I'm completely lost on this differential equation problem.
A frozen grape that is currently 20F is placed into a glass of 68F wine. After 4 minutes the temperature of the grape has risen 4F. However, the temperature of the wine also changes as time passes. The temperature of the wine is given by
$$ T_m = 68 + \frac{20-T}{2} $$
If the rate of change of the temperature of the grape is proportional to the difference between the temperature of the grape and the temperature of the wine, set up and solve a differential equation to find the temperature of the grape at a given time t.
MY Thoughts
I know this is Newton's Law of Cooling, and have found the following information
$$ \frac{dT_g}{dt} = k(T_m - T_g)$$
which when solved becomes
$$ -\ln(T_m - T_g) = kt+C $$
and then $$ T_g = A e^{-kt} + T_m, \quad \text{where} \quad A= e^{-c} -10$$
I also know the following points
$$ T_g(0) = 20, T_m(0) = 68, T_g(4) = 24 $$ and by plugging in $ T_m(0) $ into the equation for $ T_m $ that $T(0) = 20$
which can be used to find that $$ 20 = A+T_m = A+78-\frac{T(0)}{2} = A+78-10 $$ which implies $A = -48$.
But I cannot figure out how to find $k$ or anything. I cannot find an equation for $T_m$ in terms of time either. Any help would be appreciated
You solved your equation as if $T_m $ is constant, which it isn't.
The equation for the grape is $$ T'=k\left ( 68+\frac {20-T}2-T\right), $$which simplifies to $$ T'=k\left ( 78-\frac {3T}2\right). $$
Then $$T (t)=52+c\,e^{-3kt/2}. $$ From $T (0)=20$, you get $20=52+c $, so $c=-32$. Then, from $T (4)=20+4=24$, $$24=52-32e^{-6k}. $$ So $$k=-\frac16\,\ln \frac 78. $$