Suppose you make hot chocolate and it starts off just too hot to drink, so you put it in the freezer temporarily to cool it off. if the original temperature is 190 degrees fahrenheit and the freezer is 0 degrees fahrenheit, what will the temperature of the hot chocolate be after 2 minutes? (k = 0.12)

191.27 degrees
189.21 degrees
149.46 degrees
188.73 degrees

After 2 minutes the temperature of the hot chocolate will be 149.46 degrees Fahrenheit.

Step-by-step explanation:

We are going to use the Newton's law of cooling to solve this exercise. The Newton's law of cooling states that the amount of heat lost by a body is proportional to the difference of temperature between the body and the enviroment. We are going to use the following function :

Where ''T(t)'' is the temperature of the body that depends of the variable ''t'' which is time.

Where is the temperature of the surroundings

In this case is the temperature of the freezer

Where is the initial temperature of the body which is cooling. In this case, is the temperature of the hot chocolate

And where ''k'' is a constant. In this case, is a data of the exercise

If we replace all the values in the equation and replacing

⇒

We find that the temperature of the hot chocolate after 2 minutes is 149.46 degrees Fahrenheit

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.