Question

You got your water for your tea boiling to 212 degree f. you prefer to drink your tea at a temperature of 100 degree f. if you set it aside in a room where the temperature is 72 degree f, how long until you can drink it? use newton�s law of cooling to answer the question. the k-value is known to be negative 0.118, relative to time in minutes.

Answer 1

To determine the time required to cool down the cup of tea to the desired temperature, we can use Newton's Law of Cooling, which states that the rate of heat loss of a body is proportional to the difference in temperature between the body and its surroundings.

Given that the temperature of the water is 212°F, and the desired temperature is 100°F, the initial temperature difference is (212 - 100) = 112°F. The room temperature is 72°F, which will be the temperature of the surroundings.

The equation for Newton's Law of Cooling is:

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

where T(t) is the temperature of the tea at time t, T_s is the temperature of the surroundings, T_0 is the initial temperature of the tea, k is the cooling rate constant, and e is the natural logarithm base.

Substituting the given values, we get:

T(t) = 72 + (212 - 100 - 72) * e^(-0.118t)

Simplifying the equation, we get:

T(t) = 72 + 112 * e^(-0.118t)

To find the time required for the tea to cool down to 100°F, we need to solve for t when T(t) = 100. Substituting 100 for T(t), we get:

100 = 72 + 112 * e^(-0.118t)

Simplifying the equation, we get:

e^(-0.118t) = 0.1964

Taking the natural logarithm of both sides, we get:

-0.118t = ln(0.1964)

Solving for t, we get:

t = 14.9 minutes (approximately)

Therefore, it will take about 14.9 minutes for the cup of tea to cool down to the desired temperature of 100°F in the given conditions.