$$0.02083 = [1.10 e^-(2.05)^2 Fo_cyl] \times [1.05 e^-(1.52)^2 Fo_slab]$$ But $Fo_cyl = \frac\alpha tR^2$, $Fo_slab = \frac\alpha tL^2 = Fo_cyl \times \fracR^2L^2 = Fo_cyl \times \frac0.04^20.06^2 = 0.444 Fo_cyl$
$$Q = \dotm_w (4180)(85 - 50) \Rightarrow \dotm_w = \frac1326004180 \times 35 = 0.906 \text kg/s$$ Introduction To Food Engineering Solutions Manual
For a short cylinder, use product solution: $$\fracT_0 - T_\inftyT_i - T_\infty = \left(\fracT_center,cyl - T_\inftyT_i - T_\infty\right) infinite\ cyl \times \left(\fracT center,slab - T_\inftyT_i - T_\infty\right)_infinite\ slab$$ out - T_m
$$Q = \dotm m c p,m (T_m,out - T_m,in)$$ $$Q = (0.5)(3900)(72 - 4) = 0.5 \times 3900 \times 68 = 132,600 \text W$$ Introduction To Food Engineering Solutions Manual
$$Q = U A \Delta T_lm \Rightarrow A = \fracQU \Delta T_lm = \frac1326001500 \times 26.13$$ $$A = \frac13260039195 = 3.383 \text m^2$$