Asked by Tsunayoshi
After a snow storm, your driveway of area 50m^2 is covered with a layer of ice 10cm high at a temperature of 0C. You decide to use sunlight falling at 200 W/m^2 to melt the ice? Is it good idea? (Density of ice is 920kg/m^3 and 1 cal = 4.184 J.
Answers
Answered by
bobpursley
heat to melt ice : mass*Hf
= volume*density*Hf
= 50m^2*.1m*920kg/m^3)(4.18KJ/kg
= 19200kJ
energy available: 200W/m^2*50m^2*8hrs*3600sec/hr
= 28800kJ
The truth is, that intense sunlight doesn't exist through the day, as the sun rises and sets, it is much less, and the ice reflects a lot of energy.
check my calculations.
= volume*density*Hf
= 50m^2*.1m*920kg/m^3)(4.18KJ/kg
= 19200kJ
energy available: 200W/m^2*50m^2*8hrs*3600sec/hr
= 28800kJ
The truth is, that intense sunlight doesn't exist through the day, as the sun rises and sets, it is much less, and the ice reflects a lot of energy.
check my calculations.
Answered by
Tsunayoshi
How long will it take to melt the ice?
Answered by
Tsunayoshi
How did you get the 8hrs and 3600 sec/hr??
Answered by
Damon
60 sec/ min *60 min/hr = 3600 seconds/hour
He assumed 8 hours of daylight but as he said you will not get the full force for 8 hours.
He assumed 8 hours of daylight but as he said you will not get the full force for 8 hours.
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