High School

A cup of tea is cooling in a room with a constant temperature of 70 degrees Fahrenheit. If the initial temperature of the tea is 200 degrees Fahrenheit and the rate \( R(t) = 6.89e^{-0.053t} \) degrees Fahrenheit per minute, what is the temperature, to the nearest degree, after 4 minutes?

A. Temperature: 175 degrees Fahrenheit
B. Temperature: 130 degrees Fahrenheit
C. Temperature: 95 degrees Fahrenheit
D. Temperature: 70 degrees Fahrenheit

Answer :

The temperature of the cooling tea can be calculated using an exponential decay formula, resulting in approximately 94.83°F, which, when rounded, is 95°F after 4 minutes. Option c) is correct.

To find the temperature of the tea after 4 minutes, we need to integrate the cooling rate over the period of time we are interested in. However, the provided rate of cooling is a simple exponential decay, and it's more straightforward to use the formula for exponential decay, which is given by:

T(t) = T0 + (Tambient - T0) ∙ (1 - e-kt)

Where T(t) is the temperature of the tea at time t, T0 is the initial temperature of the tea, Tambient is the ambient temperature, k is the decay constant, and t is the time in minutes.

The question gives us T0 = 200°F, Tambient = 70°F, k = 0.053, and we want to know T(t) when t = 4 minutes.

Plugging the values into the formula, we get:

T(4) = 200°F + (70°F - 200°F) ∙ (1 - e-0.053∙4)

Calculating this gives us:

T(4) ≈ 200°F - 130°F ∙ (1 - e-0.212)

T(4) ≈ 200°F - 130°F ∙ 0.809

T(4) ≈ 200°F - 105.17°F

T(4) ≈ 94.83°F (Rounded to the nearest degree 95°F)

Thus, the correct answer would be:

c. Temperature: 95 degrees Fahrenheit