Use Newton's Law of Cooling, `T = C + (T_0 - C)e^(kt)`, to solve the problem. Thanks?!

I get `35^@ "F"`, but you'll have to fix the equation.

Well, for one, the equation is wrong... Let's derive it to show it... The change in temperature over time is given by:

`(dT)/(dt) = -k(T - C)`

where `C` is the surrounding ambient temperature (a constant) and `T` is the current temperature. `k` is the rate constant of temperature decay.

Separation of variables gives the integration to be:

`int_(T_0)^(T) 1/(T - C)dT = -kint_(0)^(t)dt`

`ln((T - C)/(T_0 - C)) = -kt`

where `T_0` is the initial temperature and `t_0 = 0`.

Thus, the equation should be:

`color(green)(T = C + (T_0 - C)e^(-kt))`

Here, we have the surrounding temperature to be `C = 0^@ "F"`, and the initial temperature of the coffee to be `102^@ "F"`. Those are both constants.

So we have:

`barul|stackrel(" ")(" "T(t) = 102e^(-kt)" ")|` in `""^@ "F"`

We are told that `T = 52.5^@ "F"` at `"t = 8 min"`, which allows us to find the rate constant `k`. Then we can find `T` at `t = "13 min"`.

`T("8 min") = 52.5^@ "F" = (102^@ "F")e^(-k cdot "8 min")`

`0.5147 = e^(-8k)`

`ln0.5147 = -8k`

For this first-order process then, the rate constant (which is a constant for a constant surrounding temperature `C`) is:

`k = -(ln0.5147)/8 "min"^(-1) = "0.0830 min"^(-1)`

Now we can find `T` at any other time.

`color(blue)(T("13 min")) = 102^@ "F" cdot e^(-"0.0830 min"^(-1) cdot "13 min")`

`= 34.7^@ "F"`

`~~ color(blue)(35^@ "F")`