A can of soda at 79^∘ F is placed in a re frigerator that maintains a constant temperature of 36

step 1 about question 59 so we are given that a can of soda at 79 degrees Fahrenheit is placed in a ref

A can of soda at 79^∘ F is placed in a re frigerator that...

A can of soda at $79^{\circ} \mathrm{F}$ is placed in a $\mathrm{re}$ frigerator that maintains a constant temperature of $36^{\circ} \mathrm{F}$ The temperature $T$ of the soda $t$ minutes after it is placed in the refrigerator is given by $$ T(t)=36+43 e^{-0.058 t} $$ a. Find the temperature, to the nearest degree, of the soda 10 minutes after it is placed in the refrigerator. b. When, to the nearest minute, will the temperature of the soda be $45^{\circ} \mathrm{F} ?$

A can of soda at $79^{\circ} \mathrm{F}$ is placed in a $\mathrm{re}$ frigerator that maintains a constant temperature of $36^{\circ} \mathrm{F}$ The temperature $T$ of the soda $t$ minutes after it is placed in the refrigerator is given by
$$
T(t)=36+43 e^{-0.058 t}
$$
a. Find the temperature, to the nearest degree, of the soda
10 minutes after it is placed in the refrigerator.
b. When, to the nearest minute, will the temperature of the soda be $45^{\circ} \mathrm{F} ?$

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Key Concepts

Newton's Law of Cooling

This principle states that the rate at which an object's temperature changes is proportional to the difference between its own temperature and that of its surroundings. It forms the basis for the exponential function used to model the cooling or heating of an object when placed in an environment with a constant temperature.

Exponential Decay Function

An exponential decay function describes a process where a quantity decreases at a rate proportional to its current value. In the context of cooling, it is used to model how the temperature difference between an object and its environment decreases exponentially over time.

Substitution and Evaluation

Substitution is the process of replacing the variable in an equation with a given value to compute a corresponding outcome. When dealing with time-dependent functions, it involves plugging in a specific time value to calculate the temperature at that moment.

Using Logarithms to Solve for Time

When an equation involves an exponential function and you need to solve for the time variable, logarithms are used to isolate and solve for the exponent. This technique is crucial for finding the time at which the temperature reaches a certain value.

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