Question

A freezer has a temperature of 14 degrees Fahrenheit. An ice-cube tray full of water is placed in the freezer. The function f(t)=Ce(−kt)+14 represents the situation, where t is time in minutes, C is a constant, and k=0.045. After 15 minutes the water has a temperature of 40 degrees. What was the initial temperature of the water? Round your answer to the nearest degree, and do not include units

Answer 1

Answer:

  65

Step-by-step explanation:

You can use f(15) = 40 to solve for C, then find f(0), the initial temperature.

  40 = f(15)

  40 = Ce^(-0.045·15) +14 = .50916C +14

  26 = .50916C

  26/.50916 = C ≈ 51.065

Then f(0) is ...

  f(0) = 51.065·e^0 +14 = 65.065 ≈ 65

The initial temperature of the water was 65 degrees Fahrenheit.

Answer 2

Answer:

23

Step-by-step explanation:

data

$f(15)= 40\\t= 15\\ C=?\\t_0=0$

$t_0=0$ because it's the initial time, when we start counting

To know how much the temperature is worth in the initial time$t_0=0$ we must find out the value of the constant C with the data we have of the situation at 15 minutes

$f(t)= Ce^{(-kt)}+14\\t=15\\f(15)= Ce^{(-k(15))}+14\\40= Ce^{-((0.045)(15))}+14\\40-14= Ce^{-0.675}\\\frac{26}{e^{-0.675}} = C \\8.89=C$

Find the initial temperature by replacing the data given and obtained

$f(t)= Ce^{(-kt)}+14\\t=0\\f(0)= Ce^{(-k(0))}+14\\f(0)= 8.89(1)+14\\f(0)= 22.89\\f(0)= 23$