Question

Cup of tea cooling room that has a constant temperature of 70 degrees Fahrenheit (F). If the initial temperature of the tea is 200 degrees Fahrenheit and the rate R(t) is 6.89e⁻⁰.⁰⁵³ᵗ degrees Fahrenheit per minute, what is the temperature, to the nearest degree, of the after 4 minutes?

a. Temperature: 175 degrees Fahrenheit
b. Temperature: 130 degrees Fahrenheit
c. Temperature: 95 degrees Fahrenheit
d. Temperature: 70 degrees Fahrenheit

Answer 1

After 4 minutes, the temperature of the tea will be approximately 130 degrees Fahrenheit. Option B is answer.

Step-by-step explanation:

Given Information:

Room temperature (T_room) = 70°F

Initial tea temperature (T_0) = 200°F

Cooling rate (R(t)) = 6.89e^(-0.053t) °F/min

Time (t) = 4 minutes

Model the Cooling Process:

Use Newton's Law of Cooling:

dT/dt = k(T - T_room)

Where:

dT/dt is the rate of change of temperature.

k is a constant representing the cooling rate.

T is the tea temperature at time t.

T_room is the room temperature.

Solve the Differential Equation:

Separate variables and integrate:

ln|T - T_room| = -kt + C

Solve for C using the initial condition (T(0) = T_0):

C = ln|T_0 - T_room|

Substitute C and express T:

T(t) = T_room + (T_0 - T_room) * e^(-kt)

Calculate Temperature After 4 Minutes:

Plug in the values:

T(4) = 70 + (200 - 70) * e^(-0.053*4)

T(4) ≈ 130°F

Therefore, the temperature of the tea after 4 minutes will be approximately 130 degrees Fahrenheit.

Option B is answer.