(3))} +70[/tex]

⇒ [tex]100-70=50e^{-3k}[/tex]

⇒ [tex]\frac{30}{50} = e^{-3k}[/tex]

⇒ [tex]0.6=e^{-3k}[/tex]

Taking ln on both sides,

⇒ [tex]-3k=ln 0.6[/tex]

⇒ [tex]k=\frac{-ln0.6}{3}[/tex]

At t = 5 minutes,

[tex]f(t) = 50e^{-(5)(\frac{-ln0.6}{3} )} +70[/tex]

⇒ [tex]f(t) = 50e^{\frac{5}{3}(ln0.6) } +70[/tex]

⇒ [tex]f(t) = 50e^{(1.66)(ln0.6) } +70[/tex]

⇒ [tex]f(t) = 50e^{-0.8479 } +70[/tex]

⇒ [tex]f(t) = 50(0.4283)+70[/tex]

⇒ [tex]f(t) = 21.4156+70[/tex]

⇒ [tex]f(t) = 91.4156[/tex] ≈ [tex]91.40[/tex]

Hence we can conclude that the temperature of the tea, in degrees Fahrenheit, after 5 minutes is 91.40° F.

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