The temperature of a cup of Starbucks coffee at time t (in minutes) is T(t)= 70 + c e^(-kt) . Initially, the temperature of the coffee was 200 degrees F. Three minutes later, it was 180 degrees. When will the temperature of the coffee be 150 degrees F?
The back of the book says it is 8.7 minutes, but I can't figure out how to get that answer.
Thank you for your help.
t(0)=200=70+ce^0
solve for c (130)
then
t(3)=70+130e^-k3=180
solve for k
e^-k3=11o/130
-k3=.167
k=.0557
150=70+130e^-(..0557)
-.0557t=.486
t=8.7 min
Thank you so much.
To find the time at which the temperature of the coffee will be 150 degrees F, we need to substitute the given information into the equation for the temperature.
Given:
Initial temperature: T(0) = 200 degrees F
Temperature after 3 minutes: T(3) = 180 degrees F
We can use these equations to set up a system of two equations and solve for the constants c and k.
Step 1: Substitute the given information into the temperature equation:
T(0) = 70 + c * e^(-k * 0) = 200
Simplifying this equation gives: c = 130
T(3) = 70 + c * e^(-k * 3) = 180
Step 2: Substitute the value of c into the second equation:
70 + 130 * e^(-k * 3) = 180
Step 3: Solve for k:
130 * e^(-3k) = 110
Divide both sides by 130:
e^(-3k) = 110/130 = 0.846
Take the natural logarithm (ln) of both sides:
ln(e^(-3k)) = ln(0.846)
-3k = ln(0.846)
Solve for k:
k = ln(0.846) / -3 ≈ -0.123
Step 4: Substitute the values of c and k into the temperature equation:
T(t) = 70 + 130 * e^(-0.123t)
Now we need to find the time at which the temperature will be 150 degrees F:
T(t) = 150
Substitute this into the equation:
70 + 130 * e^(-0.123t) = 150
Subtract 70 from both sides:
130 * e^(-0.123t) = 80
Divide both sides by 130:
e^(-0.123t) = 80/130 = 0.615
Take the natural logarithm (ln) of both sides:
ln(e^(-0.123t)) = ln(0.615)
-0.123t = ln(0.615)
Solve for t:
t = ln(0.615) / -0.123 ≈ 8.7 minutes
Therefore, the temperature of the coffee will be approximately 150 degrees F after 8.7 minutes.