Question

HELPP

If you pour a cup of coffee that is 200F, and set it on a desk in a room that is 68F, and 10 minutes later it is 145F, what will the temperature be 15 minutes after you originally poured it?

Answer 1

127 F

Explanation:

we will use newton's law of cooling to understand the process. Since the coffee here is losing heat so the heat loss will be proportional to the temperature difference

Answer 2

To solve this problem, we can use Newton's law of cooling, which states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. We can use the formula:

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

where:

T(t) is the temperature of the coffee at time t

T_s is the temperature of the surroundings (68F)

T_0 is the initial temperature of the coffee (200F)

k is a constant that depends on the specific heat transfer properties of the coffee and its container

e is the mathematical constant e, approximately 2.71828

We can use the fact that the temperature is 145F after 10 minutes to find the value of k:

145 = 68 + (200 - 68) * e^(-10k)

e^(-10k) = (145 - 68) / (200 - 68) = 77 / 132

-10k = ln(77/132)

k = -ln(77/132) / 10

Now we can use this value of k to find the temperature of the coffee after 15 minutes:

T(15) = 68 + (200 - 68) * e^(-k*15)

T(15) = 68 + (200 - 68) * e^(-15ln(77/132)/10)

T(15) = 111.72

Therefore, the temperature of the coffee 15 minutes after it was poured would be approximately 111.72F.