Answer
$W/A=-187.3\ Btu/ft^2$ (in)
$T_{min}=32°F$
Work Step by Step
From the material balance:
$\frac{dm}{dt}=-\dot{m}_e$
From the energy balance:
$-\dot{W}=\frac{dU}{dt}+\dot{m}_eh_e$
$\dot{W}=-\frac{d(mu)}{dt}+\frac{dm}{dt}h_e$
Integrating over time:
$W=(m_2-m_1)h_e-m_2u_2+m_1u_1$
Since the ice is totally removed $m_2=0$
$W=m_1(u_1-h_e)$
With the mass being given by: $m_1=\frac{t.A}v$ and since $u_{if}=h_{if}\rightarrow u_1-h_e=-u_{if}$
$W=-\frac{t.A}vu_{if}$
Given $u_{if}=144\ Btu/lbm,\ t=\frac{0.25}{12}ft,\ v=0.01602\ ft^3/lbm$
$W/A=-187.3\ Btu/ft^2$ (in)
From the second law, the defroster temperature can't be lower than the temperature of the ice: $T_{min}=32°F$
