Answer
The required area is $~~33~m^2$
Work Step by Step
We can find the required heat energy:
$Q = cm\Delta T$
$Q = (200~kg)(4187~J/kg\cdot K)(20~K)$
$Q = 1.6748\times 10^7~J$
We can find the heat energy provided per $m^2$ in 1.0 hour:
$(700~W/m^2)(3600~s) = 2.52\times 10^6~J/m^2$
We can find the amount of this provided energy that is used to heat the water based on the efficiency:
$(0.20)(2.52\times 10^6~J) = 5.04\times 10^5~J/m^2$
We can find the required area:
$Area = \frac{1.6748\times 10^7~J}{5.04\times 10^5~J/m^2} = 33~m^2$
The required area is $~~33~m^2$
