A cup of coffee is heated to 180°F and placed in a room that maintains a temperature of 65°F. The temperature of the coffee after t minutes is given by T(t) = 65 + 115e^−0.042t.(a) Find the temperature, to the nearest degree, of the coffee 10 minutes after it is placed in the room.
_ °F ?
(b) Use a graphing utility to determine when, to the nearest tenth of a minute, the temperature of the coffee will reach 110°F.
After ? min
let's just plug in t = 10
T(10) = 65 + 115 e^-.42
= 140.56
= 141° to the nearest degree.
b)
we want 110 = 65 + 115 e^(-.042t)
e^(-.042t) = (110-65)/115 = .391304...
take ln of both sides
-.042t = ln .391304..
t = 22.34 minutes or 22.3 minutes
(a) To find the temperature of the coffee 10 minutes after it is placed in the room, we substitute t = 10 into the equation T(t) = 65 + 115e^(-0.042t):
T(10) = 65 + 115e^(-0.042*10)
T(10) ≈ 65 + 115e^(-0.42)
Using a calculator, we can approximate the value of e^(-0.42) as 0.657 and calculate:
T(10) ≈ 65 + 115 * 0.657
T(10) ≈ 65 + 75.255
T(10) ≈ 140.255
Therefore, the temperature of the coffee 10 minutes after it is placed in the room is approximately 140°F.
(b) To determine when the temperature of the coffee will reach 110°F, we set T(t) equal to 110 and solve for t:
T(t) = 65 + 115e^(-0.042t)
110 = 65 + 115e^(-0.042t)
Subtract 65 from both sides:
45 = 115e^(-0.042t)
Divide both sides by 115:
0.391 = e^(-0.042t)
Take the natural logarithm of both sides:
ln(0.391) = -0.042t
Divide both sides by -0.042:
t = ln(0.391) / -0.042
Using a calculator, we find:
t ≈ 22.271
Therefore, the temperature of the coffee will reach 110°F after approximately 22.3 minutes.
(a) To find the temperature of the coffee 10 minutes after it is placed in the room, we need to substitute t = 10 into the equation T(t) = 65 + 115e^(-0.042t):
T(10) = 65 + 115e^(-0.042*10)
Using a calculator, we can simplify this as:
T(10) ≈ 65 + 115e^(-0.42)
Evaluating the exponential part, we get:
T(10) ≈ 65 + 115 * 0.65757
T(10) ≈ 65 + 75.71055
T(10) ≈ 140.71
Therefore, the temperature of the coffee 10 minutes after it is placed in the room is approximately 141°F.
(b) To determine when the temperature of the coffee will reach 110°F, we need to set T(t) equal to 110 and solve for t:
110 = 65 + 115e^(-0.042t)
Rearranging the equation, we get:
115e^(-0.042t) = 110 - 65
115e^(-0.042t) = 45
Now, divide both sides by 115:
e^(-0.042t) = 45/115
e^(-0.042t) ≈ 0.3913
To solve for t, we can take the natural logarithm (ln) of both sides:
ln(e^(-0.042t)) ≈ ln(0.3913)
-0.042t ≈ ln(0.3913)
Divide both sides by -0.042:
t ≈ ln(0.3913) / -0.042
Using a calculator, we can find:
t ≈ -9.802
Since time cannot be negative, we discard this solution.
Therefore, using a graphing utility, we can determine that the temperature of the coffee will reach 110°F after approximately 9.8 minutes.