Answer
a) $3.4W$
b)$2.3\frac{^oC}{s}$
c) no
d) $86^oC$
e) conduction, convection, evaporation
Work Step by Step
$A=40\times10^{-4}m^2$
$m=4.5\times10^{-4}kg$
$\epsilon=0.85$
$c=0.80\frac{kcal}{kg\cdot K}$
a) $Q=\epsilon \sigma A(\Delta T^4)t$
$=(0.85) (1000\frac{W}{m^2}) (40\times10^{-4}m^2)cos(90^o)=3.4W$
b) $\frac{\Delta T}{\Delta t}=\frac{\Delta Q}{mct}=\frac{3.4W}{(4.5\times10^{-4}kg)(0.80\frac{kcal}{kg\cdot K})}\times\frac{1kcal}{4186J}=2.25\frac{^oC}{s}$
c) No. Heat will be dissipated, the sun will not shine on the leaf for may hours at 90 degrees. Furthemore, in just 1 hour, the leaf's temperature would reach $8122^oC$ and we do not see any burning leaves.
d) $\frac{\Delta Q}{\Delta t}_s=\frac{\Delta Q}{\Delta t}_r$
$\epsilon_s \sigma A(\Delta T^4)=\epsilon_r \sigma A(\Delta T^4)$
$T_1=(\frac{1000\cos(90^o)}{2\sigma}+T_2^4)^\frac{1}{4}=359K=86^oC$
e) The leaf can dissipate heat through conduction, convection through air, and evaporation.
