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")#