A \"solar cooker\" consists of a curved reflecting mirror thatfocuses sunlight o

Question

A "solar cooker" consists of a curved reflecting mirror thatfocuses sunlight onto the object to be heated. The solar power perunit area reaching the Earth at the location of a 0.50 m diametersolar cooker is 600 W/m2. Assuming that 50% of theincident energy is converted to thermal energy, how long would ittake to boil away 3.0 L of water initiallyat 40°C? (Neglect the specific heat ofthe container.)
min
A "solar cooker" consists of a curved reflecting mirror thatfocuses sunlight onto the object to be heated. The solar power perunit area reaching the Earth at the location of a 0.50 m diametersolar cooker is 600 W/m2. Assuming that 50% of theincident energy is converted to thermal energy, how long would ittake to boil away 3.0 L of water initiallyat 40°C? (Neglect the specific heat ofthe container.)
min

Explanation / Answer

It's actually not possible to answer this question without knowingthe total surface area of the solar cooker - it's notr2 since it's curved and not flat like a circle!The amount of energy that would be absorbed by the curved sideswould not neccesarily be 600 W/m2, either - so we'rejust going to pretend that it DOES have the surface area of a flatcircle and solve that way, since that seems to be what the problemwants. Surface area: A = r2 = (0.50 m /2)2 = 0.1963 m2 Total power: P = I (intensity) * A(area) = (600W/m2)(0.1963 m2) = 117.8 W Remember that watts are joules per second, so Energy: 50% * Wt * (60 sec/min) where t isminutes. Now we just need the energy to boil 3 liters of water at 40°C.Remember that 1 liter of water is roughly equal to one kilogram ofwater, so you have about three kilograms of water. Set up: Specific heat of liquid water (amount of energy needed to raise onemass unit of water by one degree of temperature) = 4186J/(kg°C) Heat of vaporization of water (amount of energy needed to boil onemass unit of water) = 2257 J/kg The equation is Q = cmT + mLvap. The two parts of the equation represent the energy needed to getthe water to 100 degrees C, and the amount of energy needed to boilit. c represents specific heat, m represents mass, and Lvaprepresents the heat of vaporization. Q = (4186 J/(kg°C))(3 kg)(100°C - 40°C) + (3 kg)(2257 x103 J/kg) = 7533480 J Now set the energy absorbed equal to the energy needed: Q = Wt/(60 sec/min) 7533480 J = t * (117.8 J/sec) * (60 sec/min) * 50% t = 2132 min This is about 2132/60 = 35.5 hours Edit: used the wrong constant. Enthalpy of vaporization was givenin kJ, not J, and diameter is given, not radius.

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