A patch of ice at a temperature t0oc has an area 847 cm2...

A patch of ice at a temperature T=0oC has an area 84.7 cm2. When exposed to sunlight, 27.6 g of ice melt in one hour. Determine the amount of solar energy per second on each cm2 of the patch, knowing that only 35% of the incident solar energy is absorbed by the ice

Transcript

00:01 We have a question and this it is given that solar radiation on a frozen lake is 600 watt per meter square okay now okay this is like this watt per meter is square and 70 % of this is absorbed by ice so 70 % so 600 into 70 by 100 so this would be 420 watt per meter square is absorbed by ice now how much time will it take for 2 .5 centimeter thick layer of ice to melt the ice and water underneath are at 0 degrees centigrade though a thickness of water is 2 .5 centimeter okay thickness of ice sorry okay now this is thickness and if a is the area of the lake so volume of the lake will be area into the thickness which is given as 2 .5 centimeter so this is 2 .5 centimeter means 2 .5 into power or minus 2 meters so if area is a it will be 2 .5 into 10 as to power minus 2 into a meters this is the now we have to find mass of ice so mass of ice will be equal to mass of ice will be equal to density is mass by volume so volume of ice into so density of ice into volume therefore density of ice will be equal to 917 kilogram per meter cube kg per meter cube into volume is 2 .5 into 10th of power minus 2 into area and this will be meter cube definitely meter cube so this is the mass in kg this will be in kilograms now we'll be using the formula for heat transfer okay now we have the area given so heat is 400 20 watt per meter square so total heat will be into area this is the power that is 420 into a watt is jule per second that module per second of energy it required okay so now if you talk about this is for one second and one more thing to be given that the heat required for water for ice to melt will be q equal to m l simply m because our temperature are the same only ice is melting the mass of ice we have already calculated and this should be we have already calculated mass of ice 917 into 2 .5 into 10 to power minus 2 into a and let in it is 334 kilojoule per kg okay so into 10 to power 3 jule per kg that much heat is required now let us calculate it 917 917 into 2 .5 into 334 and multiply the 10 it will be 765 695 0 jule and into a into area so this much jule of heat is required and available jule is 420 into a per second so time will be equal to 7 656950 a divided by 420 into a that's what time would be then divide by 420 k 18 ,220.

05:27 18 ,230.

05:29 30 .3 .33 seconds.

05:33 If you divide bit 3 ,600, it will be in hours...

Question

A patch of ice at a temperature T=0oC has an area 84.7 cm2. When exposed to sunlight, 27.6 g of ice melt in one hour. Determine the amount of solar energy per second on each cm2 of the patch, knowing that only 35% of the incident solar energy is absorbed by the ice

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