A solar cooker consists of a curved reflecting surface that concentrates sunlight onto the object to Earth's surface at the location is `6W//m^(2)`. The cooker face the sun and has a face diameter of `0.600m`. Assume `40.0%` of the incident energy is tranferred to `0.500L` of water in an open container, initially at `20.0^(@)C`. Over what time interval does the water completely boil away? (Ignore the heat capacity of the container.)

Explanation

If we point the axis of the reflecting surface toward the sun, the power incident on the solar collector is
`P_(i)=IA=(600W//m^(2))[pi(0.300m)^(2)]=170W`
For a `40.0%` efficient reflector, the collected power is `P=(0.400)(170W)=67.9W =67.9 J//s`
The total energy required to increase the temperature of the water to the boiling point and to evaporate it is
`Q=mcDeltaT+mL`,
`Q=(0.500kg)(4180 J//kg^(@)C)(80.0^(@)C)+(0.500kg)(2.26xx10^(6)J//kg)`
`Q=1.30xx10^(6)J`
The time required is
`Deltat=(Q)/(P)=(1.30xx10^(6)J)/(67.9J//s)=1.91xx10^(4)s=5.31h`