The vector field of the equation T'(t)=-1/50 (T[t]-20) does not depend on t. The slope depends only on the y value. You get a vector of that slope for example with {1, - 1/50 (y - 20)}. Try with
VectorPlot[{1, - 1/50 (y - 20)}, {t, -3, 100}, {y, 0, 40}]
or, even better, with
StreamPlot[{1, - 1/50 (y - 20)}, {t, -3, 100}, {y, 0, 40}]
