Question

At 7:00 A.M., Alicia pours a cup of tea whose temperature is 200°F. The tea starts to cool to room temperature (72°F). At 7:02 A.M. the temperature of the tea is 197°F. At 7:15 A.M. the temperature of the tea is 180°F. Alicia will drink the tea when its temperature is 172°F.

Part A

Which of the following shows an exponential cooling equation that models the temperature of the tea?

A. y = 72(0.989)x + 128

B. y = 128(0.989)x + 72

C. y = 200(0.989)x + 72

D. y = 128(0.989)x + 200

Part B

When can Ella drink the tea?


Ella can drink the tea after
7:22
7:18

Answer 1

Answer: A) Option B

B) 22.3 minutes.

Explanation:

The temperature decreases in an exponential decrease, this means that we can write the equation as:

T(x) = A*r^x + B

Where A is the difference between the initial temperature and the room temperature, B is the room temperature, and r is a positive number smaller than 1, that says "how fast" the temperature decreases (and x is the variable, in units of time)

We know that the initial temperature

A = 200° - 72° = 128°

B = 72°

T(x) = 128°r^x + 72°

The only option with those two values is option B.

T(x) = 128(0.989)^x + 72

We should check that when x = 2m, the temperature must be 197°

T(2m) = 128*(0.989)^2 + 72 = 197.2° (that we can round down to 197°)

So this equation is correct

Now we want to find the time such the temperature is 172°F

then:

172° = 128°(0.989)^x + 72°

100° = 128°(0.989)^x

100/128 = (0.989)^x

x = ln(100/128)/ln(0.989) = 22.3 minutes.

Let's check in our equation:

T(22.3) = 128°(0.989)^22.3 + 72 = 172°