1 Answer | Add Yours
Power available in sunlight = 1 KJ/s (per unit area, m^2)
Area of the solar cooker = 1 m^2
Specific Heat Capacity of water = 4190 J/Kg/C
Heat required to boil 1 lt of water from 20 degrees Celsius = mass of water x specific heat capacity x temperature change
= 1 kg x 4190 J/Kg/C x (100-20) Celsius = 3,35,200 J = 335.2 KJ
Assuming 100% capacity of heat transfer from the reflector to the food
Rate of heat generation by the solar cooker = 1 KJ/s/m^2 x 1 m^2 = 1000 J/s
Total time required to boil the water = heat required/heat generation rate
= 335.2 KJ/1 KJ/s = 335.2 sec or 5 minutes and 35 seconds (about) or Option C.
In reality, there would be a number of power losses and the actual heat transfer efficiency would be less than 100% and it will take more time than that calculated above.
Hope this helps.
We’ve answered 319,647 questions. We can answer yours, too.Ask a question